{"id":6678,"date":"2018-03-13T20:21:23","date_gmt":"2018-03-13T20:21:23","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-albany-chemistry\/?post_type=chapter&#038;p=6678"},"modified":"2024-01-24T19:35:36","modified_gmt":"2024-01-24T19:35:36","slug":"linear-equations-with-one-variable","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-albany-chemistry\/chapter\/linear-equations-with-one-variable\/","title":{"raw":"0.5 Linear Equations with One Variable","rendered":"0.5 Linear Equations with One Variable"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n\r\n[embed]https:\/\/youtu.be\/pL3wnjWYfN8[\/embed]\r\n<h3>Quick reference<\/h3>\r\n<ul>\r\n \t<li>Addition Property of Equalities\r\n<ul>\r\n \t<li>For all real numbers <i>a<\/i>, <i>b<\/i>, and <i>c<\/i>: If [latex]a=b[\/latex], then [latex]a+c=b+c[\/latex].<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Multiplication Property of Equalities\r\n<ul>\r\n \t<li>For all real numbers <i>a<\/i>, <i>b<\/i>, and <i>c<\/i>: If <i>a<\/i> = <i>b<\/i>, then [latex]a\\cdot{c}=b\\cdot{c}[\/latex]\u00a0(or <i>ab<\/i> = <i>ac<\/i>).<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Absolute Value\r\n<ul>\r\n \t<li>For any positive number <i>a<\/i>, the solution of [latex]\\left|x\\right|=a[\/latex]\u00a0is\r\n<p style=\"text-align: center;\">[latex]x=a[\/latex] or [latex]x=\u2212a[\/latex]<\/p>\r\n<i>x <\/i>can be a single variable or any algebraic expression.<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>The Distributive Property of Multiplication\r\n<ul>\r\n \t<li>For all real numbers <i>a, b,<\/i> and <i>c<\/i>,\u00a0[latex]a(b+c)=ab+ac[\/latex].What this means is that when a number multiplies an expression inside parentheses, you can distribute the multiplication to each term of the expression individually.<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 id=\"title1\">Solve an algebraic equation using the addition property of equality<\/h2>\r\nFirst, let's define some important terminology:\r\n<ul>\r\n \t<li><strong>variables:\u00a0<\/strong> variables are symbols that stand for an unknown quantity, they are often represented with letters, like <i>x<\/i>, <i>y<\/i>, or <i>z<\/i>.<\/li>\r\n \t<li><strong>coefficient:\u00a0<\/strong>Sometimes a variable is multiplied by a number. This number is called the coefficient of the variable. For example, the coefficient of 3<i>x <\/i>is 3.<\/li>\r\n \t<li><strong>term:\u00a0<\/strong>a single number, or variables and numbers connected by multiplication. -4, 6x and [latex]x^2[\/latex] are all terms<\/li>\r\n \t<li><strong>expression: <\/strong>groups of terms connected by addition and subtraction.\u00a0 [latex]2x^2-5[\/latex] is an expression<\/li>\r\n \t<li><strong>equation: <\/strong>\u00a0an equation is a mathematical statement that two expressions are equal. An equation will always contain an equal sign with an expression on each side.\u00a0Think of an equal sign as meaning \"the same as.\" Some examples of equations are\u00a0[latex]y = mx +b[\/latex], \u00a0[latex]\\frac{3}{4}r = v^{3} - r[\/latex], and \u00a0[latex]2(6-d) + f(3 +k) = \\frac{1}{4}d[\/latex]<\/li>\r\n<\/ul>\r\nThe following figure shows how coefficients, variables, terms, and expressions all come together to make equations. In the equation [latex]2x-3^2=10x[\/latex], the variable is [latex]x[\/latex], a coefficient is [latex]10[\/latex], a term is [latex]10x[\/latex], an expression is [latex]2x-3^2[\/latex].\r\n\r\n[caption id=\"attachment_4693\" align=\"aligncenter\" width=\"424\"]<img class=\"wp-image-4693\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2651\/2018\/02\/14194601\/Screen-Shot-2016-06-08-at-2.45.15-PM-300x242.png\" alt=\"Equation made of coefficients, variables, terms and expressions.\" width=\"424\" height=\"342\" \/> Equation made of coefficients, variables, terms and expressions.[\/caption]\r\n<h3>Using the Addition Property of Equality<\/h3>\r\nAn important property of equations is that you can add the same quantity to both sides of an equation and still maintain an equivalent equation.\r\n<div class=\"textbox shaded\">\r\n<h3>Addition Property of Equality<\/h3>\r\nFor all real numbers <i>a<\/i>, <i>b<\/i>, and <i>c<\/i>: If [latex]a=b[\/latex], then [latex]a+c=b+c[\/latex].\r\n\r\nIf two expressions are equal to each other, and you add the same value to both sides of the equation, the equation will remain equal.\r\n\r\n<\/div>\r\n<h3>Solve algebraic equations using the addition property of equality<\/h3>\r\nWhen you solve an equation, you find the value of the variable that makes the equation true. In order to solve the equation, you <b>isolate the variable<\/b>. Isolating the variable means rewriting an equivalent equation in which the variable is on one side of the equation and everything else is on the other side of the equation.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Examples<\/h3>\r\nSolve [latex]x-6=8[\/latex].\r\n\r\n[reveal-answer q=\"577240\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"577240\"]\r\n\r\nThis equation means that if you begin with some unknown number, <i>x<\/i>, and subtract 6, you will end up with 8. You are trying to figure out the value of the variable <i>x.<\/i>\r\n\r\nUsing the Addition Property of Equality, add 6 to both sides of the equation to isolate the variable. You choose to add 6 because 6 is being subtracted from the variable.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\begin{array}{r}x-6\\,\\,\\,=\\,\\,\\,\\,8\\\\\\,\\,\\,\\,\\,\\,\\,\\underline{+\\,6\\,\\,\\,\\,\\,\\,\\,\\,+6}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x\\,\\,=\\, 14\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]x=14[\/latex]\r\n\r\n[\/hidden-answer]\r\nSolve [latex]x+5=27[\/latex].\r\n\r\n[reveal-answer q=\"579240\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"579240\"]\r\n\r\nThis equation means that if you begin with some unknown number, <i>x<\/i>, and add 5, you will end up with 27. You are trying to figure out the value of the variable <i>x.<\/i>\r\n\r\nUsing the Addition Property of Equality, subtract 5 from both sides of the equation to isolate the variable. You choose to subtract 5, as 5 is being added from the variable.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\begin{array}{r}x+5\\,\\,=\\,\\,27\\\\\\,\\,\\,\\,\\,\\,\\,\\underline{-5\\,\\,\\,\\,\\,\\,\\,\\,-5}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x\\,\\,=\\, 22\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]x=22[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nSince subtraction can be written as addition (adding the opposite), the <b>addition property of equality<\/b> can be used for subtraction as well. So just as you can add the same value to each side of an equation without changing the meaning of the equation, you can subtract the same value from each side of an equation.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Examples<\/h3>\r\nSolve [latex]x+10=-65[\/latex]. Check your solution.\r\n\r\n[reveal-answer q=\"684455\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"684455\"]\r\n\r\nSolve:\r\n<p style=\"text-align: center;\">[latex]x+10=-65[\/latex]<\/p>\r\nSince 10 is being added to the variable, subtract 10 from both sides. Note that subtracting 10 is the same as adding [latex]\u201310[\/latex].\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\begin{array}{r}x+10\\,\\,=\\,\\,\\,\\,-65\\\\\\,\\,\\,\\,\\,\\underline{-10\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-10}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x\\,\\,\\,\\,=\\,\\,\\,-75\\end{array}[\/latex]<\/p>\r\nTo check, substitute the solution, [latex]\u201375[\/latex] for <i>x <\/i>in the original equation, then simplify.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\begin{array}{r}\\,\\,\\,\\,\\,x+10\\,\\,\\,=-65\\\\-75+\\,10\\,\\,\\,=-65\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-65\\,\\,\\,=-65\\end{array}[\/latex]<\/p>\r\nThis equation is true, so the solution is correct.\r\n<h4>Answer<\/h4>\r\n[latex]x=\u201375[\/latex] is the solution to the equation [latex]x+10=\u201365[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\nSolve [latex]x-4=-32[\/latex]. Check your solution.\r\n\r\n[reveal-answer q=\"624455\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"624455\"]\r\n\r\nSolve:\r\n<p style=\"text-align: center;\">[latex]x-4=-32[\/latex]<\/p>\r\nSince 4 is being subtracted from\u00a0the variable, add 4 to\u00a0both sides.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\begin{array}{r}x-4\\,\\,=\\,\\,\\,\\,-32\\\\\\,\\,\\,\\,\\,\\underline{+4\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+4}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x\\,\\,\\,\\,=\\,\\,\\,-28\\end{array}[\/latex]<\/p>\r\nCheck:\r\n\r\nTo check, substitute the solution, [latex]\u201328[\/latex] for <i>x <\/i>in the original equation, then simplify.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\begin{array}{r}\\,\\,\\,\\,\\,x-4\\,\\,\\,=-32\\\\-28-\\,4\\,\\,\\,=-32\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-32\\,\\,\\,=-32\\end{array}[\/latex]<\/p>\r\nThis equation is true, so the solution is correct.\r\n<h4>Answer<\/h4>\r\n[latex]x=\u201328[\/latex] is the solution to the equation [latex]x-4=\u201332[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIt is always a good idea to check your answer whether you are\u00a0requested to or not.\r\n\r\nThe examples above are sometimes called <b>one-step equations<\/b> because they require only one step to solve. In these examples, you either added or subtracted a <b>constant<\/b> from both sides of the equation to isolate the variable and solve the equation.\r\n\r\nWith any equation, you can check your solution by substituting the value for the variable in the original equation. In other words, you evaluate the original equation using your solution. If you get a true statement, then your solution is correct.\r\n<h2 id=\"title3\">Solving One-Step Equations Containing Absolute Values with Addition<\/h2>\r\nThe <b>absolute value<\/b> of a number or expression describes its distance from 0 on a number line. Since the absolute value expresses only the distance, not the direction of the number on a number line, it is always expressed as a positive number or 0.\r\n\r\nFor example, [latex]\u22124[\/latex] and 4 both have an absolute value of 4 because they are each 4 units from 0 on a number line\u2014though they are located in opposite directions from 0 on the number line.\r\n\r\nWhen solving absolute value <b>equations<\/b> and <b>inequalities<\/b>, you have to consider both the behavior of absolute value and the properties of equality and inequality.\r\n\r\nBecause both positive and negative values have a positive absolute value, solving absolute value equations means finding the solution for both the positive and the negative values.\r\n\r\nLet\u2019s first look at a very basic example.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\left| x \\right|=5[\/latex]<\/p>\r\nThis equation is read \u201cthe absolute value of <i>x <\/i>is equal to five.\u201d The solution is the value(s) that are five units away from 0 on a number line.\r\n\r\nYou might think of 5 right away; that is one solution to the equation. Notice that [latex]\u22125[\/latex] is also a solution because [latex]\u22125[\/latex] is 5 units away from 0 in the opposite direction. So, the solution to this equation [latex] \\displaystyle \\left| x \\right|=5[\/latex] is [latex]x = \u22125[\/latex] or [latex]x = 5[\/latex].\r\n<div class=\"textbox shaded\">\r\n<h3>Solving Equations of the Form [latex]|x|=a[\/latex]<\/h3>\r\nFor any positive number <i>a<\/i>, the solution of [latex]\\left|x\\right|=a[\/latex]\u00a0is\r\n<p style=\"text-align: center;\">[latex]x=a[\/latex] or [latex]x=\u2212a[\/latex]<\/p>\r\n<i>x <\/i>can be a single variable or any algebraic expression.\r\n\r\n<\/div>\r\nYou can solve a more complex absolute value problem in a similar fashion.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve [latex] \\displaystyle \\left| x+5\\right|=15[\/latex].\r\n[reveal-answer q=\"624457\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"624457\"]\r\n\r\nThis equation asks you to find what number plus 5 has an absolute value of 15. Since 15 and [latex]\u221215[\/latex] both have an absolute value of 15, the absolute value equation is true when the quantity [latex]x + 5[\/latex]\u00a0is 15 <i>or<\/i>\u00a0[latex]x + 5[\/latex] is [latex]\u221215[\/latex], since [latex]|15|=15[\/latex] and [latex]|\u221215|=15[\/latex]. So, you need to find out what value for <i>x<\/i> will make this expression equal to 15 as well as what value for <em>x<\/em> will make the expression equal to [latex]\u221215[\/latex]. Solving the two equations you get\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\begin{array}{l}x+5=15\\,\\,\\,\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,\\,\\,\\,{x+5=-15}\\\\\\underline{\\,\\,\\,\\,\\,-5\\,\\,\\,\\,-5}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-5\\,\\,\\,\\,\\,\\,\\,\\,\\,-5}\\\\x\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\,\\,10\\,\\,\\,\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,\\,\\,\\,x\\,\\,\\,\\,\\,\\,\\,\\,=-20\\end{array}[\/latex]<\/p>\r\nYou can check these two solutions in the absolute value equation to see if [latex]x=10[\/latex] and [latex]x=\u221220[\/latex] are correct.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\begin{array}{r}\\,\\,\\left| x+5 \\right|=15\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left| x+5 \\right|=15\\\\\\left| 10+5 \\right|=15\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left| -20+5 \\right|=15\\\\\\,\\,\\,\\,\\,\\,\\,\\left| 15 \\right|=15\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left| -15 \\right|=15\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,15=15\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,15=15\\end{array}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<h2 id=\"title2\">Solve algebraic equations using the multiplication property of equality<\/h2>\r\nJust as you can add or subtract the same exact quantity on both sides of an equation, you can also multiply both sides of an equation by the same quantity to write an equivalent equation. Let\u2019s look at a numeric equation, [latex]5\\cdot3=15[\/latex], to start. If you multiply both sides of this equation by 2, you will still have a true equation.\r\n\r\n[latex]\\begin{array}{r}5\\cdot 3=15\\,\\,\\,\\,\\,\\,\\, \\\\ 5\\cdot3\\cdot2=15\\cdot2 \\\\ 30=30\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]\r\n\r\nThis characteristic of equations is generalized in the <b>multiplication property of equality<\/b>.\r\n<div class=\"textbox shaded\">\r\n<h3>Multiplication Property of Equality<\/h3>\r\nFor all real numbers <i>a<\/i>, <i>b<\/i>, and <i>c<\/i>: If <i>a<\/i> = <i>b<\/i>, then [latex]a\\cdot{c}=b\\cdot{c}[\/latex]\u00a0(or <i>ab<\/i> = <i>ac<\/i>).\r\n\r\nIf two expressions are equal to each other and you multiply both sides by the same number, the resulting expressions will also be equivalent.\r\n\r\n<\/div>\r\nWhen the equation involves multiplication or division, you can \u201cundo\u201d these operations by using the inverse operation to isolate the variable. When the operation is multiplication or division, your goal is to change the coefficient to 1, the multiplicative identity.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve [latex]3x=24[\/latex]. When you are done, check your solution.\r\n\r\n[reveal-answer q=\"42404\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"42404\"]Divide both sides of the equation by 3 to isolate the variable (have a coefficient of 1).\u00a0Dividing by 3 is the same as having multiplied by [latex] \\frac{1}{3}[\/latex].\r\n\r\n[latex]\\begin{array}{r}\\underline{3x}=\\underline{24}\\\\3\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3\\,\\\\x=8\\,\\,\\,\\end{array}[\/latex]\r\n\r\nCheck by substituting your solution, 8, for the variable in the original equation.\r\n\r\n[latex]\\begin{array}{r}3x=24 \\\\ 3\\cdot8=24 \\\\ 24=24\\end{array}[\/latex]\r\n\r\nThe solution is correct!\r\n<h4>Answer<\/h4>\r\n[latex]x=8[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nYou can also multiply the coefficient by the multiplicative inverse (reciprocal) in order to change the coefficient to 1.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve [latex] \\frac{1}{2 }{ x }={ 8}[\/latex] for x.\r\n[reveal-answer q=\"128018\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"128018\"]The only difference between this and the previous example is that the coefficient on <em>x<\/em> is a fraction. In the last example we used the reciprocal of 3 to isolate the <em>x<\/em> (disguised as division). Now we can use the reciprocal of [latex]\\frac{1}{2}[\/latex], which is 2.\r\n\r\nMultiply both sides by 2:\r\n\r\n[latex]\\begin{array}{r}\\left(2\\right)\\frac{1}{2 }{ x }=\\left(2\\right){ 8}\\\\\\left(2\\right)\\frac{1}{2 } = 1\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\left(2\\right)8 = 16\\,\\,\\,\\,\\,\\\\{ x }=16\\,\\,\\,\\,\\,\\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next example, we will solve a one-step equation using the multiplication property of equality. You will see that the variable is part of a fraction in the given equation, and using the multiplication property of equality allows us to remove the variable from the fraction. Remember that fractions imply division, so you can think of this as the variable <em>k<\/em> is being divided by 10. To \"undo\" the division, you can use multiplication to isolate <em>k<\/em>. Lastly, note that there is a negative term in the equation, so it will be important to think about the sign of each term as you work through the problem. Stop after each step you take to make sure all the terms have the correct sign.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve [latex]-\\frac{7}{2}=\\frac{k}{10}[\/latex] for <em>k<\/em>.\r\n\r\n[reveal-answer q=\"471772\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"471772\"]We want to isolate the <em>k<\/em>, which is being divided by 10. The first thing we should do is multiply both sides by 10.\r\n\r\n[latex]\\begin{array}{l}\\text{ multiply the left side by 10 }\\\\\\\\\\left(10\\right)-\\frac{7}{2}=\\frac{10\\cdot7}{2} = \\frac{70}{2} = 35\\\\\\\\\\text{ now multiply the right side by 10 }\\\\\\\\\\,\\,\\,\\frac{k}{10}\\left(10\\right) = \\frac{k\\cdot10}{10} = k\\\\\\\\\\text{ now replace your results in the equation }\\\\35=k\\end{array}[\/latex]\r\n<h4>Answer<\/h4>\r\nWe write the k on the left side as a matter of convention.\r\n\r\n[latex]k=35[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Solving One-Step Equations Containing Absolute Values With Multiplication<\/h2>\r\nRemember that absolute value refers to the distance from zero. You can use the same technique of first isolating the absolute value, then setting up and solving two equations to solve an absolute value equation involving multiplication.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve [latex] \\displaystyle \\left| 2x\\right|=6[\/latex].\r\n[reveal-answer q=\"624455\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"624455\"]\r\n\r\nThis equation asks you to find what number times 2 has an absolute value of 6.\r\n\r\nSince 6 and [latex]\u22126[\/latex] both have an absolute value of 6, the absolute value equation is true when the quantity [latex]2x[\/latex]\u00a0is 6 <i>or<\/i>\u00a0[latex]2x[\/latex] is [latex]\u22126[\/latex], since [latex]|6|=6[\/latex] and [latex]|\u22126|=6[\/latex].\r\n\r\nSo, you need to find out what value for <i>x<\/i> will make this expression equal to 6 as well as what value for <em>x<\/em> will make the expression equal to [latex]\u22126[\/latex].\r\n\r\nSolving the two equations you get\r\n<p style=\"text-align: center;\">[latex]2x=6\\text{ or }2x=-6[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{2x}{2}=\\frac{6}{2}\\text{ or }\\frac{2x}{2}=\\frac{-6}{2}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]x=3\\text{ or }x=-3[\/latex]<\/p>\r\nYou can check these two solutions in the absolute value equation to see if [latex]x=3[\/latex] and [latex]x=\u22123[\/latex] are correct.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\begin{array}{r}\\,\\,\\left|3\\cdot2 \\right|=6\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left| -3\\cdot2 \\right|=6\\\\\\left| 6 \\right|=6\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left| -6 \\right|=6\\\\\\end{array}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve [latex] \\displaystyle\\frac{1}{3}\\left|k\\right|=12[\/latex].\r\n[reveal-answer q=\"604455\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"604455\"]\r\n\r\nNotice how this example is different from the last; [latex] \\displaystyle\\frac{1}{3}[\/latex] is outside the absolute value grouping symbols. This means we need to isolate the absolute value first, then apply the definition of absolute value.\r\n\r\nFirst, isolate the absolute value term by multiplying by the inverse of [latex] \\displaystyle\\frac{1}{3}[\/latex]:\r\n\r\n[latex]\\begin{array}{r}\\frac{1}{3}\\left|k\\right|=12\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\left(3\\right)\\frac{1}{3}\\left|k\\right|=\\left(3\\right)12\\\\\\left|k\\right|=36\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]\r\n\r\nApply the definition of absolute value:\r\n\r\n[latex] \\displaystyle{k }=36\\text{ or }{k }=-36[\/latex]\r\n\r\nYou can check these two solutions in the absolute value equation to see if [latex]x=36[\/latex] and [latex]x=\u221236[\/latex] are correct.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\begin{array}{r}\\,\\,\\frac{1}{3}\\left|36 \\right|=12\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{1}{3}\\left|-36 \\right|=12\\\\\\left| 12 \\right|=12\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left| -12 \\right|=12\\\\\\end{array}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<h2>Use properties of equality to isolate variables and solve algebraic equations<\/h2>\r\n[caption id=\"attachment_4415\" align=\"aligncenter\" width=\"300\"]<img class=\"size-medium wp-image-4415\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2651\/2018\/02\/14194606\/Screen-Shot-2016-05-27-at-1.13.50-PM-300x296.png\" alt=\"steps leading to a gold ball\" width=\"300\" height=\"296\" \/> Steps With an End In Sight[\/caption]\r\n\r\nThere are some <b>equations<\/b> that you can solve in your head quickly. For example\u2014what is the value of <i>y<\/i> in the equation [latex]2y=6[\/latex]? Chances are you didn\u2019t need to get out a pencil and paper to calculate that [latex]y=3[\/latex]. You only needed to do one thing to get the answer: divide 6 by 2.\r\n\r\nOther equations are more complicated. Solving [latex]\\displaystyle 4\\left( \\frac{1}{3}t+\\frac{1}{2}\\right)=6[\/latex] without writing anything down is difficult! That\u2019s because this equation contains not just a <b>variable<\/b> but also fractions and <b>terms<\/b> inside parentheses. This is a <b>multi-step equation<\/b>, one that takes several steps to solve. Although multi-step equations take more time and more operations, they can still be simplified and solved by applying basic algebraic rules.\r\n\r\nYou can think of an equation as a balance scale, with the goal being to rewrite the equation so that it is easier to solve but still balanced. The <b>addition property of equality<\/b> and the <b>multiplication property of equality<\/b> explain how you can keep the scale, or the equation, balanced. Whenever you perform an operation to one side of the equation, if you perform the same exact operation to the other side, you\u2019ll keep both sides of the equation equal.\r\n\r\nIf the equation is in the form [latex]ax+b=c[\/latex], where <i>x<\/i> is the variable, you can solve the equation as before. First \u201cundo\u201d the addition and subtraction, and then \u201cundo\u201d the multiplication and division.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve [latex]3y+2=11[\/latex].\r\n\r\n[reveal-answer q=\"843520\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"843520\"]\r\n\r\nSubtract 2 from both sides of the equation to get the term with the variable by itself.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\begin{array}{r}3y+2\\,\\,\\,=\\,\\,11\\\\\\underline{\\,\\,\\,\\,\\,\\,\\,-2\\,\\,\\,\\,\\,\\,\\,\\,-2}\\\\3y\\,\\,\\,\\,=\\,\\,\\,\\,\\,9\\end{array}[\/latex]<\/p>\r\nDivide both sides of the equation by 3 to get a coefficient of 1 for the variable.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\,\\,\\,\\,\\,\\,\\underline{3y}\\,\\,\\,\\,=\\,\\,\\,\\,\\,\\underline{9}\\\\3\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,9\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,y\\,\\,\\,\\,=\\,\\,\\,\\,3\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]y=3[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve [latex]3x+5x+4-x+7=88[\/latex].\r\n\r\n[reveal-answer q=\"455516\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"455516\"]\r\n\r\nThere are three like terms [latex]3x[\/latex], [latex]5x[\/latex],\u00a0and\u00a0[latex]\u2013x[\/latex]\u00a0involving a variable.\u00a0Combine these like terms.\u00a04 and 7 are also like terms and can be added.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\,\\,3x+5x+4-x+7=\\,\\,\\,88\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,7x+4+7=\\,\\,\\,88\\end{array}[\/latex]<\/p>\r\nThe equation is now in the form\u00a0[latex]ax+b=c[\/latex], so we can solve as before.\r\n<p style=\"text-align: center;\">[latex]7x+11\\,\\,\\,=\\,\\,\\,88[\/latex]<\/p>\r\nSubtract 11 from both sides.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}7x+11\\,\\,\\,=\\,\\,\\,88\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-11\\,\\,\\,\\,\\,\\,\\,-11}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,7x\\,\\,\\,=\\,\\,\\,77\\end{array}[\/latex]<\/p>\r\nDivide both sides by 7.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{7x}\\,\\,\\,=\\,\\,\\,\\underline{77}\\\\7\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,7\\,\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x\\,\\,\\,=\\,\\,\\,11\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]x=11[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nSome equations may have the variable on both sides of the equal sign, as in this equation: [latex]4x-6=2x+10[\/latex].\r\n\r\nTo solve this equation, we need to \u201cmove\u201d one of the variable terms.\u00a0This can make it difficult to decide which side to work with. It doesn\u2019t matter which term gets moved, [latex]4x[\/latex] or [latex]2x[\/latex], however, to avoid negative coefficients, you can move the smaller term.\r\n<div class=\"textbox exercises\">\r\n<h3>Examples<\/h3>\r\nSolve:\u00a0[latex]4x-6=2x+10[\/latex]\r\n\r\n[reveal-answer q=\"457216\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"457216\"]\r\n\r\nChoose the variable term to move\u2014to avoid negative terms choose [latex]2x[\/latex]\r\n<p style=\"text-align: center;\">[latex]\\,\\,\\,4x-6=2x+10\\\\\\underline{-2x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-2x}\\\\\\,\\,\\,2x-6=10[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Now add\u00a06 to both\u00a0sides to isolate the term with the variable.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x-6=10\\\\\\underline{\\,\\,\\,\\,+6\\,\\,\\,+6}\\\\2x=16\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Now divide each side by 4 to isolate the variable x.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\frac{2x}{2}=\\frac{16}{2}\\\\\\\\x=8\\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Solving Multi-Step Equations With Absolute Value<\/h2>\r\nWe can apply the same techniques we used for solving a one-step equation which contains absolute value to an equation that will take more than one step to solve. \u00a0Let's start with an example where the first step is to write two equations, one equal to positive 26 and one equal to negative 26.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve for <i>p<\/i>.\u00a0[latex]\\left|2p\u20134\\right|=26[\/latex]\r\n\r\n[reveal-answer q=\"371950\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"371950\"]\r\n\r\nWrite the two equations that will give an absolute value of 26.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle 2p-4=26\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,\\,\\,2p-4=\\,-26[\/latex]<\/p>\r\nSolve each equation for <i>p <\/i>by isolating the variable<i>.<\/i>\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\begin{array}{r}2p-4=26\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2p-4=\\,-26\\\\\\underline{\\,\\,\\,\\,\\,\\,+4\\,\\,\\,\\,+4}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{\\,\\,\\,\\,\\,\\,\\,+4\\,\\,\\,\\,\\,\\,\\,+4}\\\\\\underline{2p}\\,\\,\\,\\,\\,\\,=\\underline{30}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{2p}\\,\\,\\,\\,\\,=\\,\\underline{-22}\\\\2\\,\\,\\,\\,\\,\\,\\,=\\,2\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2\\,\\,\\,\\,\\,\\,\\,=\\,\\,\\,\\,\\,\\,\\,\\,2\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,p=15\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,p=\\,-11\\end{array}[\/latex]<\/p>\r\nCheck the solutions in the original equation.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\begin{array}{r}\\,\\,\\,\\,\\,\\left| 2p-4 \\right|=26\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left| 2p-4 \\right|=26\\\\\\left| 2(15)-4 \\right|=26\\,\\,\\,\\,\\,\\,\\,\\left| 2(-11)-4 \\right|=26\\\\\\,\\,\\,\\,\\,\\left| 30-4 \\right|=26\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left| -22-4 \\right|=26\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left| 26 \\right|=26\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left| -26 \\right|=26\\end{array}[\/latex]<\/p>\r\nBoth solutions check!\r\n<h4>Answer<\/h4>\r\n[latex]p=15[\/latex] or [latex]p=-11[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nNow let's look at an example where you need to do an algebraic step or two before you can write your two equations. The goal here is to get the absolute value on one side of the equation by itself. Then we can proceed as we did in the previous example.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve for <i>w<\/i>. [latex]3\\left|4w\u20131\\right|\u20135=10[\/latex]\r\n\r\n[reveal-answer q=\"303228\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"303228\"]\r\n\r\nIsolate the term with the absolute value by adding 5 to both sides.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}3\\left|4w-1\\right|-5=10\\\\\\underline{\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+5\\,\\,\\,+5}\\\\ 3\\left|4w-1\\right|=15\\end{array}[\/latex]<\/p>\r\nDivide both sides by 3.\u00a0Now the absolute value is isolated.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r} \\underline{3\\left|4w-1\\right|}=\\underline{15}\\\\3\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3\\,\\,\\\\\\left|4w-1\\right|=\\,\\,5\\end{array}[\/latex]<\/p>\r\nWrite the two equations that will give an absolute value of 5 and solve them.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\begin{array}{r}4w-1=5\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,4w-1=-5\\\\\\underline{\\,\\,\\,\\,\\,\\,\\,+1\\,\\,+1}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{\\,\\,\\,\\,\\,\\,\\,\\,\\,+1\\,\\,\\,\\,\\,+1}\\\\\\,\\,\\,\\,\\,\\underline{4w}=\\underline{6}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{4w}\\,\\,\\,\\,\\,\\,\\,=\\underline{-4}\\\\4\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,4\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,4\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,4\\,\\,\\\\\\,\\,\\,\\,\\,\\,\\,\\,w=\\frac{3}{2}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,w=-1\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,w=\\frac{3}{2}\\,\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,\\,-1\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nCheck the solutions in the original equation.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\begin{array}{r}\\,\\,\\,\\,\\,3\\left| 4w-1\\, \\right|-5=10\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3\\left| 4w-1\\, \\right|-5=10\\\\\\\\3\\left| 4\\left( \\frac{3}{2} \\right)-1\\, \\right|-5=10\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3\\left| 4w-1\\, \\right|-5=10\\\\\\\\\\,\\,\\,\\,\\,\\,3\\left| \\frac{12}{2}-1\\, \\right|-5=10\\,\\,\\,\\,\\,\\,\\,3\\left| 4(-1)-1\\, \\right|-5=10\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,3\\left| 6-1\\, \\right|-5=10\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3\\left| -4-1\\, \\right|-5=10\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3\\left(5\\right)-5=10\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3\\left| -5 \\right|-5=10\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,15-5=10\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,15-5=10\\\\10=10\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,10=10\\end{array}[\/latex]<\/p>\r\nBoth solutions check\r\n<h4>Answer<\/h4>\r\n[latex]w=-1\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,w=\\frac{3}{2}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>The Distributive Property<\/h2>\r\nAs we solve linear equations, we often need to do some work to write\u00a0the linear equations in a form we are familiar with solving.\u00a0This section will focus on manipulating an equation we are asked to solve in such a way that we can use the skills we learned for solving multi-step equations to ultimately arrive at the solution.\r\n\r\nParentheses can\u00a0make solving a problem difficult, if not impossible. To get rid of these unwanted parentheses we have to use the distributive property. Using this property we multiply the number in front of the parentheses by each term inside of the parentheses.\r\n<div class=\"textbox shaded\">\r\n<h3>The Distributive Property of Multiplication<\/h3>\r\nFor all real numbers <i>a, b,<\/i> and <i>c<\/i>,\u00a0[latex]a(b+c)=ab+ac[\/latex].\r\n\r\nWhat this means is that when a number multiplies an expression inside parentheses, you can distribute the multiplication to each term of the expression individually. Then, you can follow the steps we have already practiced\u00a0to <b>isolate the variable<\/b>\u00a0and solve the equation.\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve for [latex]a[\/latex]. [latex]4\\left(2a+3\\right)=28[\/latex]\r\n\r\n[reveal-answer q=\"372387\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"372387\"]\r\n\r\nApply the distributive property to expand [latex]4\\left(2a+3\\right)[\/latex] to [latex]8a+12[\/latex]\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}4\\left(2a+3\\right)=28\\\\ 8a+12=28\\end{array}[\/latex]<\/p>\r\nSubtract 12\u00a0from both sides to isolate\u00a0the variable term.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}8a+12\\,\\,\\,=\\,\\,\\,28\\\\ \\underline{-12\\,\\,\\,\\,\\,\\,-12}\\\\ 8a\\,\\,\\,=\\,\\,\\,16\\end{array}[\/latex]<\/p>\r\nDivide both terms by 8 to get a coefficient of 1.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\underline{8a}=\\underline{16}\\\\8\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,8\\\\a\\,=\\,\\,2\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]a=2[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next example, you will see that there are parentheses on both sides of the equal sign, so you will need to use the distributive property twice. Notice that you are going to need to distribute a negative number, so be careful with negative signs!\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve for [latex]t[\/latex].\u00a0\u00a0[latex]2\\left(4t-5\\right)=-3\\left(2t+1\\right)[\/latex]\r\n\r\n[reveal-answer q=\"302387\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"302387\"]\r\n\r\nApply the distributive property to expand [latex]2\\left(4t-5\\right)[\/latex] to [latex]8t-10[\/latex] and [latex]-3\\left(2t+1\\right)[\/latex] to[latex]-6t-3[\/latex]. Be careful in this step\u2014you are distributing a negative number, so keep track of the sign of each number after you multiply.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2\\left(4t-5\\right)=-3\\left(2t+1\\right)\\,\\,\\,\\,\\,\\, \\\\ 8t-10=-6t-3\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nAdd [latex]-6t[\/latex] to both sides to begin combining like terms.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}8t-10=-6t-3\\\\ \\underline{+6t\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+6t}\\,\\,\\,\\,\\,\\,\\,\\\\ 14t-10=\\,\\,\\,\\,-3\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nAdd 10 to both sides of the equation to isolate <em>t<\/em>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}14t-10=-3\\\\ \\underline{+10\\,\\,\\,+10}\\\\ 14t=\\,\\,\\,7\\,\\end{array}[\/latex]<\/p>\r\nThe last step is to divide both sides by 14 to completely isolate <em>t<\/em>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}14t=7\\,\\,\\,\\,\\\\\\frac{14t}{14}=\\frac{7}{14}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]t=\\frac{1}{2}[\/latex]\r\n\r\nWe simplified the fraction [latex]\\frac{7}{14}[\/latex] into [latex]\\frac{1}{2}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nSometimes, you will encounter a multi-step equation with fractions. If you prefer not working with fractions, you can use the multiplication property of equality to multiply both sides of the equation by a common denominator of all of the fractions in the equation. This will clear all the fractions out of the equation. See the example below.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve \u00a0[latex]\\frac{1}{2}x-3=2-\\frac{3}{4}x[\/latex] by clearing the fractions in the equation first.\r\n\r\n[reveal-answer q=\"129951\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"129951\"]\r\n\r\nMultiply both sides of the equation by 4, the common denominator of the fractional coefficients.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\frac{1}{2}x-3=2-\\frac{3}{4}x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\ 4\\left(\\frac{1}{2}x-3\\right)=4\\left(2-\\frac{3}{4}x\\right)\\end{array}[\/latex]<\/p>\r\nUse the distributive property to expand the expressions on both sides.\u00a0Multiply.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}4\\left(\\frac{1}{2}x\\right)-4\\left(3\\right)=4\\left(2\\right)-4\\left(-\\frac{3}{4}x\\right)\\\\\\\\ \\frac{4}{2}x-12=8-\\frac{12}{4}x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\\\\\ 2x-12=8-3x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\end{array}[\/latex]<\/p>\r\nAdd 3<em>x<\/em> to both sides to move the variable terms to only one side. Add 12 to both sides to move the variable\u00a0terms to only one side.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x-12=8-3x\\, \\\\\\underline{+3x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+3x}\\\\ 5x-12=8\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nAdd 12 to both sides to move the <b>constant<\/b> terms to the other side.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}5x-12=8\\,\\,\\\\ \\underline{\\,\\,\\,\\,\\,\\,+12\\,+12} \\\\5x=20\\end{array}[\/latex]<\/p>\r\nDivide to isolate the variable.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\underline{5x}=\\underline{5}\\\\ 5\\,\\,\\,\\,\\,\\,\\,\\,\\,5\\\\ x=4\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]x=4[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nOf course, if you like to work with fractions, you can just apply your knowledge of operations with fractions and solve.\r\n\r\nRegardless of which method you use to solve equations containing variables, you will get the same answer. You can choose the method you find the easiest! Remember to check your answer by substituting your solution into the original equation.\r\n\r\nSometimes, you will encounter a multi-step equation with decimals. If you prefer not working with decimals, you can use the multiplication property of equality to multiply both sides of the equation by a factor of 10 that will help clear the decimals. See the example below.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve [latex]3y+10.5=6.5+2.5y[\/latex] by clearing the decimals in the equation first.\r\n\r\n[reveal-answer q=\"159951\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"159951\"]\r\n\r\nSince the smallest decimal place represented in the equation is 0.10, we want to multiply by 10 to make 1.0\u00a0and clear\u00a0the decimals from the equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}3y+10.5=6.5+2.5y\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\ 10\\left(3y+10.5\\right)=10\\left(6.5+2.5y\\right)\\end{array}[\/latex]<\/p>\r\nUse the distributive property to expand the expressions on both sides.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}10\\left(3y\\right)+10\\left(10.5\\right)=10\\left(6.5\\right)+10\\left(2.5y\\right)\\end{array}[\/latex]<\/p>\r\nMultiply.\r\n<p style=\"text-align: center;\">[latex]30y+105=65+25y[\/latex]<\/p>\r\nMove the smaller variable term, [latex]25y[\/latex], by subtracting it from both sides.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}30y+105=65+25y\\,\\,\\\\ \\underline{-25y\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-25y} \\\\5y+105=65\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nSubtract 105 from both sides to isolate the term with the variable.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}5y+105=65\\,\\,\\,\\\\ \\underline{\\,\\,\\,\\,\\,\\,-105\\,-105} \\\\5y=-40\\end{array}[\/latex]<\/p>\r\nDivide both sides by 5 to isolate the <em>y<\/em>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\underline{5y}=\\underline{-40}\\\\ 5\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,5\\\\ \\,\\,\\,x=-8\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]x=-8[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nHere are some steps to follow when you solve multi-step equations.\r\n<div class=\"textbox shaded\">\r\n<h3>Solving Multi-Step Equations<\/h3>\r\n1. (Optional) Multiply to clear any fractions or decimals.\r\n\r\n2. Simplify each side by clearing parentheses and combining like terms.\r\n\r\n3. Add or subtract to isolate the variable term\u2014you may have to move a term with the variable.\r\n\r\n4. Multiply or divide to isolate the variable.\r\n\r\n5. Check the solution.\r\n\r\n<\/div>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"General Chemistry Lecture 0.5 Linear Equations with One Variable\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/pL3wnjWYfN8?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h3>Quick reference<\/h3>\n<ul>\n<li>Addition Property of Equalities\n<ul>\n<li>For all real numbers <i>a<\/i>, <i>b<\/i>, and <i>c<\/i>: If [latex]a=b[\/latex], then [latex]a+c=b+c[\/latex].<\/li>\n<\/ul>\n<\/li>\n<li>Multiplication Property of Equalities\n<ul>\n<li>For all real numbers <i>a<\/i>, <i>b<\/i>, and <i>c<\/i>: If <i>a<\/i> = <i>b<\/i>, then [latex]a\\cdot{c}=b\\cdot{c}[\/latex]\u00a0(or <i>ab<\/i> = <i>ac<\/i>).<\/li>\n<\/ul>\n<\/li>\n<li>Absolute Value\n<ul>\n<li>For any positive number <i>a<\/i>, the solution of [latex]\\left|x\\right|=a[\/latex]\u00a0is\n<p style=\"text-align: center;\">[latex]x=a[\/latex] or [latex]x=\u2212a[\/latex]<\/p>\n<p><i>x <\/i>can be a single variable or any algebraic expression.<\/li>\n<\/ul>\n<\/li>\n<li>The Distributive Property of Multiplication\n<ul>\n<li>For all real numbers <i>a, b,<\/i> and <i>c<\/i>,\u00a0[latex]a(b+c)=ab+ac[\/latex].What this means is that when a number multiplies an expression inside parentheses, you can distribute the multiplication to each term of the expression individually.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<h2 id=\"title1\">Solve an algebraic equation using the addition property of equality<\/h2>\n<p>First, let&#8217;s define some important terminology:<\/p>\n<ul>\n<li><strong>variables:\u00a0<\/strong> variables are symbols that stand for an unknown quantity, they are often represented with letters, like <i>x<\/i>, <i>y<\/i>, or <i>z<\/i>.<\/li>\n<li><strong>coefficient:\u00a0<\/strong>Sometimes a variable is multiplied by a number. This number is called the coefficient of the variable. For example, the coefficient of 3<i>x <\/i>is 3.<\/li>\n<li><strong>term:\u00a0<\/strong>a single number, or variables and numbers connected by multiplication. -4, 6x and [latex]x^2[\/latex] are all terms<\/li>\n<li><strong>expression: <\/strong>groups of terms connected by addition and subtraction.\u00a0 [latex]2x^2-5[\/latex] is an expression<\/li>\n<li><strong>equation: <\/strong>\u00a0an equation is a mathematical statement that two expressions are equal. An equation will always contain an equal sign with an expression on each side.\u00a0Think of an equal sign as meaning &#8220;the same as.&#8221; Some examples of equations are\u00a0[latex]y = mx +b[\/latex], \u00a0[latex]\\frac{3}{4}r = v^{3} - r[\/latex], and \u00a0[latex]2(6-d) + f(3 +k) = \\frac{1}{4}d[\/latex]<\/li>\n<\/ul>\n<p>The following figure shows how coefficients, variables, terms, and expressions all come together to make equations. In the equation [latex]2x-3^2=10x[\/latex], the variable is [latex]x[\/latex], a coefficient is [latex]10[\/latex], a term is [latex]10x[\/latex], an expression is [latex]2x-3^2[\/latex].<\/p>\n<div id=\"attachment_4693\" style=\"width: 434px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-4693\" class=\"wp-image-4693\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2651\/2018\/02\/14194601\/Screen-Shot-2016-06-08-at-2.45.15-PM-300x242.png\" alt=\"Equation made of coefficients, variables, terms and expressions.\" width=\"424\" height=\"342\" \/><\/p>\n<p id=\"caption-attachment-4693\" class=\"wp-caption-text\">Equation made of coefficients, variables, terms and expressions.<\/p>\n<\/div>\n<h3>Using the Addition Property of Equality<\/h3>\n<p>An important property of equations is that you can add the same quantity to both sides of an equation and still maintain an equivalent equation.<\/p>\n<div class=\"textbox shaded\">\n<h3>Addition Property of Equality<\/h3>\n<p>For all real numbers <i>a<\/i>, <i>b<\/i>, and <i>c<\/i>: If [latex]a=b[\/latex], then [latex]a+c=b+c[\/latex].<\/p>\n<p>If two expressions are equal to each other, and you add the same value to both sides of the equation, the equation will remain equal.<\/p>\n<\/div>\n<h3>Solve algebraic equations using the addition property of equality<\/h3>\n<p>When you solve an equation, you find the value of the variable that makes the equation true. In order to solve the equation, you <b>isolate the variable<\/b>. Isolating the variable means rewriting an equivalent equation in which the variable is on one side of the equation and everything else is on the other side of the equation.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Examples<\/h3>\n<p>Solve [latex]x-6=8[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q577240\">Show Solution<\/span><\/p>\n<div id=\"q577240\" class=\"hidden-answer\" style=\"display: none\">\n<p>This equation means that if you begin with some unknown number, <i>x<\/i>, and subtract 6, you will end up with 8. You are trying to figure out the value of the variable <i>x.<\/i><\/p>\n<p>Using the Addition Property of Equality, add 6 to both sides of the equation to isolate the variable. You choose to add 6 because 6 is being subtracted from the variable.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{r}x-6\\,\\,\\,=\\,\\,\\,\\,8\\\\\\,\\,\\,\\,\\,\\,\\,\\underline{+\\,6\\,\\,\\,\\,\\,\\,\\,\\,+6}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x\\,\\,=\\, 14\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=14[\/latex]<\/p>\n<\/div>\n<\/div>\n<p>Solve [latex]x+5=27[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q579240\">Show Solution<\/span><\/p>\n<div id=\"q579240\" class=\"hidden-answer\" style=\"display: none\">\n<p>This equation means that if you begin with some unknown number, <i>x<\/i>, and add 5, you will end up with 27. You are trying to figure out the value of the variable <i>x.<\/i><\/p>\n<p>Using the Addition Property of Equality, subtract 5 from both sides of the equation to isolate the variable. You choose to subtract 5, as 5 is being added from the variable.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{r}x+5\\,\\,=\\,\\,27\\\\\\,\\,\\,\\,\\,\\,\\,\\underline{-5\\,\\,\\,\\,\\,\\,\\,\\,-5}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x\\,\\,=\\, 22\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=22[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Since subtraction can be written as addition (adding the opposite), the <b>addition property of equality<\/b> can be used for subtraction as well. So just as you can add the same value to each side of an equation without changing the meaning of the equation, you can subtract the same value from each side of an equation.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Examples<\/h3>\n<p>Solve [latex]x+10=-65[\/latex]. Check your solution.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q684455\">Show Solution<\/span><\/p>\n<div id=\"q684455\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solve:<\/p>\n<p style=\"text-align: center;\">[latex]x+10=-65[\/latex]<\/p>\n<p>Since 10 is being added to the variable, subtract 10 from both sides. Note that subtracting 10 is the same as adding [latex]\u201310[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{r}x+10\\,\\,=\\,\\,\\,\\,-65\\\\\\,\\,\\,\\,\\,\\underline{-10\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-10}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x\\,\\,\\,\\,=\\,\\,\\,-75\\end{array}[\/latex]<\/p>\n<p>To check, substitute the solution, [latex]\u201375[\/latex] for <i>x <\/i>in the original equation, then simplify.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{r}\\,\\,\\,\\,\\,x+10\\,\\,\\,=-65\\\\-75+\\,10\\,\\,\\,=-65\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-65\\,\\,\\,=-65\\end{array}[\/latex]<\/p>\n<p>This equation is true, so the solution is correct.<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=\u201375[\/latex] is the solution to the equation [latex]x+10=\u201365[\/latex].<\/p>\n<\/div>\n<\/div>\n<p>Solve [latex]x-4=-32[\/latex]. Check your solution.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q624455\">Show Solution<\/span><\/p>\n<div id=\"q624455\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solve:<\/p>\n<p style=\"text-align: center;\">[latex]x-4=-32[\/latex]<\/p>\n<p>Since 4 is being subtracted from\u00a0the variable, add 4 to\u00a0both sides.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{r}x-4\\,\\,=\\,\\,\\,\\,-32\\\\\\,\\,\\,\\,\\,\\underline{+4\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+4}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x\\,\\,\\,\\,=\\,\\,\\,-28\\end{array}[\/latex]<\/p>\n<p>Check:<\/p>\n<p>To check, substitute the solution, [latex]\u201328[\/latex] for <i>x <\/i>in the original equation, then simplify.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{r}\\,\\,\\,\\,\\,x-4\\,\\,\\,=-32\\\\-28-\\,4\\,\\,\\,=-32\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-32\\,\\,\\,=-32\\end{array}[\/latex]<\/p>\n<p>This equation is true, so the solution is correct.<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=\u201328[\/latex] is the solution to the equation [latex]x-4=\u201332[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>It is always a good idea to check your answer whether you are\u00a0requested to or not.<\/p>\n<p>The examples above are sometimes called <b>one-step equations<\/b> because they require only one step to solve. In these examples, you either added or subtracted a <b>constant<\/b> from both sides of the equation to isolate the variable and solve the equation.<\/p>\n<p>With any equation, you can check your solution by substituting the value for the variable in the original equation. In other words, you evaluate the original equation using your solution. If you get a true statement, then your solution is correct.<\/p>\n<h2 id=\"title3\">Solving One-Step Equations Containing Absolute Values with Addition<\/h2>\n<p>The <b>absolute value<\/b> of a number or expression describes its distance from 0 on a number line. Since the absolute value expresses only the distance, not the direction of the number on a number line, it is always expressed as a positive number or 0.<\/p>\n<p>For example, [latex]\u22124[\/latex] and 4 both have an absolute value of 4 because they are each 4 units from 0 on a number line\u2014though they are located in opposite directions from 0 on the number line.<\/p>\n<p>When solving absolute value <b>equations<\/b> and <b>inequalities<\/b>, you have to consider both the behavior of absolute value and the properties of equality and inequality.<\/p>\n<p>Because both positive and negative values have a positive absolute value, solving absolute value equations means finding the solution for both the positive and the negative values.<\/p>\n<p>Let\u2019s first look at a very basic example.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\left| x \\right|=5[\/latex]<\/p>\n<p>This equation is read \u201cthe absolute value of <i>x <\/i>is equal to five.\u201d The solution is the value(s) that are five units away from 0 on a number line.<\/p>\n<p>You might think of 5 right away; that is one solution to the equation. Notice that [latex]\u22125[\/latex] is also a solution because [latex]\u22125[\/latex] is 5 units away from 0 in the opposite direction. So, the solution to this equation [latex]\\displaystyle \\left| x \\right|=5[\/latex] is [latex]x = \u22125[\/latex] or [latex]x = 5[\/latex].<\/p>\n<div class=\"textbox shaded\">\n<h3>Solving Equations of the Form [latex]|x|=a[\/latex]<\/h3>\n<p>For any positive number <i>a<\/i>, the solution of [latex]\\left|x\\right|=a[\/latex]\u00a0is<\/p>\n<p style=\"text-align: center;\">[latex]x=a[\/latex] or [latex]x=\u2212a[\/latex]<\/p>\n<p><i>x <\/i>can be a single variable or any algebraic expression.<\/p>\n<\/div>\n<p>You can solve a more complex absolute value problem in a similar fashion.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve [latex]\\displaystyle \\left| x+5\\right|=15[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q624457\">Show Solution<\/span><\/p>\n<div id=\"q624457\" class=\"hidden-answer\" style=\"display: none\">\n<p>This equation asks you to find what number plus 5 has an absolute value of 15. Since 15 and [latex]\u221215[\/latex] both have an absolute value of 15, the absolute value equation is true when the quantity [latex]x + 5[\/latex]\u00a0is 15 <i>or<\/i>\u00a0[latex]x + 5[\/latex] is [latex]\u221215[\/latex], since [latex]|15|=15[\/latex] and [latex]|\u221215|=15[\/latex]. So, you need to find out what value for <i>x<\/i> will make this expression equal to 15 as well as what value for <em>x<\/em> will make the expression equal to [latex]\u221215[\/latex]. Solving the two equations you get<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{l}x+5=15\\,\\,\\,\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,\\,\\,\\,{x+5=-15}\\\\\\underline{\\,\\,\\,\\,\\,-5\\,\\,\\,\\,-5}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-5\\,\\,\\,\\,\\,\\,\\,\\,\\,-5}\\\\x\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\,\\,10\\,\\,\\,\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,\\,\\,\\,x\\,\\,\\,\\,\\,\\,\\,\\,=-20\\end{array}[\/latex]<\/p>\n<p>You can check these two solutions in the absolute value equation to see if [latex]x=10[\/latex] and [latex]x=\u221220[\/latex] are correct.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{r}\\,\\,\\left| x+5 \\right|=15\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left| x+5 \\right|=15\\\\\\left| 10+5 \\right|=15\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left| -20+5 \\right|=15\\\\\\,\\,\\,\\,\\,\\,\\,\\left| 15 \\right|=15\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left| -15 \\right|=15\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,15=15\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,15=15\\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<h2 id=\"title2\">Solve algebraic equations using the multiplication property of equality<\/h2>\n<p>Just as you can add or subtract the same exact quantity on both sides of an equation, you can also multiply both sides of an equation by the same quantity to write an equivalent equation. Let\u2019s look at a numeric equation, [latex]5\\cdot3=15[\/latex], to start. If you multiply both sides of this equation by 2, you will still have a true equation.<\/p>\n<p>[latex]\\begin{array}{r}5\\cdot 3=15\\,\\,\\,\\,\\,\\,\\, \\\\ 5\\cdot3\\cdot2=15\\cdot2 \\\\ 30=30\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>This characteristic of equations is generalized in the <b>multiplication property of equality<\/b>.<\/p>\n<div class=\"textbox shaded\">\n<h3>Multiplication Property of Equality<\/h3>\n<p>For all real numbers <i>a<\/i>, <i>b<\/i>, and <i>c<\/i>: If <i>a<\/i> = <i>b<\/i>, then [latex]a\\cdot{c}=b\\cdot{c}[\/latex]\u00a0(or <i>ab<\/i> = <i>ac<\/i>).<\/p>\n<p>If two expressions are equal to each other and you multiply both sides by the same number, the resulting expressions will also be equivalent.<\/p>\n<\/div>\n<p>When the equation involves multiplication or division, you can \u201cundo\u201d these operations by using the inverse operation to isolate the variable. When the operation is multiplication or division, your goal is to change the coefficient to 1, the multiplicative identity.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve [latex]3x=24[\/latex]. When you are done, check your solution.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q42404\">Show Solution<\/span><\/p>\n<div id=\"q42404\" class=\"hidden-answer\" style=\"display: none\">Divide both sides of the equation by 3 to isolate the variable (have a coefficient of 1).\u00a0Dividing by 3 is the same as having multiplied by [latex]\\frac{1}{3}[\/latex].<\/p>\n<p>[latex]\\begin{array}{r}\\underline{3x}=\\underline{24}\\\\3\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3\\,\\\\x=8\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Check by substituting your solution, 8, for the variable in the original equation.<\/p>\n<p>[latex]\\begin{array}{r}3x=24 \\\\ 3\\cdot8=24 \\\\ 24=24\\end{array}[\/latex]<\/p>\n<p>The solution is correct!<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=8[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>You can also multiply the coefficient by the multiplicative inverse (reciprocal) in order to change the coefficient to 1.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve [latex]\\frac{1}{2 }{ x }={ 8}[\/latex] for x.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q128018\">Show Solution<\/span><\/p>\n<div id=\"q128018\" class=\"hidden-answer\" style=\"display: none\">The only difference between this and the previous example is that the coefficient on <em>x<\/em> is a fraction. In the last example we used the reciprocal of 3 to isolate the <em>x<\/em> (disguised as division). Now we can use the reciprocal of [latex]\\frac{1}{2}[\/latex], which is 2.<\/p>\n<p>Multiply both sides by 2:<\/p>\n<p>[latex]\\begin{array}{r}\\left(2\\right)\\frac{1}{2 }{ x }=\\left(2\\right){ 8}\\\\\\left(2\\right)\\frac{1}{2 } = 1\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\left(2\\right)8 = 16\\,\\,\\,\\,\\,\\\\{ x }=16\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next example, we will solve a one-step equation using the multiplication property of equality. You will see that the variable is part of a fraction in the given equation, and using the multiplication property of equality allows us to remove the variable from the fraction. Remember that fractions imply division, so you can think of this as the variable <em>k<\/em> is being divided by 10. To &#8220;undo&#8221; the division, you can use multiplication to isolate <em>k<\/em>. Lastly, note that there is a negative term in the equation, so it will be important to think about the sign of each term as you work through the problem. Stop after each step you take to make sure all the terms have the correct sign.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve [latex]-\\frac{7}{2}=\\frac{k}{10}[\/latex] for <em>k<\/em>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q471772\">Show Solution<\/span><\/p>\n<div id=\"q471772\" class=\"hidden-answer\" style=\"display: none\">We want to isolate the <em>k<\/em>, which is being divided by 10. The first thing we should do is multiply both sides by 10.<\/p>\n<p>[latex]\\begin{array}{l}\\text{ multiply the left side by 10 }\\\\\\\\\\left(10\\right)-\\frac{7}{2}=\\frac{10\\cdot7}{2} = \\frac{70}{2} = 35\\\\\\\\\\text{ now multiply the right side by 10 }\\\\\\\\\\,\\,\\,\\frac{k}{10}\\left(10\\right) = \\frac{k\\cdot10}{10} = k\\\\\\\\\\text{ now replace your results in the equation }\\\\35=k\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>We write the k on the left side as a matter of convention.<\/p>\n<p>[latex]k=35[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Solving One-Step Equations Containing Absolute Values With Multiplication<\/h2>\n<p>Remember that absolute value refers to the distance from zero. You can use the same technique of first isolating the absolute value, then setting up and solving two equations to solve an absolute value equation involving multiplication.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve [latex]\\displaystyle \\left| 2x\\right|=6[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q624455\">Show Solution<\/span><\/p>\n<div id=\"q624455\" class=\"hidden-answer\" style=\"display: none\">\n<p>This equation asks you to find what number times 2 has an absolute value of 6.<\/p>\n<p>Since 6 and [latex]\u22126[\/latex] both have an absolute value of 6, the absolute value equation is true when the quantity [latex]2x[\/latex]\u00a0is 6 <i>or<\/i>\u00a0[latex]2x[\/latex] is [latex]\u22126[\/latex], since [latex]|6|=6[\/latex] and [latex]|\u22126|=6[\/latex].<\/p>\n<p>So, you need to find out what value for <i>x<\/i> will make this expression equal to 6 as well as what value for <em>x<\/em> will make the expression equal to [latex]\u22126[\/latex].<\/p>\n<p>Solving the two equations you get<\/p>\n<p style=\"text-align: center;\">[latex]2x=6\\text{ or }2x=-6[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{2x}{2}=\\frac{6}{2}\\text{ or }\\frac{2x}{2}=\\frac{-6}{2}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x=3\\text{ or }x=-3[\/latex]<\/p>\n<p>You can check these two solutions in the absolute value equation to see if [latex]x=3[\/latex] and [latex]x=\u22123[\/latex] are correct.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{r}\\,\\,\\left|3\\cdot2 \\right|=6\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left| -3\\cdot2 \\right|=6\\\\\\left| 6 \\right|=6\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left| -6 \\right|=6\\\\\\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve [latex]\\displaystyle\\frac{1}{3}\\left|k\\right|=12[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q604455\">Show Solution<\/span><\/p>\n<div id=\"q604455\" class=\"hidden-answer\" style=\"display: none\">\n<p>Notice how this example is different from the last; [latex]\\displaystyle\\frac{1}{3}[\/latex] is outside the absolute value grouping symbols. This means we need to isolate the absolute value first, then apply the definition of absolute value.<\/p>\n<p>First, isolate the absolute value term by multiplying by the inverse of [latex]\\displaystyle\\frac{1}{3}[\/latex]:<\/p>\n<p>[latex]\\begin{array}{r}\\frac{1}{3}\\left|k\\right|=12\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\left(3\\right)\\frac{1}{3}\\left|k\\right|=\\left(3\\right)12\\\\\\left|k\\right|=36\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Apply the definition of absolute value:<\/p>\n<p>[latex]\\displaystyle{k }=36\\text{ or }{k }=-36[\/latex]<\/p>\n<p>You can check these two solutions in the absolute value equation to see if [latex]x=36[\/latex] and [latex]x=\u221236[\/latex] are correct.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{r}\\,\\,\\frac{1}{3}\\left|36 \\right|=12\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{1}{3}\\left|-36 \\right|=12\\\\\\left| 12 \\right|=12\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left| -12 \\right|=12\\\\\\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<h2>Use properties of equality to isolate variables and solve algebraic equations<\/h2>\n<div id=\"attachment_4415\" style=\"width: 310px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-4415\" class=\"size-medium wp-image-4415\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2651\/2018\/02\/14194606\/Screen-Shot-2016-05-27-at-1.13.50-PM-300x296.png\" alt=\"steps leading to a gold ball\" width=\"300\" height=\"296\" \/><\/p>\n<p id=\"caption-attachment-4415\" class=\"wp-caption-text\">Steps With an End In Sight<\/p>\n<\/div>\n<p>There are some <b>equations<\/b> that you can solve in your head quickly. For example\u2014what is the value of <i>y<\/i> in the equation [latex]2y=6[\/latex]? Chances are you didn\u2019t need to get out a pencil and paper to calculate that [latex]y=3[\/latex]. You only needed to do one thing to get the answer: divide 6 by 2.<\/p>\n<p>Other equations are more complicated. Solving [latex]\\displaystyle 4\\left( \\frac{1}{3}t+\\frac{1}{2}\\right)=6[\/latex] without writing anything down is difficult! That\u2019s because this equation contains not just a <b>variable<\/b> but also fractions and <b>terms<\/b> inside parentheses. This is a <b>multi-step equation<\/b>, one that takes several steps to solve. Although multi-step equations take more time and more operations, they can still be simplified and solved by applying basic algebraic rules.<\/p>\n<p>You can think of an equation as a balance scale, with the goal being to rewrite the equation so that it is easier to solve but still balanced. The <b>addition property of equality<\/b> and the <b>multiplication property of equality<\/b> explain how you can keep the scale, or the equation, balanced. Whenever you perform an operation to one side of the equation, if you perform the same exact operation to the other side, you\u2019ll keep both sides of the equation equal.<\/p>\n<p>If the equation is in the form [latex]ax+b=c[\/latex], where <i>x<\/i> is the variable, you can solve the equation as before. First \u201cundo\u201d the addition and subtraction, and then \u201cundo\u201d the multiplication and division.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve [latex]3y+2=11[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q843520\">Show Solution<\/span><\/p>\n<div id=\"q843520\" class=\"hidden-answer\" style=\"display: none\">\n<p>Subtract 2 from both sides of the equation to get the term with the variable by itself.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{r}3y+2\\,\\,\\,=\\,\\,11\\\\\\underline{\\,\\,\\,\\,\\,\\,\\,-2\\,\\,\\,\\,\\,\\,\\,\\,-2}\\\\3y\\,\\,\\,\\,=\\,\\,\\,\\,\\,9\\end{array}[\/latex]<\/p>\n<p>Divide both sides of the equation by 3 to get a coefficient of 1 for the variable.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\,\\,\\,\\,\\,\\,\\underline{3y}\\,\\,\\,\\,=\\,\\,\\,\\,\\,\\underline{9}\\\\3\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,9\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,y\\,\\,\\,\\,=\\,\\,\\,\\,3\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]y=3[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve [latex]3x+5x+4-x+7=88[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q455516\">Show Solution<\/span><\/p>\n<div id=\"q455516\" class=\"hidden-answer\" style=\"display: none\">\n<p>There are three like terms [latex]3x[\/latex], [latex]5x[\/latex],\u00a0and\u00a0[latex]\u2013x[\/latex]\u00a0involving a variable.\u00a0Combine these like terms.\u00a04 and 7 are also like terms and can be added.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\,\\,3x+5x+4-x+7=\\,\\,\\,88\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,7x+4+7=\\,\\,\\,88\\end{array}[\/latex]<\/p>\n<p>The equation is now in the form\u00a0[latex]ax+b=c[\/latex], so we can solve as before.<\/p>\n<p style=\"text-align: center;\">[latex]7x+11\\,\\,\\,=\\,\\,\\,88[\/latex]<\/p>\n<p>Subtract 11 from both sides.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}7x+11\\,\\,\\,=\\,\\,\\,88\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-11\\,\\,\\,\\,\\,\\,\\,-11}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,7x\\,\\,\\,=\\,\\,\\,77\\end{array}[\/latex]<\/p>\n<p>Divide both sides by 7.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{7x}\\,\\,\\,=\\,\\,\\,\\underline{77}\\\\7\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,7\\,\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x\\,\\,\\,=\\,\\,\\,11\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=11[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<p>Some equations may have the variable on both sides of the equal sign, as in this equation: [latex]4x-6=2x+10[\/latex].<\/p>\n<p>To solve this equation, we need to \u201cmove\u201d one of the variable terms.\u00a0This can make it difficult to decide which side to work with. It doesn\u2019t matter which term gets moved, [latex]4x[\/latex] or [latex]2x[\/latex], however, to avoid negative coefficients, you can move the smaller term.<\/p>\n<div class=\"textbox exercises\">\n<h3>Examples<\/h3>\n<p>Solve:\u00a0[latex]4x-6=2x+10[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q457216\">Show Solution<\/span><\/p>\n<div id=\"q457216\" class=\"hidden-answer\" style=\"display: none\">\n<p>Choose the variable term to move\u2014to avoid negative terms choose [latex]2x[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\,\\,\\,4x-6=2x+10\\\\\\underline{-2x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-2x}\\\\\\,\\,\\,2x-6=10[\/latex]<\/p>\n<p style=\"text-align: left;\">Now add\u00a06 to both\u00a0sides to isolate the term with the variable.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x-6=10\\\\\\underline{\\,\\,\\,\\,+6\\,\\,\\,+6}\\\\2x=16\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">Now divide each side by 4 to isolate the variable x.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\frac{2x}{2}=\\frac{16}{2}\\\\\\\\x=8\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Solving Multi-Step Equations With Absolute Value<\/h2>\n<p>We can apply the same techniques we used for solving a one-step equation which contains absolute value to an equation that will take more than one step to solve. \u00a0Let&#8217;s start with an example where the first step is to write two equations, one equal to positive 26 and one equal to negative 26.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve for <i>p<\/i>.\u00a0[latex]\\left|2p\u20134\\right|=26[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q371950\">Show Solution<\/span><\/p>\n<div id=\"q371950\" class=\"hidden-answer\" style=\"display: none\">\n<p>Write the two equations that will give an absolute value of 26.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle 2p-4=26\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,\\,\\,2p-4=\\,-26[\/latex]<\/p>\n<p>Solve each equation for <i>p <\/i>by isolating the variable<i>.<\/i><\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{r}2p-4=26\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2p-4=\\,-26\\\\\\underline{\\,\\,\\,\\,\\,\\,+4\\,\\,\\,\\,+4}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{\\,\\,\\,\\,\\,\\,\\,+4\\,\\,\\,\\,\\,\\,\\,+4}\\\\\\underline{2p}\\,\\,\\,\\,\\,\\,=\\underline{30}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{2p}\\,\\,\\,\\,\\,=\\,\\underline{-22}\\\\2\\,\\,\\,\\,\\,\\,\\,=\\,2\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2\\,\\,\\,\\,\\,\\,\\,=\\,\\,\\,\\,\\,\\,\\,\\,2\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,p=15\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,p=\\,-11\\end{array}[\/latex]<\/p>\n<p>Check the solutions in the original equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{r}\\,\\,\\,\\,\\,\\left| 2p-4 \\right|=26\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left| 2p-4 \\right|=26\\\\\\left| 2(15)-4 \\right|=26\\,\\,\\,\\,\\,\\,\\,\\left| 2(-11)-4 \\right|=26\\\\\\,\\,\\,\\,\\,\\left| 30-4 \\right|=26\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left| -22-4 \\right|=26\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left| 26 \\right|=26\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left| -26 \\right|=26\\end{array}[\/latex]<\/p>\n<p>Both solutions check!<\/p>\n<h4>Answer<\/h4>\n<p>[latex]p=15[\/latex] or [latex]p=-11[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<p>Now let&#8217;s look at an example where you need to do an algebraic step or two before you can write your two equations. The goal here is to get the absolute value on one side of the equation by itself. Then we can proceed as we did in the previous example.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve for <i>w<\/i>. [latex]3\\left|4w\u20131\\right|\u20135=10[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q303228\">Show Solution<\/span><\/p>\n<div id=\"q303228\" class=\"hidden-answer\" style=\"display: none\">\n<p>Isolate the term with the absolute value by adding 5 to both sides.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}3\\left|4w-1\\right|-5=10\\\\\\underline{\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+5\\,\\,\\,+5}\\\\ 3\\left|4w-1\\right|=15\\end{array}[\/latex]<\/p>\n<p>Divide both sides by 3.\u00a0Now the absolute value is isolated.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r} \\underline{3\\left|4w-1\\right|}=\\underline{15}\\\\3\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3\\,\\,\\\\\\left|4w-1\\right|=\\,\\,5\\end{array}[\/latex]<\/p>\n<p>Write the two equations that will give an absolute value of 5 and solve them.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{r}4w-1=5\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,4w-1=-5\\\\\\underline{\\,\\,\\,\\,\\,\\,\\,+1\\,\\,+1}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{\\,\\,\\,\\,\\,\\,\\,\\,\\,+1\\,\\,\\,\\,\\,+1}\\\\\\,\\,\\,\\,\\,\\underline{4w}=\\underline{6}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{4w}\\,\\,\\,\\,\\,\\,\\,=\\underline{-4}\\\\4\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,4\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,4\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,4\\,\\,\\\\\\,\\,\\,\\,\\,\\,\\,\\,w=\\frac{3}{2}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,w=-1\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,w=\\frac{3}{2}\\,\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,\\,-1\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Check the solutions in the original equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{r}\\,\\,\\,\\,\\,3\\left| 4w-1\\, \\right|-5=10\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3\\left| 4w-1\\, \\right|-5=10\\\\\\\\3\\left| 4\\left( \\frac{3}{2} \\right)-1\\, \\right|-5=10\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3\\left| 4w-1\\, \\right|-5=10\\\\\\\\\\,\\,\\,\\,\\,\\,3\\left| \\frac{12}{2}-1\\, \\right|-5=10\\,\\,\\,\\,\\,\\,\\,3\\left| 4(-1)-1\\, \\right|-5=10\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,3\\left| 6-1\\, \\right|-5=10\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3\\left| -4-1\\, \\right|-5=10\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3\\left(5\\right)-5=10\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3\\left| -5 \\right|-5=10\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,15-5=10\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,15-5=10\\\\10=10\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,10=10\\end{array}[\/latex]<\/p>\n<p>Both solutions check<\/p>\n<h4>Answer<\/h4>\n<p>[latex]w=-1\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,w=\\frac{3}{2}[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<h2>The Distributive Property<\/h2>\n<p>As we solve linear equations, we often need to do some work to write\u00a0the linear equations in a form we are familiar with solving.\u00a0This section will focus on manipulating an equation we are asked to solve in such a way that we can use the skills we learned for solving multi-step equations to ultimately arrive at the solution.<\/p>\n<p>Parentheses can\u00a0make solving a problem difficult, if not impossible. To get rid of these unwanted parentheses we have to use the distributive property. Using this property we multiply the number in front of the parentheses by each term inside of the parentheses.<\/p>\n<div class=\"textbox shaded\">\n<h3>The Distributive Property of Multiplication<\/h3>\n<p>For all real numbers <i>a, b,<\/i> and <i>c<\/i>,\u00a0[latex]a(b+c)=ab+ac[\/latex].<\/p>\n<p>What this means is that when a number multiplies an expression inside parentheses, you can distribute the multiplication to each term of the expression individually. Then, you can follow the steps we have already practiced\u00a0to <b>isolate the variable<\/b>\u00a0and solve the equation.<\/p>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve for [latex]a[\/latex]. [latex]4\\left(2a+3\\right)=28[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q372387\">Show Solution<\/span><\/p>\n<div id=\"q372387\" class=\"hidden-answer\" style=\"display: none\">\n<p>Apply the distributive property to expand [latex]4\\left(2a+3\\right)[\/latex] to [latex]8a+12[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}4\\left(2a+3\\right)=28\\\\ 8a+12=28\\end{array}[\/latex]<\/p>\n<p>Subtract 12\u00a0from both sides to isolate\u00a0the variable term.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}8a+12\\,\\,\\,=\\,\\,\\,28\\\\ \\underline{-12\\,\\,\\,\\,\\,\\,-12}\\\\ 8a\\,\\,\\,=\\,\\,\\,16\\end{array}[\/latex]<\/p>\n<p>Divide both terms by 8 to get a coefficient of 1.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\underline{8a}=\\underline{16}\\\\8\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,8\\\\a\\,=\\,\\,2\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]a=2[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<p>In the next example, you will see that there are parentheses on both sides of the equal sign, so you will need to use the distributive property twice. Notice that you are going to need to distribute a negative number, so be careful with negative signs!<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve for [latex]t[\/latex].\u00a0\u00a0[latex]2\\left(4t-5\\right)=-3\\left(2t+1\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q302387\">Show Solution<\/span><\/p>\n<div id=\"q302387\" class=\"hidden-answer\" style=\"display: none\">\n<p>Apply the distributive property to expand [latex]2\\left(4t-5\\right)[\/latex] to [latex]8t-10[\/latex] and [latex]-3\\left(2t+1\\right)[\/latex] to[latex]-6t-3[\/latex]. Be careful in this step\u2014you are distributing a negative number, so keep track of the sign of each number after you multiply.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2\\left(4t-5\\right)=-3\\left(2t+1\\right)\\,\\,\\,\\,\\,\\, \\\\ 8t-10=-6t-3\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Add [latex]-6t[\/latex] to both sides to begin combining like terms.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}8t-10=-6t-3\\\\ \\underline{+6t\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+6t}\\,\\,\\,\\,\\,\\,\\,\\\\ 14t-10=\\,\\,\\,\\,-3\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Add 10 to both sides of the equation to isolate <em>t<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}14t-10=-3\\\\ \\underline{+10\\,\\,\\,+10}\\\\ 14t=\\,\\,\\,7\\,\\end{array}[\/latex]<\/p>\n<p>The last step is to divide both sides by 14 to completely isolate <em>t<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}14t=7\\,\\,\\,\\,\\\\\\frac{14t}{14}=\\frac{7}{14}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]t=\\frac{1}{2}[\/latex]<\/p>\n<p>We simplified the fraction [latex]\\frac{7}{14}[\/latex] into [latex]\\frac{1}{2}[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<p>Sometimes, you will encounter a multi-step equation with fractions. If you prefer not working with fractions, you can use the multiplication property of equality to multiply both sides of the equation by a common denominator of all of the fractions in the equation. This will clear all the fractions out of the equation. See the example below.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve \u00a0[latex]\\frac{1}{2}x-3=2-\\frac{3}{4}x[\/latex] by clearing the fractions in the equation first.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q129951\">Show Solution<\/span><\/p>\n<div id=\"q129951\" class=\"hidden-answer\" style=\"display: none\">\n<p>Multiply both sides of the equation by 4, the common denominator of the fractional coefficients.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\frac{1}{2}x-3=2-\\frac{3}{4}x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\ 4\\left(\\frac{1}{2}x-3\\right)=4\\left(2-\\frac{3}{4}x\\right)\\end{array}[\/latex]<\/p>\n<p>Use the distributive property to expand the expressions on both sides.\u00a0Multiply.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}4\\left(\\frac{1}{2}x\\right)-4\\left(3\\right)=4\\left(2\\right)-4\\left(-\\frac{3}{4}x\\right)\\\\\\\\ \\frac{4}{2}x-12=8-\\frac{12}{4}x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\\\\\ 2x-12=8-3x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\end{array}[\/latex]<\/p>\n<p>Add 3<em>x<\/em> to both sides to move the variable terms to only one side. Add 12 to both sides to move the variable\u00a0terms to only one side.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x-12=8-3x\\, \\\\\\underline{+3x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+3x}\\\\ 5x-12=8\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Add 12 to both sides to move the <b>constant<\/b> terms to the other side.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}5x-12=8\\,\\,\\\\ \\underline{\\,\\,\\,\\,\\,\\,+12\\,+12} \\\\5x=20\\end{array}[\/latex]<\/p>\n<p>Divide to isolate the variable.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\underline{5x}=\\underline{5}\\\\ 5\\,\\,\\,\\,\\,\\,\\,\\,\\,5\\\\ x=4\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=4[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<p>Of course, if you like to work with fractions, you can just apply your knowledge of operations with fractions and solve.<\/p>\n<p>Regardless of which method you use to solve equations containing variables, you will get the same answer. You can choose the method you find the easiest! Remember to check your answer by substituting your solution into the original equation.<\/p>\n<p>Sometimes, you will encounter a multi-step equation with decimals. If you prefer not working with decimals, you can use the multiplication property of equality to multiply both sides of the equation by a factor of 10 that will help clear the decimals. See the example below.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve [latex]3y+10.5=6.5+2.5y[\/latex] by clearing the decimals in the equation first.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q159951\">Show Solution<\/span><\/p>\n<div id=\"q159951\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since the smallest decimal place represented in the equation is 0.10, we want to multiply by 10 to make 1.0\u00a0and clear\u00a0the decimals from the equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}3y+10.5=6.5+2.5y\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\ 10\\left(3y+10.5\\right)=10\\left(6.5+2.5y\\right)\\end{array}[\/latex]<\/p>\n<p>Use the distributive property to expand the expressions on both sides.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}10\\left(3y\\right)+10\\left(10.5\\right)=10\\left(6.5\\right)+10\\left(2.5y\\right)\\end{array}[\/latex]<\/p>\n<p>Multiply.<\/p>\n<p style=\"text-align: center;\">[latex]30y+105=65+25y[\/latex]<\/p>\n<p>Move the smaller variable term, [latex]25y[\/latex], by subtracting it from both sides.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}30y+105=65+25y\\,\\,\\\\ \\underline{-25y\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-25y} \\\\5y+105=65\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Subtract 105 from both sides to isolate the term with the variable.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}5y+105=65\\,\\,\\,\\\\ \\underline{\\,\\,\\,\\,\\,\\,-105\\,-105} \\\\5y=-40\\end{array}[\/latex]<\/p>\n<p>Divide both sides by 5 to isolate the <em>y<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\underline{5y}=\\underline{-40}\\\\ 5\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,5\\\\ \\,\\,\\,x=-8\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=-8[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<p>Here are some steps to follow when you solve multi-step equations.<\/p>\n<div class=\"textbox shaded\">\n<h3>Solving Multi-Step Equations<\/h3>\n<p>1. (Optional) Multiply to clear any fractions or decimals.<\/p>\n<p>2. Simplify each side by clearing parentheses and combining like terms.<\/p>\n<p>3. Add or subtract to isolate the variable term\u2014you may have to move a term with the variable.<\/p>\n<p>4. Multiply or divide to isolate the variable.<\/p>\n<p>5. Check the solution.<\/p>\n<\/div>\n","protected":false},"author":80958,"menu_order":5,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-6678","chapter","type-chapter","status-publish","hentry"],"part":6254,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-albany-chemistry\/wp-json\/pressbooks\/v2\/chapters\/6678","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-albany-chemistry\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-albany-chemistry\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-albany-chemistry\/wp-json\/wp\/v2\/users\/80958"}],"version-history":[{"count":16,"href":"https:\/\/courses.lumenlearning.com\/suny-albany-chemistry\/wp-json\/pressbooks\/v2\/chapters\/6678\/revisions"}],"predecessor-version":[{"id":9053,"href":"https:\/\/courses.lumenlearning.com\/suny-albany-chemistry\/wp-json\/pressbooks\/v2\/chapters\/6678\/revisions\/9053"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-albany-chemistry\/wp-json\/pressbooks\/v2\/parts\/6254"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-albany-chemistry\/wp-json\/pressbooks\/v2\/chapters\/6678\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-albany-chemistry\/wp-json\/wp\/v2\/media?parent=6678"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-albany-chemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=6678"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-albany-chemistry\/wp-json\/wp\/v2\/contributor?post=6678"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-albany-chemistry\/wp-json\/wp\/v2\/license?post=6678"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}