{"id":6741,"date":"2020-10-08T16:59:38","date_gmt":"2020-10-08T16:59:38","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-beginalgebra\/?post_type=chapter&p=6741"},"modified":"2026-02-01T05:17:38","modified_gmt":"2026-02-01T05:17:38","slug":"4-2-solving-2x2-system-of-linear-equations-by-substitution","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/chapter\/4-2-solving-2x2-system-of-linear-equations-by-substitution\/","title":{"raw":"4.2: Solving a 2x2 System of Linear Equations by Substitution","rendered":"4.2: Solving a 2×2 System of Linear Equations by Substitution"},"content":{"raw":"
\r\n

section 4.2 Learning Objectives<\/h3>\r\n
\r\n
<\/div>\r\n<\/div>\r\n
\r\n
\r\n\r\n4.2:\u00a0 Solving a 2x2 System of Linear Equations by Substitution\u00a0<\/strong>\r\n\r\n<\/div>\r\n
\r\n
    \r\n \t
  • Solve systems of linear equations using substitution<\/li>\r\n \t
  • Recognize when systems of linear equations have no solution or an infinite number of solutions<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

    Solve a system of equations using the substitution method<\/h2>\r\nIn the previous section, we verified that ordered pairs were potential solutions to systems, and we used graphs to classify how many solutions a system of two linear equations had. What if we are not given a point of intersection, or it is not obvious from a graph? Can we still find a solution to the system? Of course you can, using algebra!\r\n\r\nIn this section we will learn the substitution\u00a0method for finding a solution to a system of linear equations in two variables. We have used substitution in different ways throughout this course.\u00a0 For example, when we were using formulas, we substituted values that we knew into the formula to solve for values that we did not know. \u00a0The idea is similar when applied to solving systems, there are just a few different steps in the process. You will first solve for one variable, and then substitute that expression into the other equation. Let's start with an example to see what this means.\r\n\r\n \r\n
    \r\n

    Example 1<\/h3>\r\nFind the value of [latex]x[\/latex] for this system.\r\n

    [latex]\\begin{array}{r}4x+3y=\u221214\\\\y=2\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"478211\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"478211\"]The problem asks to solve for x<\/i>. Note that we won't need to solve for [latex]y[\/latex] since the second equation already gives us the value for [latex]y[\/latex] as [latex]y=2[\/latex]. So, you can substitute 2 into the first equation for [latex]y[\/latex].\r\n

    [latex]\\begin{array}{r}4x+3y=\u221214 \\\\ 4x+3\\left(2\\right)=\u221214\\end{array}[\/latex]<\/p>\r\nSimplify and solve the equation for [latex]x[\/latex].<\/i>\r\n

    [latex]\\begin{array}{r}4x+6=\u221214\\\\ \\underline{\\hspace{.25in}-\\hspace{.03in}6 \\hspace{.23in}-6}\\\\ 4x\\hspace{.31in}=\u221220\\end{array}[\/latex]<\/p>\r\n

    [latex] {\\hspace{.31in}\\displaystyle \\frac{4x}{4}=\\frac{-20}{4}}[\/latex]<\/p>\r\n

    [latex]\\hspace{.3in}x=\u22125[\/latex]<\/p>\r\n\r\n

    Answer<\/h4>\r\n[latex]x=\u22125[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nYou can substitute a value for a variable even if it is an expression. Here\u2019s an example.\r\n
    \r\n

    Example 2<\/h3>\r\nSolve the system of equations for [latex]x[\/latex]\u00a0<\/i>and [latex]y[\/latex].\r\n

    [latex]\\begin{array}{l}y+x=3\\\\x=y+5\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"300993\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"300993\"]The goal of the substitution method is to rewrite one of the equations in terms of a single variable. The second equation tells us that [latex]x=y+5[\/latex], so it makes sense to substitute [latex]y+5[\/latex] into the first equation for [latex]x[\/latex].\r\n

    [latex]\\begin{array}{r}y+x=3\\\\y+\\left(y+5\\right)=3\\end{array}[\/latex]<\/i><\/p>\r\nSimplify and solve the equation for [latex]y[\/latex].<\/i>\r\n

    [latex]\\begin{array}{r}2y+5=3\\,\\,\\,\\,\\,\\\\{\\underline{\\hspace{.25in}\u2212\\hspace{.03in}5 \\hspace{.07in}\u22125}}\\,\\,\\,\\\\2y\\hspace{.33in}=\u22122\\end{array}[\/latex]<\/p>\r\n

    [latex]\\hspace{.34in} \\displaystyle \\frac{2y}{2}=\\frac{-2}{2}[\/latex]<\/p>\r\n

    [latex]\\hspace{.4in}y=\u22121[\/latex]<\/p>\r\nNow find [latex]x[\/latex]\u00a0<\/i>by\u00a0substituting this value for [latex]y[\/latex] into either equation. We will use the first equation here.\r\n

    [latex]y+x=3[\/latex]<\/p>\r\n

    [latex]\\begin{array}{r} -1+x=3\\hspace{.03in}\\\\ \\underline{+1\\hspace{.36in}+1}\\\\x=4\\hspace{.03in}\\end{array}[\/latex]<\/p>\r\nFinally, we can check the solution\u00a0[latex]x=4[\/latex], [latex]y=\u22121[\/latex]\u00a0by substituting these values into each of the original equations.\r\n

    [latex]\\begin{array}{r}y+x=3\\\\\u22121+4=3\\\\3=3\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\r\n \r\n

    [latex]\\begin{array}{r}x=y+5\\\\4=\u22121+5\\\\4=4\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\r\n\r\n

    Answer<\/h4>\r\n[latex]x=4[\/latex] and [latex]y=\u22121[\/latex]\r\n\r\nEquivalently, the solution is [latex](4,\u22121)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nRemember, a solution to a system of equations must be a solution to each of the equations within the system. The ordered pair [latex](4,\u22121)[\/latex] does work for both equations, so you know that it is a solution to the system.\r\n\r\nLet\u2019s look at another example whose substitution involves the distributive property.\r\n
    \r\n

    Example 3<\/h3>\r\nSolve the system for [latex]x[\/latex]\u00a0<\/i>and [latex]y[\/latex].\r\n

    [latex]\\begin{array}{r}y = 3x + 6\\\\\u22122x + 4y = 4\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"240040\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"240040\"]Choose an equation to use for the substitution.\r\n\r\nThe first equation tells you how to express [latex]y[\/latex] in terms of [latex]x[\/latex], so it makes sense to substitute [latex]3x+6[\/latex] into the second equation for [latex]y[\/latex].\r\n

    [latex]\\begin{array}{r}\u22122x+4y=4\\\\\u22122x+4\\left(3x+6\\right)=4\\end{array}[\/latex]<\/i><\/p>\r\nSimplify and solve the equation for [latex]x[\/latex].<\/i>\r\n

    [latex]\u22122x+12x+24=4\\hspace{.8in}[\/latex]<\/p>\r\n

    [latex]\\begin{array}{r}10x+24=4\\,\\,\\,\\,\\,\\,\\,\\\\\\underline{\\hspace{.38in}\u2212\\hspace{.03in}24\\hspace{.08in}\u221224\\,}\\\\10x\\hspace{.38in}=\u221220\\end{array}[\/latex]<\/p>\r\n

    [latex]\\hspace{.93in}\\displaystyle \\frac{10x}{10}=\\frac{-20}{10}\\hspace{.5in}[\/latex]<\/p>\r\n

    [latex]\\hspace{1.01in}x=-2\\hspace{.47in}[\/latex]<\/p>\r\nTo find [latex]y[\/latex], substitute this value for [latex]x[\/latex]back into one of the original equations.\r\n

    [latex]\\begin{array}{l}y=3x+6\\\\y=3\\left(\u22122\\right)+6\\\\y =\u22126+6\\\\y=0\\end{array}[\/latex]<\/p>\r\nCheck the solution [latex]x=\u22122[\/latex], [latex]y=0[\/latex]\u00a0by substituting them into each of the original equations.\r\n

    [latex]\\begin{array}{l}y=3x+6\\\\0=3\\left(\u22122\\right)+6\\\\0=\u22126+6\\\\0=0\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\r\n \r\n

    [latex]\\begin{array}{r}\u22122x+4y=4\\\\\u22122\\left(\u22122\\right)+4\\left(0\\right)=4\\\\4+0=4\\\\4=4\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\r\n\r\n

    Answer<\/h4>\r\n[latex]x=\u22122[\/latex] and [latex]y=0[\/latex]\r\n\r\nEquivalently, the solution is (\u22122, 0).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the examples above, one of the equations was already given to us in terms of the variable x<\/i> or y<\/i>. This allowed us to quickly substitute into the other equation and solve for one of the unknowns.\r\n\r\nSometimes you may have to rewrite one of the equations in terms of one of the variables first before you can substitute. In\u00a0the example below, you will first need to isolate one of the variables before you can substitute it into the other equation.\r\n
    \r\n

    Example 4<\/h3>\r\nSolve for [latex]x[\/latex] and [latex]y[\/latex].\r\n

    [latex]\\begin{array}{r}2x+3y=22\\\\3x+y=19\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"344538\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"344538\"]Choose an equation to use for the substitution. The second equation,\r\n\r\n[latex]3x+y=19[\/latex], can easily be solved for [latex]y[\/latex], so it makes sense to start there.\r\n\r\nSolve [latex]3x+y=19[\/latex] for [latex]y[\/latex].\r\n

    [latex]\\begin{array}{r}3x+y=19\\hspace{.42in}\\\\ \\underline{-3x \\hspace{.41in}-3x}\\hspace{.31in}\\\\ \\hspace{.78in}y=19-3x\\end{array}[\/latex]<\/p>\r\nSubstitute [latex]19\u20133x[\/latex] for [latex]y[\/latex] in the first equation.\r\n

    <\/i>[latex]\\begin{array}{r}2x+3y=22\\\\2x+3(19\u20133x)=22\\end{array}[\/latex]<\/i><\/p>\r\nSimplify and solve the equation for [latex]x[\/latex].<\/i>\r\n

    [latex]2x+57-9x=22\\hspace{.36in}[\/latex]<\/p>\r\n

    [latex]\\begin{array}{r}-7x+57=22\\,\\,\\,\\,\\,\\\\{\\underline{\\hspace{.39in}-\\hspace{.04in}57\\hspace{.09in}-57}}\\,\\,\\\\-7x\\hspace{.41in}=-35\\end{array}[\/latex]<\/p>\r\n

    [latex]\\displaystyle \\hspace{.7in}\\frac{-7x}{-7}=\\frac{-35}{-7}\\hspace{.28in}[\/latex]<\/p>\r\n

    [latex]\\hspace{.7in}x=5\\hspace{.3in}[\/latex]<\/p>\r\nSubstitute [latex]x=5[\/latex] back into one of the original equations to solve for y.<\/i>\r\n

    [latex]\\begin{array}{r}3x+y=19\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\3\\left(5\\right)+y=19\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\15+y=19\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\ \\underline{\u221215\\hspace{.3in}-15}\\,\\,\\,\\,\\,\\hspace{.2in}\\\\y=4\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nCheck both solutions by substituting them into each of the original equations.\r\n

    [latex]\\begin{array}{r}2x+3y=22\\\\2(5)+3\\left(4\\right)=22\\\\10+12=22\\\\22=22\\\\\\text{TRUE}\\\\\\\\3x+y=19\\\\3\\left(5\\right)+4= 19\\\\19=19\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\r\n\r\n

    Answer<\/h4>\r\n[latex]x=5[\/latex] and [latex]y=4[\/latex]\r\n\r\nEquivalently, the solution is (5, 4).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, you will be given an example of solving a systems of two equations using the substitution method.\r\n\r\nhttps:\/\/youtu.be\/MIXL35YRzRw\r\n\r\nIf you had chosen the other equation to start with in the previous example, you would still be able to find the same solution. \u00a0It is really a matter of preference because sometimes solving for a variable will result in having to work with fractions. \u00a0As you become more experienced with algebra, you will be able to anticipate what choices will lead to more desirable outcomes.\r\n

    Recognize systems of equations that have no solution or an infinite number of solutions<\/h2>\r\nWhen we learned methods for solving linear equations in one variable, we found that some equations didn't have any solutions, and others had an infinite number of solutions. \u00a0We saw this behavior again\u00a0when we started describing solutions to systems of equations in two variables.\r\n\r\nRecall this example from Module 1 for solving linear equations in one variable:\r\n

    Solve for x<\/i>.\u00a0[latex]12+2x\u20138=7x+5\u20135x[\/latex]<\/p>\r\n

    [latex] \\displaystyle \\begin{array}{l}12+2x-8=7x+5-5x\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2x+4=2x+5\\end{array}[\/latex]<\/p>\r\n

    [latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2x+4=2x+5\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-2x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-2x\\,\\,\\,\\,\\,\\,\\,\\,}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,4= \\,5\\end{array}[\/latex]<\/p>\r\nThis false statement implies there are no solutions<\/strong> to this equation. \u00a0In the same way, you may see an outcome like this when you use the substitution method to find a solution to a system of linear equations in two variables. In the next example, you will see an example of a system of two equations that does not have a solution.\r\n\r\n \r\n

    \r\n

    Example 5<\/h3>\r\nSolve for x<\/i> and y<\/i>.\r\n

    [latex]\\begin{array}{l}y=5x+4\\\\10x\u22122y=4\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"787022\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"787022\"]Since the first equation is [latex]y=5x+4[\/latex], you can substitute [latex]5x+4[\/latex] in for y<\/i> in the second equation.\r\n

    [latex]\\begin{array}{r}10x\u22122y=4\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\10x\u20132\\left(5x+4\\right)=4\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nExpand the expression on the left.\r\n

    [latex]10x\u201310x\u20138=4[\/latex]<\/p>\r\nCombine like terms on the left side of equation. But since [latex]10x-10x=0[\/latex], we are left with\r\n

    [latex]\u22128=4[\/latex].<\/p>\r\nSince this is a false statement, we conclude that there is no solution.\r\n

    Answer<\/h4>\r\nNo Solution\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nYou get the false statement [latex]\u22128=4[\/latex]. What does this mean? The graph of this system sheds some light on what is happening.\r\n\r\n\"Two\r\n\r\nThe lines are parallel, they never intersect and there is no solution to this system of linear equations. Note that the result [latex]\u22128=4[\/latex] is not<\/b> a solution. It is simply a false statement and it indicates that there is no<\/b> solution.\r\n\r\nWe have also seen linear equations in one variable and systems of equations in two variables that have an infinite number of solutions. \u00a0In the next example, you will see what happens when you apply the substitution method to a system with an infinite number of solutions.\r\n
    \r\n

    Example 6<\/h3>\r\nSolve for x and y.\r\n

    [latex]\\begin{array}{r}\\,\\,\\,y=\u2212\\frac{1}{2}x+\\frac{1}{3}\\\\3x+6y=2\\end{array}[\/latex]<\/p>\r\n

    [reveal-answer q=\"683508\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"683508\"]<\/p>\r\nWe can substitute [latex]y=\u2212\\frac{1}{2}x+\\frac{1}{3}[\/latex]\u00a0for\u00a0[latex]y[\/latex] in the second equation. This gives you find the following:\r\n

    \r\n

    [latex]\\begin{array}{r}3x+6y=2\\\\3x+6\\left(-\\frac{1}{2}x+\\frac{1}{3}\\right)=2\\\\3x-3x+2=2\\\\2=2\\end{array}[\/latex]<\/p>\r\nThis true statement implies that there are an infinite number of solutions, since the statement is true for all values of [latex]x[\/latex]. The discussion below describes why this is the case and what exactly it means.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nThis time you get a true statement: [latex]2=2[\/latex]. But what does this type of answer mean? Again, graphing can help you make sense of this system.\r\n

    \"Two<\/p>\r\nThis system consists of two equations that both represent the same line<\/em>. Every point along the line will be a solution to the system, and that\u2019s why the substitution method yields a true statement. In this case, there are an infinite number of solutions.\r\n\r\nIn the previous section, we learned that we can write the solution set in this case as the following:\r\n

    [latex]\\{(x,y)\\hspace{.01in}|\\hspace{.01in}y=-\\frac{1}{2}x+\\frac{1}{3}\\}[\/latex]<\/p>\r\nIn the following video you will see an example of solving a system that has an infinite number of solutions.\r\n\r\nhttps:\/\/youtu.be\/Pcqb109yK5Q\r\n\r\nIn the next video you will see an example of solving a system of equations that has no solutions.\r\n\r\nhttps:\/\/youtu.be\/kTtKfh5gFUc\r\n\r\n \r\n

    Summary<\/h2>\r\nThe substitution method is one way of solving systems of equations. To use the substitution method, use one equation to find an expression for one of the variables in terms of the other variable. Then substitute that expression in place of that variable in the second equation. You can then solve this equation as it will now have only one variable. Solving using the substitution method will yield one of three results: a single value for each variable within the system (indicating one solution), an untrue statement (indicating no solutions), or a true statement (indicating an infinite number of solutions).","rendered":"
    \n

    section 4.2 Learning Objectives<\/h3>\n
    \n
    <\/div>\n<\/div>\n
    \n
    \n

    4.2:\u00a0 Solving a 2×2 System of Linear Equations by Substitution\u00a0<\/strong><\/p>\n<\/div>\n

    \n
      \n
    • Solve systems of linear equations using substitution<\/li>\n
    • Recognize when systems of linear equations have no solution or an infinite number of solutions<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n

      Solve a system of equations using the substitution method<\/h2>\n

      In the previous section, we verified that ordered pairs were potential solutions to systems, and we used graphs to classify how many solutions a system of two linear equations had. What if we are not given a point of intersection, or it is not obvious from a graph? Can we still find a solution to the system? Of course you can, using algebra!<\/p>\n

      In this section we will learn the substitution\u00a0method for finding a solution to a system of linear equations in two variables. We have used substitution in different ways throughout this course.\u00a0 For example, when we were using formulas, we substituted values that we knew into the formula to solve for values that we did not know. \u00a0The idea is similar when applied to solving systems, there are just a few different steps in the process. You will first solve for one variable, and then substitute that expression into the other equation. Let’s start with an example to see what this means.<\/p>\n

       <\/p>\n

      \n

      Example 1<\/h3>\n

      Find the value of [latex]x[\/latex] for this system.<\/p>\n

      [latex]\\begin{array}{r}4x+3y=\u221214\\\\y=2\\end{array}[\/latex]<\/p>\n

      Show Solution<\/span><\/p>\n
      The problem asks to solve for x<\/i>. Note that we won’t need to solve for [latex]y[\/latex] since the second equation already gives us the value for [latex]y[\/latex] as [latex]y=2[\/latex]. So, you can substitute 2 into the first equation for [latex]y[\/latex].<\/p>\n

      [latex]\\begin{array}{r}4x+3y=\u221214 \\\\ 4x+3\\left(2\\right)=\u221214\\end{array}[\/latex]<\/p>\n

      Simplify and solve the equation for [latex]x[\/latex].<\/i><\/p>\n

      [latex]\\begin{array}{r}4x+6=\u221214\\\\ \\underline{\\hspace{.25in}-\\hspace{.03in}6 \\hspace{.23in}-6}\\\\ 4x\\hspace{.31in}=\u221220\\end{array}[\/latex]<\/p>\n

      [latex]{\\hspace{.31in}\\displaystyle \\frac{4x}{4}=\\frac{-20}{4}}[\/latex]<\/p>\n

      [latex]\\hspace{.3in}x=\u22125[\/latex]<\/p>\n

      Answer<\/h4>\n

      [latex]x=\u22125[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n

      You can substitute a value for a variable even if it is an expression. Here\u2019s an example.<\/p>\n

      \n

      Example 2<\/h3>\n

      Solve the system of equations for [latex]x[\/latex]\u00a0<\/i>and [latex]y[\/latex].<\/p>\n

      [latex]\\begin{array}{l}y+x=3\\\\x=y+5\\end{array}[\/latex]<\/p>\n

      Show Solution<\/span><\/p>\n
      The goal of the substitution method is to rewrite one of the equations in terms of a single variable. The second equation tells us that [latex]x=y+5[\/latex], so it makes sense to substitute [latex]y+5[\/latex] into the first equation for [latex]x[\/latex].<\/p>\n

      [latex]\\begin{array}{r}y+x=3\\\\y+\\left(y+5\\right)=3\\end{array}[\/latex]<\/i><\/p>\n

      Simplify and solve the equation for [latex]y[\/latex].<\/i><\/p>\n

      [latex]\\begin{array}{r}2y+5=3\\,\\,\\,\\,\\,\\\\{\\underline{\\hspace{.25in}\u2212\\hspace{.03in}5 \\hspace{.07in}\u22125}}\\,\\,\\,\\\\2y\\hspace{.33in}=\u22122\\end{array}[\/latex]<\/p>\n

      [latex]\\hspace{.34in} \\displaystyle \\frac{2y}{2}=\\frac{-2}{2}[\/latex]<\/p>\n

      [latex]\\hspace{.4in}y=\u22121[\/latex]<\/p>\n

      Now find [latex]x[\/latex]\u00a0<\/i>by\u00a0substituting this value for [latex]y[\/latex] into either equation. We will use the first equation here.<\/p>\n

      [latex]y+x=3[\/latex]<\/p>\n

      [latex]\\begin{array}{r} -1+x=3\\hspace{.03in}\\\\ \\underline{+1\\hspace{.36in}+1}\\\\x=4\\hspace{.03in}\\end{array}[\/latex]<\/p>\n

      Finally, we can check the solution\u00a0[latex]x=4[\/latex], [latex]y=\u22121[\/latex]\u00a0by substituting these values into each of the original equations.<\/p>\n

      [latex]\\begin{array}{r}y+x=3\\\\\u22121+4=3\\\\3=3\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\n

       <\/p>\n

      [latex]\\begin{array}{r}x=y+5\\\\4=\u22121+5\\\\4=4\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\n

      Answer<\/h4>\n

      [latex]x=4[\/latex] and [latex]y=\u22121[\/latex]<\/p>\n

      Equivalently, the solution is [latex](4,\u22121)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n

      Remember, a solution to a system of equations must be a solution to each of the equations within the system. The ordered pair [latex](4,\u22121)[\/latex] does work for both equations, so you know that it is a solution to the system.<\/p>\n

      Let\u2019s look at another example whose substitution involves the distributive property.<\/p>\n

      \n

      Example 3<\/h3>\n

      Solve the system for [latex]x[\/latex]\u00a0<\/i>and [latex]y[\/latex].<\/p>\n

      [latex]\\begin{array}{r}y = 3x + 6\\\\\u22122x + 4y = 4\\end{array}[\/latex]<\/p>\n

      Show Solution<\/span><\/p>\n
      Choose an equation to use for the substitution.<\/p>\n

      The first equation tells you how to express [latex]y[\/latex] in terms of [latex]x[\/latex], so it makes sense to substitute [latex]3x+6[\/latex] into the second equation for [latex]y[\/latex].<\/p>\n

      [latex]\\begin{array}{r}\u22122x+4y=4\\\\\u22122x+4\\left(3x+6\\right)=4\\end{array}[\/latex]<\/i><\/p>\n

      Simplify and solve the equation for [latex]x[\/latex].<\/i><\/p>\n

      [latex]\u22122x+12x+24=4\\hspace{.8in}[\/latex]<\/p>\n

      [latex]\\begin{array}{r}10x+24=4\\,\\,\\,\\,\\,\\,\\,\\\\\\underline{\\hspace{.38in}\u2212\\hspace{.03in}24\\hspace{.08in}\u221224\\,}\\\\10x\\hspace{.38in}=\u221220\\end{array}[\/latex]<\/p>\n

      [latex]\\hspace{.93in}\\displaystyle \\frac{10x}{10}=\\frac{-20}{10}\\hspace{.5in}[\/latex]<\/p>\n

      [latex]\\hspace{1.01in}x=-2\\hspace{.47in}[\/latex]<\/p>\n

      To find [latex]y[\/latex], substitute this value for [latex]x[\/latex]back into one of the original equations.<\/p>\n

      [latex]\\begin{array}{l}y=3x+6\\\\y=3\\left(\u22122\\right)+6\\\\y =\u22126+6\\\\y=0\\end{array}[\/latex]<\/p>\n

      Check the solution [latex]x=\u22122[\/latex], [latex]y=0[\/latex]\u00a0by substituting them into each of the original equations.<\/p>\n

      [latex]\\begin{array}{l}y=3x+6\\\\0=3\\left(\u22122\\right)+6\\\\0=\u22126+6\\\\0=0\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\n

       <\/p>\n

      [latex]\\begin{array}{r}\u22122x+4y=4\\\\\u22122\\left(\u22122\\right)+4\\left(0\\right)=4\\\\4+0=4\\\\4=4\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\n

      Answer<\/h4>\n

      [latex]x=\u22122[\/latex] and [latex]y=0[\/latex]<\/p>\n

      Equivalently, the solution is (\u22122, 0).<\/p>\n<\/div>\n<\/div>\n<\/div>\n

      In the examples above, one of the equations was already given to us in terms of the variable x<\/i> or y<\/i>. This allowed us to quickly substitute into the other equation and solve for one of the unknowns.<\/p>\n

      Sometimes you may have to rewrite one of the equations in terms of one of the variables first before you can substitute. In\u00a0the example below, you will first need to isolate one of the variables before you can substitute it into the other equation.<\/p>\n

      \n

      Example 4<\/h3>\n

      Solve for [latex]x[\/latex] and [latex]y[\/latex].<\/p>\n

      [latex]\\begin{array}{r}2x+3y=22\\\\3x+y=19\\end{array}[\/latex]<\/p>\n

      Show Solution<\/span><\/p>\n
      Choose an equation to use for the substitution. The second equation,<\/p>\n

      [latex]3x+y=19[\/latex], can easily be solved for [latex]y[\/latex], so it makes sense to start there.<\/p>\n

      Solve [latex]3x+y=19[\/latex] for [latex]y[\/latex].<\/p>\n

      [latex]\\begin{array}{r}3x+y=19\\hspace{.42in}\\\\ \\underline{-3x \\hspace{.41in}-3x}\\hspace{.31in}\\\\ \\hspace{.78in}y=19-3x\\end{array}[\/latex]<\/p>\n

      Substitute [latex]19\u20133x[\/latex] for [latex]y[\/latex] in the first equation.<\/p>\n

      <\/i>[latex]\\begin{array}{r}2x+3y=22\\\\2x+3(19\u20133x)=22\\end{array}[\/latex]<\/i><\/p>\n

      Simplify and solve the equation for [latex]x[\/latex].<\/i><\/p>\n

      [latex]2x+57-9x=22\\hspace{.36in}[\/latex]<\/p>\n

      [latex]\\begin{array}{r}-7x+57=22\\,\\,\\,\\,\\,\\\\{\\underline{\\hspace{.39in}-\\hspace{.04in}57\\hspace{.09in}-57}}\\,\\,\\\\-7x\\hspace{.41in}=-35\\end{array}[\/latex]<\/p>\n

      [latex]\\displaystyle \\hspace{.7in}\\frac{-7x}{-7}=\\frac{-35}{-7}\\hspace{.28in}[\/latex]<\/p>\n

      [latex]\\hspace{.7in}x=5\\hspace{.3in}[\/latex]<\/p>\n

      Substitute [latex]x=5[\/latex] back into one of the original equations to solve for y.<\/i><\/p>\n

      [latex]\\begin{array}{r}3x+y=19\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\3\\left(5\\right)+y=19\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\15+y=19\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\ \\underline{\u221215\\hspace{.3in}-15}\\,\\,\\,\\,\\,\\hspace{.2in}\\\\y=4\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n

      Check both solutions by substituting them into each of the original equations.<\/p>\n

      [latex]\\begin{array}{r}2x+3y=22\\\\2(5)+3\\left(4\\right)=22\\\\10+12=22\\\\22=22\\\\\\text{TRUE}\\\\\\\\3x+y=19\\\\3\\left(5\\right)+4= 19\\\\19=19\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\n

      Answer<\/h4>\n

      [latex]x=5[\/latex] and [latex]y=4[\/latex]<\/p>\n

      Equivalently, the solution is (5, 4).<\/p>\n<\/div>\n<\/div>\n<\/div>\n

      In the following video, you will be given an example of solving a systems of two equations using the substitution method.<\/p>\n