{"id":359,"date":"2015-09-18T22:47:26","date_gmt":"2015-09-18T22:47:26","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=359"},"modified":"2015-11-12T18:38:00","modified_gmt":"2015-11-12T18:38:00","slug":"solving-linear-equations-in-one-variable","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/solving-linear-equations-in-one-variable\/","title":{"raw":"Solving Linear Equations in One Variable","rendered":"Solving Linear Equations in One Variable"},"content":{"raw":"A linear equation<\/strong> is an equation of a straight line, written in one variable. The only power of the variable is 1. Linear equations in one variable may take the form [latex]ax+b=0[\/latex] and are solved using basic algebraic operations.\r\n\r\nWe begin by classifying linear equations in one variable as one of three types: identity, conditional, or inconsistent. An identity equation<\/strong> is true for all values of the variable. Here is an example of an identity equation.\r\n
[latex]3x=2x+x[\/latex]<\/div>\r\nThe solution set<\/strong> consists of all values that make the equation true. For this equation, the solution set is all real numbers because any real number substituted for [latex]x[\/latex] will make the equation true.\r\n\r\nA conditional equation<\/strong> is true for only some values of the variable. For example, if we are to solve the equation [latex]5x+2=3x - 6[\/latex], we have the following:\r\n
[latex]\\begin{array}{l}5x+2\\hfill&=3x - 6\\hfill \\\\ 2x\\hfill&=-8\\hfill \\\\ x\\hfill&=-4\\hfill \\end{array}[\/latex]<\/div>\r\nThe solution set consists of one number: [latex]\\{-4\\}[\/latex]. It is the only solution and, therefore, we have solved a conditional equation.\r\n\r\nAn inconsistent equation<\/strong> results in a false statement. For example, if we are to solve [latex]5x - 15=5\\left(x - 4\\right)[\/latex], we have the following:\r\n
[latex]\\begin{array}{ll}5x - 15=5x - 20\\hfill & \\hfill \\\\ 5x - 15 - 5x=5x - 20 - 5x\\hfill & \\text{Subtract }5x\\text{ from both sides}.\\hfill \\\\ -15\\ne -20 \\hfill & \\text{False statement}\\hfill \\end{array}[\/latex]<\/div>\r\nIndeed, [latex]-15\\ne -20[\/latex]. There is no solution because this is an inconsistent equation.\r\n\r\nSolving linear equations in one variable involves the fundamental properties of equality and basic algebraic operations. A brief review of those operations follows.\r\n
\r\n

A General Note: Linear Equation in One Variable<\/h3>\r\nA linear equation in one variable can be written in the form\r\n
[latex]ax+b=0[\/latex]<\/div>\r\nwhere a<\/em> and b <\/em>are real numbers, [latex]a\\ne 0[\/latex].\r\n\r\n<\/div>\r\n
\r\n

How To: Given a linear equation in one variable, use algebra to solve it.<\/h3>\r\nThe following steps are used to manipulate an equation and isolate the unknown variable, so that the last line reads x=_________, if x <\/em>is the unknown. There is no set order, as the steps used depend on what is given:\r\n
    \r\n\t
  1. We may add, subtract, multiply, or divide an equation by a number or an expression as long as we do the same thing to both sides of the equal sign. Note that we cannot divide by zero.<\/li>\r\n\t
  2. Apply the distributive property as needed: [latex]a\\left(b+c\\right)=ab+ac[\/latex].<\/li>\r\n\t
  3. Isolate the variable on one side of the equation.<\/li>\r\n\t
  4. When the variable is multiplied by a coefficient in the final stage, multiply both sides of the equation by the reciprocal of the coefficient.<\/li>\r\n<\/ol>\r\n<\/div>\r\n
    \r\n

    Example 1: Solving an Equation in One Variable<\/h3>\r\nSolve the following equation: [latex]2x+7=19[\/latex].\r\n\r\n<\/div>\r\n
    \r\n

    Solution<\/h3>\r\nThis equation can be written in the form [latex]ax+b=0[\/latex] by subtracting [latex]19[\/latex] from both sides. However, we may proceed to solve the equation in its original form by performing algebraic operations.\r\n
    [latex]\\begin{array}{ll}2x+7=19\\hfill & \\hfill \\\\ 2x=12\\hfill & \\text{Subtract 7 from both sides}.\\hfill \\\\ x=6\\hfill & \\text{Multiply both sides by }\\frac{1}{2}\\text{ or divide by 2}.\\hfill \\end{array}[\/latex]<\/div>\r\nThe solution is [latex]x=6[\/latex].\r\n\r\n<\/div>\r\n
    \r\n

    Try It 1<\/h3>\r\nSolve the linear equation in one variable: [latex]2x+1=-9[\/latex].\r\n\r\nSolution<\/a>\r\n\r\n<\/div>\r\n
    \r\n

    Example 2: Solving an Equation Algebraically When the Variable Appears on Both Sides<\/h3>\r\nSolve the following equation: [latex]4\\left(x - 3\\right)+12=15 - 5\\left(x+6\\right)[\/latex].\r\n\r\n<\/div>\r\n
    \r\n

    Solution<\/h3>\r\nApply standard algebraic properties.\r\n
    [latex]\\begin{array}{ll}4\\left(x - 3\\right)+12=15 - 5\\left(x+6\\right)\\hfill & \\hfill \\\\ 4x - 12+12=15 - 5x - 30 \\hfill & \\text{Apply the distributive property}.\\hfill \\\\ 4x=-15 - 5x\\hfill & \\text{Combine like terms}.\\hfill \\\\ 9x=-15 \\hfill & \\text{Place }x-\\text{terms on one side and simplify}.\\hfill \\\\ x=-\\frac{15}{9}\\hfill & \\text{Multiply both sides by }\\frac{1}{9}\\text{, the reciprocal of 9}.\\hfill \\\\ x=-\\frac{5}{3}\\hfill & \\hfill \\end{array}[\/latex]<\/div>\r\n<\/div>\r\n
    \r\n

    Analysis of the Solution<\/h3>\r\nThis problem requires the distributive property to be applied twice, and then the properties of algebra are used to reach the final line, [latex]x=-\\frac{5}{3}[\/latex].\r\n\r\n<\/div>\r\n
    \r\n

    Try It 2<\/h3>\r\nSolve the equation in one variable: [latex]-2\\left(3x - 1\\right)+x=14-x[\/latex].\r\n\r\nSolution<\/a>\r\n\r\n<\/div>","rendered":"

    A linear equation<\/strong> is an equation of a straight line, written in one variable. The only power of the variable is 1. Linear equations in one variable may take the form [latex]ax+b=0[\/latex] and are solved using basic algebraic operations.<\/p>\n

    We begin by classifying linear equations in one variable as one of three types: identity, conditional, or inconsistent. An identity equation<\/strong> is true for all values of the variable. Here is an example of an identity equation.<\/p>\n

    [latex]3x=2x+x[\/latex]<\/div>\n

    The solution set<\/strong> consists of all values that make the equation true. For this equation, the solution set is all real numbers because any real number substituted for [latex]x[\/latex] will make the equation true.<\/p>\n

    A conditional equation<\/strong> is true for only some values of the variable. For example, if we are to solve the equation [latex]5x+2=3x - 6[\/latex], we have the following:<\/p>\n

    [latex]\\begin{array}{l}5x+2\\hfill&=3x - 6\\hfill \\\\ 2x\\hfill&=-8\\hfill \\\\ x\\hfill&=-4\\hfill \\end{array}[\/latex]<\/div>\n

    The solution set consists of one number: [latex]\\{-4\\}[\/latex]. It is the only solution and, therefore, we have solved a conditional equation.<\/p>\n

    An inconsistent equation<\/strong> results in a false statement. For example, if we are to solve [latex]5x - 15=5\\left(x - 4\\right)[\/latex], we have the following:<\/p>\n

    [latex]\\begin{array}{ll}5x - 15=5x - 20\\hfill & \\hfill \\\\ 5x - 15 - 5x=5x - 20 - 5x\\hfill & \\text{Subtract }5x\\text{ from both sides}.\\hfill \\\\ -15\\ne -20 \\hfill & \\text{False statement}\\hfill \\end{array}[\/latex]<\/div>\n

    Indeed, [latex]-15\\ne -20[\/latex]. There is no solution because this is an inconsistent equation.<\/p>\n

    Solving linear equations in one variable involves the fundamental properties of equality and basic algebraic operations. A brief review of those operations follows.<\/p>\n

    \n

    A General Note: Linear Equation in One Variable<\/h3>\n

    A linear equation in one variable can be written in the form<\/p>\n

    [latex]ax+b=0[\/latex]<\/div>\n

    where a<\/em> and b <\/em>are real numbers, [latex]a\\ne 0[\/latex].<\/p>\n<\/div>\n

    \n

    How To: Given a linear equation in one variable, use algebra to solve it.<\/h3>\n

    The following steps are used to manipulate an equation and isolate the unknown variable, so that the last line reads x=_________, if x <\/em>is the unknown. There is no set order, as the steps used depend on what is given:<\/p>\n

      \n
    1. We may add, subtract, multiply, or divide an equation by a number or an expression as long as we do the same thing to both sides of the equal sign. Note that we cannot divide by zero.<\/li>\n
    2. Apply the distributive property as needed: [latex]a\\left(b+c\\right)=ab+ac[\/latex].<\/li>\n
    3. Isolate the variable on one side of the equation.<\/li>\n
    4. When the variable is multiplied by a coefficient in the final stage, multiply both sides of the equation by the reciprocal of the coefficient.<\/li>\n<\/ol>\n<\/div>\n
      \n

      Example 1: Solving an Equation in One Variable<\/h3>\n

      Solve the following equation: [latex]2x+7=19[\/latex].<\/p>\n<\/div>\n

      \n

      Solution<\/h3>\n

      This equation can be written in the form [latex]ax+b=0[\/latex] by subtracting [latex]19[\/latex] from both sides. However, we may proceed to solve the equation in its original form by performing algebraic operations.<\/p>\n

      [latex]\\begin{array}{ll}2x+7=19\\hfill & \\hfill \\\\ 2x=12\\hfill & \\text{Subtract 7 from both sides}.\\hfill \\\\ x=6\\hfill & \\text{Multiply both sides by }\\frac{1}{2}\\text{ or divide by 2}.\\hfill \\end{array}[\/latex]<\/div>\n

      The solution is [latex]x=6[\/latex].<\/p>\n<\/div>\n

      \n

      Try It 1<\/h3>\n

      Solve the linear equation in one variable: [latex]2x+1=-9[\/latex].<\/p>\n

      Solution<\/a><\/p>\n<\/div>\n

      \n

      Example 2: Solving an Equation Algebraically When the Variable Appears on Both Sides<\/h3>\n

      Solve the following equation: [latex]4\\left(x - 3\\right)+12=15 - 5\\left(x+6\\right)[\/latex].<\/p>\n<\/div>\n

      \n

      Solution<\/h3>\n

      Apply standard algebraic properties.<\/p>\n

      [latex]\\begin{array}{ll}4\\left(x - 3\\right)+12=15 - 5\\left(x+6\\right)\\hfill & \\hfill \\\\ 4x - 12+12=15 - 5x - 30 \\hfill & \\text{Apply the distributive property}.\\hfill \\\\ 4x=-15 - 5x\\hfill & \\text{Combine like terms}.\\hfill \\\\ 9x=-15 \\hfill & \\text{Place }x-\\text{terms on one side and simplify}.\\hfill \\\\ x=-\\frac{15}{9}\\hfill & \\text{Multiply both sides by }\\frac{1}{9}\\text{, the reciprocal of 9}.\\hfill \\\\ x=-\\frac{5}{3}\\hfill & \\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n
      \n

      Analysis of the Solution<\/h3>\n

      This problem requires the distributive property to be applied twice, and then the properties of algebra are used to reach the final line, [latex]x=-\\frac{5}{3}[\/latex].<\/p>\n<\/div>\n

      \n

      Try It 2<\/h3>\n

      Solve the equation in one variable: [latex]-2\\left(3x - 1\\right)+x=14-x[\/latex].<\/p>\n

      Solution<\/a><\/p>\n<\/div>\n\n\t\t\t

      \n\t\t\t

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