{"id":1723,"date":"2015-11-12T18:30:46","date_gmt":"2015-11-12T18:30:46","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=1723"},"modified":"2015-11-12T18:30:46","modified_gmt":"2015-11-12T18:30:46","slug":"solving-systems-of-equations-by-substitution","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/solving-systems-of-equations-by-substitution\/","title":{"raw":"Solving Systems of Equations by Substitution","rendered":"Solving Systems of Equations by Substitution"},"content":{"raw":"

Solving a linear system in two variables by graphing works well when the solution consists of integer values, but if our solution contains decimals or fractions, it is not the most precise method. We will consider two more methods of solving a system of linear equations<\/strong> that are more precise than graphing. One such method is solving a system of equations by the substitution method<\/strong>, in which we solve one of the equations for one variable and then substitute the result into the second equation to solve for the second variable. Recall that we can solve for only one variable at a time, which is the reason the substitution method is both valuable and practical.\n<\/p>

\n

How To: Given a system of two equations in two variables, solve using the substitution method.<\/h3>\n
  1. Solve one of the two equations for one of the variables in terms of the other.<\/li>\n\t
  2. Substitute the expression for this variable into the second equation, then solve for the remaining variable.<\/li>\n\t
  3. Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an ordered pair.<\/li>\n\t
  4. Check the solution in both equations.<\/li>\n<\/ol><\/div>\n
    \n

    Example 3: Solving a System of Equations in Two Variables by Substitution<\/h3>\nSolve the following system of equations by substitution.\n
    [latex]\\begin{array}{l}\\text{ }-x+y=-5\\hfill \\\\ \\text{ }2x - 5y=1\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n
    \n

    Solution<\/h3>\nFirst, we will solve the first equation for [latex]y[\/latex].\n
    [latex]\\begin{array}{l}-x+y=-5\\hfill \\\\ \\text{ }y=x - 5\\hfill \\end{array}[\/latex]<\/div>\nNow we can substitute the expression [latex]x - 5[\/latex] for [latex]y[\/latex] in the second equation.\n
    [latex]\\begin{array}{l}\\text{ }2x - 5y=1\\hfill \\\\ 2x - 5\\left(x - 5\\right)=1\\hfill \\\\ \\text{ }2x - 5x+25=1\\hfill \\\\ \\text{ }-3x=-24\\hfill \\\\ \\text{ }x=8\\hfill \\end{array}[\/latex]<\/div>\nNow, we substitute [latex]x=8[\/latex] into the first equation and solve for [latex]y[\/latex].\n
    [latex]\\begin{array}{l}-\\left(8\\right)+y=-5\\hfill \\\\ \\text{ }y=3\\hfill \\end{array}[\/latex]<\/div>\nOur solution is [latex]\\left(8,3\\right)[\/latex].\n\nCheck the solution by substituting [latex]\\left(8,3\\right)[\/latex] into both equations.\n
    [latex]\\begin{array}{llll}-x+y=-5\\hfill & \\hfill & \\hfill & \\hfill \\\\ -\\left(8\\right)+\\left(3\\right)=-5\\hfill & \\hfill & \\hfill & \\text{True}\\hfill \\\\ 2x - 5y=1\\hfill & \\hfill & \\hfill & \\hfill \\\\ 2\\left(8\\right)-5\\left(3\\right)=1\\hfill & \\hfill & \\hfill & \\text{True}\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n
    \n

    Try It 3<\/h3>\nSolve the following system of equations by substitution.\n
    [latex]\\begin{array}{l}x=y+3\\hfill \\\\ 4=3x - 2y\\hfill \\end{array}[\/latex]<\/div>\n
    Solution<\/a><\/div>\n<\/div>\n
    \n

    Q & A<\/h3>\n

    Can the substitution method be used to solve any linear system in two variables?<\/h3>\nYes, but the method works best if one of the equations contains a coefficient of 1 or \u20131 so that we do not have to deal with fractions.<\/em>\n\n<\/div>","rendered":"

    Solving a linear system in two variables by graphing works well when the solution consists of integer values, but if our solution contains decimals or fractions, it is not the most precise method. We will consider two more methods of solving a system of linear equations<\/strong> that are more precise than graphing. One such method is solving a system of equations by the substitution method<\/strong>, in which we solve one of the equations for one variable and then substitute the result into the second equation to solve for the second variable. Recall that we can solve for only one variable at a time, which is the reason the substitution method is both valuable and practical.\n<\/p>\n

    \n

    How To: Given a system of two equations in two variables, solve using the substitution method.<\/h3>\n
      \n
    1. Solve one of the two equations for one of the variables in terms of the other.<\/li>\n
    2. Substitute the expression for this variable into the second equation, then solve for the remaining variable.<\/li>\n
    3. Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an ordered pair.<\/li>\n
    4. Check the solution in both equations.<\/li>\n<\/ol>\n<\/div>\n
      \n

      Example 3: Solving a System of Equations in Two Variables by Substitution<\/h3>\n

      Solve the following system of equations by substitution.<\/p>\n

      [latex]\\begin{array}{l}\\text{ }-x+y=-5\\hfill \\\\ \\text{ }2x - 5y=1\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n
      \n

      Solution<\/h3>\n

      First, we will solve the first equation for [latex]y[\/latex].<\/p>\n

      [latex]\\begin{array}{l}-x+y=-5\\hfill \\\\ \\text{ }y=x - 5\\hfill \\end{array}[\/latex]<\/div>\n

      Now we can substitute the expression [latex]x - 5[\/latex] for [latex]y[\/latex] in the second equation.<\/p>\n

      [latex]\\begin{array}{l}\\text{ }2x - 5y=1\\hfill \\\\ 2x - 5\\left(x - 5\\right)=1\\hfill \\\\ \\text{ }2x - 5x+25=1\\hfill \\\\ \\text{ }-3x=-24\\hfill \\\\ \\text{ }x=8\\hfill \\end{array}[\/latex]<\/div>\n

      Now, we substitute [latex]x=8[\/latex] into the first equation and solve for [latex]y[\/latex].<\/p>\n

      [latex]\\begin{array}{l}-\\left(8\\right)+y=-5\\hfill \\\\ \\text{ }y=3\\hfill \\end{array}[\/latex]<\/div>\n

      Our solution is [latex]\\left(8,3\\right)[\/latex].<\/p>\n

      Check the solution by substituting [latex]\\left(8,3\\right)[\/latex] into both equations.<\/p>\n

      [latex]\\begin{array}{llll}-x+y=-5\\hfill & \\hfill & \\hfill & \\hfill \\\\ -\\left(8\\right)+\\left(3\\right)=-5\\hfill & \\hfill & \\hfill & \\text{True}\\hfill \\\\ 2x - 5y=1\\hfill & \\hfill & \\hfill & \\hfill \\\\ 2\\left(8\\right)-5\\left(3\\right)=1\\hfill & \\hfill & \\hfill & \\text{True}\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n
      \n

      Try It 3<\/h3>\n

      Solve the following system of equations by substitution.<\/p>\n

      [latex]\\begin{array}{l}x=y+3\\hfill \\\\ 4=3x - 2y\\hfill \\end{array}[\/latex]<\/div>\n
      Solution<\/a><\/div>\n<\/div>\n
      \n

      Q & A<\/h3>\n

      Can the substitution method be used to solve any linear system in two variables?<\/h3>\n

      Yes, but the method works best if one of the equations contains a coefficient of 1 or \u20131 so that we do not have to deal with fractions.<\/em><\/p>\n<\/div>\n\n\t\t\t

      \n\t\t\t

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