So far, the systems solved by substitution have included an equation where one of the variables was isolated. This is not always the case and often a variable will need to be isolated before substitution can be used.

 

Key Lesson Marker

 

The general strategy for solving a system of equations by substitution is

  1. Isolate one variable in one equation.

  2. Substitute the isolated variable's equivalent expression into the other equation.

  3. Solve the new equation to determine the value of one variable.

  4. Substitute the known value into one of the original equations to determine the value of the other variable.

  5. Verify the solution.


 

Example 2

Solve the following system of equations. Verify the solution.

Equation 1:   «math style=¨font-family:`Times New Roman`¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»3«/mn»«mi»x«/mi»«mo»+«/mo»«mn»5«/mn»«mi»y«/mi»«mo»+«/mo»«mn»24«/mn»«mo»=«/mo»«mn»0«/mn»«/math»
Equation 2:   «math style=¨font-family:`Times New Roman`¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»2«/mn»«mi»x«/mi»«mo»+«/mo»«mn»2«/mn»«mi»y«/mi»«mo»=«/mo»«mo»-«/mo»«mn»8«/mn»«/math»


To solve a system of equations by substitution, you can isolate any variable. However, it is often more efficient to choose the variable that is easy to isolate and whose equivalent expression will be simple to work with.

Begin by isolating a variable.

The variable x was isolated using Equation 2, so substitute «math style=¨font-family:`Times New Roman`¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mo mathcolor=¨#B94A48¨»-«/mo»«mn mathcolor=¨#B94A48¨»4«/mn»«mo mathcolor=¨#B94A48¨»-«/mo»«mi mathcolor=¨#B94A48¨»y«/mi»«/math» for x in Equation 1, and solve for y.

The y-value is -6. Substitute this value into either of the original equations to determine the x-value. Recall that the equation «math style=¨font-family:`Times New Roman`¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi mathcolor=¨#B94A48¨»x«/mi»«mo mathcolor=¨#B94A48¨»=«/mo»«mo mathcolor=¨#B94A48¨»-«/mo»«mn mathcolor=¨#B94A48¨»4«/mn»«mo mathcolor=¨#B94A48¨»-«/mo»«mi mathcolor=¨#B94A48¨»y«/mi»«/math» is simply a rearranged version of the second original equation «math style=¨font-family:`Times New Roman`¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn mathcolor=¨#B94A48¨»2«/mn»«mi mathcolor=¨#B94A48¨»x«/mi»«mo mathcolor=¨#B94A48¨»+«/mo»«mn mathcolor=¨#B94A48¨»2«/mn»«mi mathcolor=¨#B94A48¨»y«/mi»«mo mathcolor=¨#B94A48¨»=«/mo»«mo mathcolor=¨#B94A48¨»-«/mo»«mn mathcolor=¨#B94A48¨»8«/mn»«/math».

The solution is (2, –6).

Verify the solution.