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Section 4.2
Solving a System
of Equations in
Two Variables by
the Substitution
Method
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The Substitution Method
Procedure for Solving a System of Equations by the
Substitution Method
1. Solve one of the two equations for one variable. If possible,
solve for a variable with a coefficient of 1 or 1.
2. Substitute the expression obtained in Step 1 for this variable in
the other equation.
3. You now have one equation with one variable. Solve this
equation to find the value for that one variable.
4. Substitute this value for the variable into one of the original
equations to obtain a value for the other variable.
5. Check the solution in each equation to verify the results.
Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc.
2
Example
Find the solution.
y = 2x – 13
– 4x – 9y = 7
2x – y = 13
– 4x – 9y = 7
This variable is the
easiest to isolate.
Solve the first equation for y.
This is the original second equation.
– 4x – 9(2x – 13) = 7
– 4x – 18x + 117 = 7
Substitute the expression into the other
equation and solve.
– 22x = – 110
x = 5 Substitute this value into one of the original
equations.
Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc.
Continued
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Example (cont)
2x – y = 13
– 4x – 9y = 7
First equation
Second equation
2x – y = 13
2(5) – y = 13
10 – y = 13
Substitute x = 5 into the first equation.
Solve for y.
–y = 3
y = –3
The solution is (5, –3).
Continued
Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc.
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Example (cont)
Check the solution (5, –3) in both original equations.
2x – y = 13
– 4x – 9y = 7
2(5) – (3) = 13
– 4(5) – 9(3) = 7
10 + 3 = 13
13 = 13
– 20 + 27 = 7

7=7
Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc.

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Example
Find the solution. x  2 y  6
2
1
x  y  3
3
3
   
2
1
3 x  3 y  3(3)
3
3
2x  y  9
y  2x  9
x  2(2x  9)  6
x  4 x  18  6
Clear the fractions
from the equation.
Multiply each term by the LCD.
Solve for y.
Substitute into the other equation.
Continued
Simplify.
Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc.
6
Example (cont)
x  2 y  6
2
1
x  y  3
3
3
x  4 x  18  6
Combine like terms and simplify.
3x  12
x  4
x  2 y  6
Substitute this value into one of the original
equations.
4  2 y  6
y  1
The solution is (4, 1).
Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc.
Continued
7
Example (cont)
Check the solution (–4, –1) in both original equations.
x  2 y  6
(4)  2(1)  6
4  2  6
6   6

2
1
x  y  3
3
3
2
1
(4)  (1)  3
3
3
8 1

 3
3
3
9
 3
3
3  3 
 
Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc.
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