Solving Linear Systems of Equations by Elimination
In some systems it is necessary to multiply each equation by a constant so
that one variable will be eliminated when the equations are added.
Example
Use elimination to find the solution of this system.
3x 
+ 
4y 
= 18 
17x 
+ 
6y 
= 52 
Solution
Step 1 Eliminate one variable.
Letâ€™s eliminate y.
â€¢ Multiply both sides
of the first equation
by 3 to make the
ycoefficient 12.

3(3x + 4y = 18)
→ 9x + 12y 
= 54 
â€¢ Multiply both sides
of the second equation
by 2 to make the
ycoefficient 12. 
2(17x + 6y = 52) →
34x  12y 
= 104 
Add the two equations.

9x  34x 
+  
12y 12y 
= = 
 
54 104 

 25x 
+ 
0y 
= 
 
50 


Simplify. The yterms have been eliminated.
Divide both sides by 25.

 25x x 
=  50 = 2 
Step 2 Substitute the value found in Step 1 into either of the original
equations and solve. 
3x + 4y 
= 18 
Substitute 2 for x in the first equation.
Multiply.
Subtract 6 from both sides.
Divide both sides by 4.
The solution is (2, 3). 
3(2) + 4y
6 + 4y
4y
y 
= 18
= 18
= 12
= 3 
Step 3 To check the solution, substitute it into each original equation.
Then simplify.
Substitute 2 for x and 3 for y into each original equation. Then simplify.
In each case, the result will be a true statement.
The details of the check are left to you. 
Note:
3x 
+ 
4y 
= 18 
17x 
+ 
6y 
= 52 
In this system, the coefficients of y
are 4 and 6, respectively.
Multiples of 4 are 4, 8, 12,...
Multiples of 6 are 6, 12, 18,...
Notice that 12 is the least common
multiple of 4 and 6. To make the
coefficients of y opposites:
â€¢ multiply both sides of the first
equation by 3.
â€¢ multiply both sides of the second
equation by 2.
