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.
Let’s eliminate y.
• Multiply both sides
of the first equation
by 3 to make the
y-coefficient 12.
|
3(3x + 4y = 18)
→ 9x + 12y |
= 54 |
| • Multiply both sides
of the second equation
by -2 to make the
y-coefficient -12. |
-2(17x + 6y = 52) →
-34x - 12y |
= 104 |
| Add the two equations.
|
| 9x - 34x |
+ - |
12y 12y |
= = |
- |
54 104 |
|
| - 25x |
+ |
0y |
= |
- |
50 |
|
|
|
Simplify. The y-terms 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:
• multiply both sides of the first
equation by 3.
• multiply both sides of the second
equation by -2.
|
