Integrated Algebra
Notes: Solving systems by elimination
Name
Changing both equations
Sometimes, we need to change BOTH equations because neither variable is
easy to eliminate as is.
EX:

2x - 3y = 14
3x + 4y = -13
What variable should we eliminate? (Look at coefficients)
Neither coefficient is related, so we need to take the LCM of either one.

If we want to eliminate the x’s, we would need the LCM of 2 and 3.
That means we would change the coefficients to +6 and -6

If we want to eliminate the y’s, we would need the LCM of 3 and 4.
That means we would change the coefficients to +12 and -12
Pick one to eliminate. I usually choose the smaller LCM.
So…..let’s get rid of the x’s
2x - 3y = 14
3x + 4y = -13

What coefficients do we need? ____ and _____

To change the first equation, multiply through by +3
To change the second equation, multiply through by -2
2x - 3y = 14
3( 2x – 3y = 14)
6x – 9y = 42
3x + 4y = -13
-2(3x + 4y = -13)
+ -6x – 8y = 26
-17y = 68
y = -4

Substitute and solve for x.
3x + 4(-4) = -13
3x – 16 = -13
3x = 3
x=1
Solution: (1,-4)
Integrated Algebra
EX: Solve by elimination.

2x - 5y = 3
5x + 6y = -11
What variable should we eliminate? (Look at coefficients) ____
Neither coefficient is related, so we need to take the LCM of either one.

If we eliminate the x’s, we need the LCM of 2 and 5, so +10 and -10

If we eliminate the y’s, we need the LCM of 5 and 6, so +30 and -30
I always pick the smaller, so I will eliminate the x’s

What coefficients do we need? ____ and _____

To change the first equation, multiply through by +5.
To change the second equation, multiply through by -2.
2x - 5y = 3
5x + 6y = -11

5(2x – 5y = 3)
-2(5x + 6y = -11)
Substitute and solve for x: 2x – 5(-1) = 3
2x + 5 = 3
2x = -2
x = -1
10x – 25y = 15
+ -10x – 12y = 22
- 37y = 37
y = -1
Solution: (-1,-1)
Integrated Algebra
TRY: Solve by elimination:
2x – 7y = 9
-3x + 4y = 6

What variable should we eliminate? (Look at coefficients) ____
Neither coefficient is related, so we need to take the LCM of either one.

If we eliminate the x’s, we need the LCM of 2 and -3, so ____________

If we eliminate the y’s, we need the LCM of -7 and 4, so ____________
Which variable do you want to eliminate?

What coefficients do we need? ____ and _____

To change the first equation, multiply through by _______.
To change the second equation, multiply through by _______.
2x – 7y = 9
-3x + 4y = 6

Substitute and solve for x:
Integrated Algebra
HOMEWORK: Solving systems by elimination
Changing both equations
Solve each system by elimination.
1. 2x – 5y = -10
7x – 3y = -6
2. 7x – 4y = 16
2x + 3y = 17
3. -5x - 2y = 12
11x – 5y = -17
4. 3x + 4y = -25
2x – 3y = 6
Name
Integrated Algebra
5.
5x + 3y = 12
4x – 5y = 17
6.
-3x + 8y = 10
5x + 6y = 80
7.
8x + 5y = 6
3x – 2y = 10
8.
5x - 6y = 48
2x + 5y = -3