Solving Systems of Linear Equations using the Elimination Method (Adding)
Two or more linear equations together form a System of Linear Equations.
You can solve systems using three different methods: Graphing, Substitution, or
Elimination.
Solve the system:
5x - 6y = -32
What form are both equations written in?!
3x + 6y = 48
Standard Form
When both equations are written in standard form, it is easy to use the Elimination Method
to solve the given system of linear equations.
Eliminate means you will be able to "get rid" of a variable so that you can solve for the one left.
EXAMPLE 1:
5x - 6y = -32
3x + 6y = 48
What can you do to eliminate one of the variables?
5x - 6y = -32
+ 3x + 6y = 48
If you add the equations, you will
eliminate the y -values.
When you see opposites, you know you can ADD to
eliminate!
8x = 16
It's now a one-step equation to solve!
8x = 16
8
( 2 ,
)
8
x = 2
Now just plug the value of x back into one of the
original equations to find the value of y
3x + 6y = 48
3(2) + 6y = 48
6 + 6y = 48
-6
It's now a two-step equation to solve!
-6
6y = 42
6y = 42
6
( 2 , 7 )
6
y = 7
The Point of Intersection is (2, 7)
Always check your answer by plugging in the ordered pair for both equations.
The point of intersection MUST WORK FOR BOTH EQUATIONS to be a solution to the system.
5x - 6y = -32
3x + 6y = 48
5(2) - 6(7) = -32
3(2) + 6(7) = 48
10 - 42 = -32
6 + 42 = 48
-32 = -32
48 = 48
Solving Systems of Linear Equations using the Elimination Method (Add) Notes, Page 1
EXAMPLE 2:
6x - 3y = 3
When you see opposites, you know you can ADD!
-6x + 5y = 3
What can you do to eliminate one of the variables?
6x - 3y = 3
If you add the equations, you will
eliminate the x -values.
+ -6x + 5y = 3
2y = 6
It's now a one-step equation to solve!
2y = 6
2
(
, 3 )
2
y = 3
Now just plug the value of y back into one of the
original equations to find the value of x
-6x + 5y = 3
-6x + 5(3) = 3
-6x + 15 = 3
It's now a two-step equation to solve!
-15 -15
-6x = -12
-6x = -12
-6
( 2 , 3 )
-6
x = 2
The Point of Intersection is (2, 3)
Always check your answer by plugging in the ordered pair for both equations.
The point of intersection MUST WORK FOR BOTH EQUATIONS to be a solution to the system.
6x - 3y = 3
-6x + 5y = 3
6(2) - 3(3) = 3
-6(2) + 5(3) = 3
12 - 9 = 3
-12 + 15 = 3
3=3
EXAMPLE 3:
3=3
x+y=3
When you see opposites, you know you can ADD!
x-y=1
What can you do to eliminate one of the variables?
x+y=3
If you add the equations, you will
eliminate the y -values.
+ x-y=1
2x = 4
It's now a one-step equation to solve!
2x = 4
2
( 2
,
)
2
x = 2
Now just plug the value of x back into one of the
original equations to find the value of y
Solving Systems of Linear Equations using the Elimination Method (Add) Notes, Page 2
x+y=3
(2) + y = 3
-2
( 2 , 1 )
It's now a one-step equation to solve!
-2
y = 1
The Point of Intersection is (2, 1)
Always check your answer by plugging in the ordered pair for both equations.
The point of intersection MUST WORK FOR BOTH EQUATIONS to be a solution to the system.
x+y=3
x-y=1
(2) + (1) = 3
(2) - (1) = 1
3=3
EXAMPLE 4:
1=1
x+y=8
When you see opposites, you know you can ADD!
x - y = 10
What can you do to eliminate one of the variables?
x+y=8
If you add the equations, you will
eliminate the y -values.
x - y = 10
2x = 18
It's now a one-step equation to solve!
2x = 18
2
( 9 ,
)
2
x = 9
Now just plug the value of x back into one of the
original equations to find the value of y
x+y=8
(9) + y = 8
-9
( 9 , -1 )
It's now a one-step equation to solve!
-9
y = -1
The Point of Intersection is (9, -1)
Always check your answer by plugging in the ordered pair for both equations.
The point of intersection MUST WORK FOR BOTH EQUATIONS to be a solution to the system.
x+y=8
x - y = 10
(9) + (-1) = 8
(9) - (-1) = 10
8=8
10 = 10
EXAMPLE 5:
EXAMPLE 6:
EXAMPLE 7:
2x + y = 7
x + 3y = 9
5x + y = 16
3x - y = 3
x - 3y = -3
-5x + 3y = 8
(2,3)
(3,2)
(2,6)
Solving Systems of Linear Equations using the Elimination Method (Add) Notes, Page 3
SOMETIMES ADDING THE EQUATIONS WON'T WORK FOR YOU TO ELIMINATE
SO THE NEXT QUESTION YOU ASK YOURSELF IS: CAN I SUBTRACT THE EQUATIONS
TO ELIMINATE A VARIABLE?
EXAMPLE 1:
x + y = 292
3x + y = 470
What can you do to eliminate one of the variables?
x + y = 292
₋ 3x + y = 470
If you SUBTRACT the equations, you will
eliminate the y -values.
When you see the same terms, you know you can
SUBTRACT to eliminate!
x + y = 292
+ -3x + -y = -470
Remember to ADD the opposite!
Subtraction becomes addition and each term changes
to it's opposite!
x + y = 292
+ -3x + -y = -470
-2x = -178
Now, what variable can be eliminated?
It's now a one-step equation to solve!
-2x = -178
-2
( 89 ,
)
-2
x = 89
Now just plug the value of x back into one of the
original equations to find the value of y
x + y = 292
(89) + y = 292
-89
( 89 , 203 )
It's now a one-step equation to solve!
-89
y = 203
The Point of Intersection is (89, 203)
Always check your answer by plugging in the ordered pair for both equations.
The point of intersection MUST WORK FOR BOTH EQUATIONS to be a solution to the system.
x + y = 292
3x + y = 470
(89) + (203) = 292
3(89) + (203) = 470
292 = 292
267 + 203 = 470
470 = 470
EXAMPLE 2:
2x + 3y = 11
₋ 2x - 9y = -1
When you see the same terms, you know you can
SUBTRACT to eliminate!
2x + 3y = 11
-2x + 9y = +1
Remember to ADD the opposite!
Subtraction becomes addition and each term changes to it's opposite!
Solving Systems of Linear Equations using the Elimination Method (Subtract) Notes, Page 4
EXAMPLE 2: (continued)
2x + 3y = 11
Now, what variable can be eliminated?
+ -2x + 9y = +1
12y = 12
It's now a one-step equation to solve!
12y = 12
12
(
, 1 )
12
y = 1
Now just plug the value of y back into one of the
original equations to find the value of x
2x + 3y = 11
2x + 3(1)= 11
2x + 3 = 11
-3
It's now a two-step equation to solve!
-3
2x = 8
2x = 8
2
( 4 , 1 )
2
x = 4
The Point of Intersection is (4, 1)
Always check your answer by plugging in the ordered pair for both equations.
The point of intersection MUST WORK FOR BOTH EQUATIONS to be a solution to the system.
2x + 3y = 11
2x - 9y = -1
2(4) + 3(1) = 11
2(4) - 9(1) = -1
8 + 3 = 11
8 - 9 = -1
11 = 11
-1 = -1
TRY THESE NOW:
1) x + 3y = 9 and x - 3y = -3
2) 2x + y = 5 and x + y = 4
3) 5x + y = 16 and -5x + 3y = 8
4) 2x + y = 7 and 3x - y = 3
5) 3x + 4y = 1 and 2x + 4y = -2
6) 11x + 2y = 44 and 8x + 2y = 32
Solving Systems of Linear Equations using the Elimination Method (Subtract) Notes, Page 5
What if you cannot add or subtract because you do not have opposites or same terms?
MULTIPLYING EQUATIONS BY ONE TERM
EXAMPLE 1:
2x + 5y = -22
Can we add to eliminate? No, we do not see opposites
10x + 3y = 22
Can we subtract to eliminate? No, we do not have same terms
You can multiply by a value to prepare for elimination
Multiplying the first equation by -5 will help us eliminate the "x" terms
EXAMPLE 1:
-5(2x + 5y = -22)
-10x - 25y = 110
10x + 3y = 22
10x + 3y = 22
-22y = 132
Now divide both sides by -22
-22y = 132
-22
(
, -6 )
-22
y = -6
Now plug in -6 for y and solve for x
10x + 3y = 22
10x + 3(-6) = 22
10x - 18 = 22
Now solve the two step equation you have
10x - 18 = 22
+18
+18
10x = 40
10x = 40
10
( 4 , -6 )
10
x = 4
ALWAYS CHECK YOUR WORK TO KNOW IF YOU'RE CORRECT!
Always check your answer by plugging in the ordered pair for both equations.
The point of intersection MUST WORK FOR BOTH EQUATIONS to be a solution to the system.
2x + 5y = -22
10x + 3y = 22
2(4) + 5(-6) = -22
10(4) + 3(-6) = 22
8 - 30 = -22
40 - 18 = 22
-22 = -22
22 = 22
Solving Systems of Linear Equations using the Elimination Method (Multiply) Notes, Page 6
EXAMPLE 2:
2x + 6y = 8
Can we add to eliminate? No, we do not see opposites
6x + 14y = 12
Can we subtract to eliminate? No, we do not have same terms
You can multiply by a value to prepare for elimination
Multiplying the first equation by -3 will help us eliminate the "x" terms
EXAMPLE 2:
(
-3(2x + 6y = 8)
-6x - 18y = -24
6x + 14y = 12
6x + 14y = 12
-4y = -12
, 3 )
Now divide both sides by -4
-4y = -12
-4
-4
y = 3
Now plug in 3 for y and solve for x
2x + 6y = 8
2x + 6(3) = 8
2x + 18 = 8
Now solve the two step equation you have
2x + 18 = 8
-18
-18
2x = -10
2x = -10
2
2
( -5 , 3 ) x = -5
Always check your answer by plugging in the ordered pair for both equations.
The point of intersection MUST WORK FOR BOTH EQUATIONS to be a solution to the system.
2x + 6y = 8
6x + 14y = 12
2(-5) + 6(3) = 8
6(-5) + 14(3) = 12
-10 + 18 = 8
-30 + 42 = 12
8=8
TRY THESE:
A. 2x + 5y = 34
x + 2y = 14
(2, 6)
12 = 12
B. 2x - y = 5
5x + 2y = 17
(3, 1)
C. -2x + 5y = 9
x + y = 13
(8, 5)
Solving Systems of Linear Equations using the Elimination Method (Multiply) Notes, Page 7
What if you cannot add or subtract because you do not have opposites or same terms AND
what if you cannot multiply one equation by a term to eliminate a variable?
EXAMPLE 3:
6x + 3y = 27
-4x + 7y = 27
Can we add to eliminate? No, we do not see opposites
Can we subtract to eliminate? No, we do not have same terms
Can we multiply by one term to eliminate? No, we cannot
You can multiply each equation by a value to prepare for elimination
YOU WILL HAVE SOME OPTIONS HERE ACCORDING TO WHAT YOU SEE!!!
OPTION 1:
Multiplying the first equation by 4 & the second equation by 6 will help us eliminate the "x" terms
OPTION 2:
Multiplying the first equation by 2 & the second equation by 3 will help us eliminate the "x" terms
OPTION 3:
Multiplying the first equation by 7 & the second equation by -3 will help us eliminate the "y" terms
OPTION 4:
Multiplying the first equation by -7 & the second equation by 3 will help us eliminate the "y" terms
I AM GOING TO USE THIS OPTION BUT YOU WILL LEARN TO CHOOSE YOUR OWN, YOUR WAY.
OPTION 2:
Multiplying the first equation by 2 & the second equation by 3 will help us eliminate the "x" terms
EXAMPLE 3:
(
2(6x + 3y = 27)
12x + 6y = 54
3(-4x + 7y = 27)
-12x + 21y = 81
27y = 135
, 5 )
Now divide both sides by 27
27y = 135
27
27
y = 5
Now plug in 5 for y and solve for x
6x + 3y = 27
6x + 3(5) = 27
6x + 15 = 27
Now solve the two step equation you have
6x + 15 = 27
-15
-15
6x = 12
6x = 12
6
( 2 , 5 )
6
x = 2
Always check your answer by plugging in the ordered pair for both equations.
The point of intersection MUST WORK FOR BOTH EQUATIONS to be a solution to the system.
6(2) + 3(5) = 27
-4(2) + 7(5) = 27
12 + 15 = 27
-8 + 35 = 27