1. Systems of Equations: Substitution Method
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Sometimes graphing can take too long to determine the point of intersection. Therefore, the substitution method is a quick method to determine the point of intersection.
Example #1: If the equations are in “y–form” then do the following: =
This is the same point we would get if we graphed the two lines!!! 4 4
Point of Intersection
Now that you solved for x, go back and solve for y.

2. Substitution can be used even if the equations are not in y–form.
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Substitution can be used even if the equations are not in y–form.
x = –3y + 1
Use substitution to rewrite the two equations as one.
Example #2:
4x – 3y = –11
Rewrite the 2nd equation putting ( ) in place of x.
x = –3y + 1
x = –3y + 1
4x – 3y = –11
x = –3(1) + 1
4(–3y + 1) – 3y = –11
Substitute what x is equal to.
x = –3 + 1
x = –2
–12y + 4 – 3y = –11
You can now solve for y and then go back and solve for x.
–15y + 4 = –11
–4 –4
–15y = –15
– –15
(–2, 1)
y = 1
Point of Intersection

3. The Substitution Method
Summary
(You may need to solve one equation for x or y to get this)
1. Look for x = or y =
2. Substitute into the OTHER equation.
3. Solve.
4. Go back and substitute this value into one of the ORIGINAL equations to solve for the other variable.
5. Put answer in coordinate form.
(x, y)
This is the point of intersection of the lines on a graph!

4. Use the Substitution Method to solve these linear systems.1.
y = -x + 8 2.
y = 3x – 7
y = x - 2
5x + y = 1
x – 2 = -x + 8
5x + (3x – 7) = 1
+x x
8x – 7 = 1
2x – 2 = 8
8x = 8
2x = 10
x = 1
x = 5
y = 3(1) - 7
y = (5) – 2
y = -4
y = 3
(1, -4)
(5, 3)

5. 1. Look for opposite coefficients of either x or y.
Sometimes the substitution method for solving systems of equations is awkward or tedious.
Another way to solve systems of equations is to eliminate one variable. This is known as the ELIMINATION METHOD.
1. Look for opposite coefficients of either x or y.
(If not, make them by multiplying one or both equations by a constant!)
2. Add the equations together.
(This should ELIMINATE a variable…if not check your work!)
3. Solve for the remaining variable.
4. Substitute this value back into one of the original equations to solve for the remaining variable.
5. Answer in coordinate form.
(x,y)

6. The ELIMINATION Method
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The ELIMINATION Method
Example 1: Solve the system of equations.
OPPOSITES!
1. Add the equations vertically.
2. Substitute the value of x into either equation. Solve for y.
2x + 3y = 8
2(1) + 3y = 8
7x = 7
2 + 3y = 8 7 7
3y = 6
x = 1
y = 2
(1, 2)
3. Write your answer as an ordered pair:
Remember: This is the point of intersection of the graphs!

7. The ELIMINATION Method
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The ELIMINATION Method
Example #2:
3(3x + 2y = 11)
3x + 2y = 11
9x + 6y = 33
–2(4x + 3y = 14)
4x + 3y = 14 +
–8x – 6y = –28
x = 5
Are there opposites anywhere?
3(5) + 2y = 11
No…we need to create them!
15 + 2y = 11
Decide which variable to eliminate…
– –15
Let’s get rid of the y.
2y = –4
What is the least common multiple of 2 & 3?
y = –2 6
(5, –2)
Point of Intersection

8. 1. 2x + y = 7 2. 5x + 2y = 6 10x + 4y = 12 2x – y = 5 –3x – 4y = 2
Use the Elimination Method to solve these linear systems. 1.
2x + y = 7 2.
5x + 2y = 6
10x + 4y = 12
2x – y = 5
–3x – 4y = 2
–3x – 4y = 2
4x = 12
7x = 14
x = 3
x = 2
5(2) + 2y = 6
2(3) + y = 7
10 + 2y = 6
6 + y = 7
2y = –4
y = 1
y = –2
(3, 1)
(2, – 2 )

9. +1
1. y = 7x – x – y = 8
y = 5x –6x + 2y = – 8
4x – 2y = 16
–6x + 2y = – 8
– 2x = 8 +1 +1
7x – 2 = 5x + 8
– – 2
-5x x +1
x = – 4
2x – 2 = 8
y = 5(5) + 8
2(–4) – y = 8
y =
2x = 10
– 8 – y = 8
y = 33
– y = 16 +1
x = 5 +1
(-4, -16)
(5, 33) +1
y = – 16
3. Explain how your answer relates to the graphs of the equations.
The solutions are the points of intersection of the equations on a graph. +1