1. System of Equations Substitution Method Day 1
Today’s Objective:
I can solve a system using substitution.

2. A snack bar sells two sizes of snack packs
A snack bar sells two sizes of snack packs. A large snack pack is $5 and a small snack pack is $3. In one day, the snack bar sold 60 snack packs for a total of $220. How many small snack packs did the snack bar sell?
Identify variables:
Let L = Large packs
Let S= small snack packs
Total # of snack packs:
L S = 60
Money earned from snack packs:
5L + 3S = 220
Substitute the variable you just solved for into the other equation AND solve.
1. Solve for 1 variable
L = 60 - S
5L + 3S = 220
5( ) + 3S = 220
60 - S
300 – 5S + 3S = 220
300 – 2S= 220
They sold 40 small snack packs
– 2S = -80
S = 40

3. x + y = 7 x – y = 3 x – y = 3 x + y = 7 x y x y
What is the solution (ordered pair) that makes both equations true?
x + y = 7
x – y = 3
x + y = 7
x – y = 3 x y x y
Ways to solve:
Guess and check
Graph them to find the solution.
Tables
New method---substitution

4. x + y = 7 x – y = 3 (5, 2) x – y = 3 + y + y x = 3 + y x + y = 7
1. Solve for 1 variable (x = or y = )
x + y = 7
x – y = 3
2. Substitute the variable you just solved for into the other equation AND solve.
3. Solve for the other variable (substitute)
4. Write the ordered pair
Step 2
Step 3
Step 1
x + y = 7
( ) + y = 7
3 + y + y = 7
3 + 2y = 7
2y = 4
y = 2
x = 3 + y
x = 3 + ( )
x = 3 + 2
x = 5
x – y = 3
+ y y
x = 3 + y
3 + y 2
Step 4
(5, 2)

5. (-4, 3) 2x + 5y = 7 x = y – 7 2( ) + 5y = 7 2y – 14 + 5y = 7
1. Solve for 1 variable (x= or y = )
2x + 5y = 7
x = y – 7
Substitute the variable you just solved for into the other equation AND solve.
3. Solve for the other variable (substitute)
4. Write the ordered pair
Step 1
Step 2
Step 3
2x + 5y = 7
2( ) + 5y = 7
2y – y = 7
7y – 14 = 7
7y = 21
y = 3
x = y – 7
x = y – 7
x = (3) – 7
x = -4
y – 7
Step 4
(-4, 3)

6. (8, 3) 2x – 3y = 7 y = x – 5 2x – 3( ) = 7 2x – 3x + 15 = 7
1. Solve for 1 variable
y = x – 5
2x – 3y = 7
Substitute the variable you just solved for into the other equation AND solve
3. Solve for the other variable (substitute)
4. Write the ordered pair
Step 1
Step 2
Step 3
2x – 3y = 7
2x – 3( ) = 7
2x – 3x + 15 = 7
-1x + 15 = 7
-1x = -8
x = 8
y = x – 5
y = x – 5
y = 8 – 5
y = 3
x – 5
Step 4
(8, 3)

7. (-2, 4) x = 6 – 2y 5x + 3y = 2 5( ) + 3y = 2 30 – 10y + 3y = 2
1. Solve for 1 variable
x = 6 - 2y
5x + 3y = 2
Substitute the variable you just solved for in to the other equation AND solve.
3. Solve for the other variable (substitute)
Step 1
4. Write the ordered pair
x = 6 – 2y
Step 2
Step 3
5x + 3y = 2
5( ) + 3y = 2
30 – 10y + 3y = 2
30 – 7y = 2
-7y = -28
y = 4
x = 6 – 2y
x = 6 – 2( )
x = -2
Homework: 1-8
(4, 1)
(-1, 3)
3. (8,3)
4. (7,2)
5. (12, 9)
6 – 2y 4
Step 4
(-2, 4)

8. 1. Solve for 1 variable
x = 6 - 2y
5x + 3y = 2
Substitute the variable you just solved for in to the other equation AND solve.
3. Solve for the other variable (substitute)
4. Write the ordered pair
Homework: 1-12
(4, 1) 7. (3,4)
(-1, 3) 8. (2,2)
3. (8,3) 9. (1,8)
4. (7,2) (-4,3)
5. (12, 9) (6,3)
6. (1,-5) (4,-5)

9. I can solve a system using substitution.
Assignment:
WS: Substitution 1 #1-5