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

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

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

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 -3 -3 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)

(-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 – 14 + 5y = 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)

(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)

(-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)

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) 10. (-4,3) 5. (12, 9) 11. (6,3) 6. (1,-5) 12. (4,-5)

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