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

2. Solution:An ordered pair that makes the equation true. Example: x + y = 7 What are some possible solutions? 5 + 2 = 7 3 + 4 = 7 -1 + 8 = 7 -10 + 17 = 7 (5,2) (3,4) (-1,8) (-10,17) The solutions are always written in ordered pairs.: How many more solutions are there?

3. System of Equations Definition: a set of two or more equations that have variables in common. Example of a system: y = 3x + 1 y = -2x - 4 Solution: ALL Any ordered pair that makes ALL the equations in a system true. Solution for this system: (-1,-2)

4. ( ) = -2( ) - 4 -2 = 2 - 4 Verify (check your answer) whether or not the given ordered pair is a solution for the following system of equations. (same system as previous slide) y = 3x + 1 y = -2x -4 ( -1,-2) ( ) = 3 ( ) + 1 -2 -2 = -3 + 1 Is (-1, -2) a solution? explain YES, because the ordered pair makes both equations true.

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

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

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

8. x – 5 = y 2x – 3y = 7 1. Solve for 1 variable ( x = or y =) Step 1 y = x – 5 2.Substitute the variable you just solved for into the other equation AND solve Step 2 2x – 3y = 7 2x – 3( ) = 7 2x – 3x + 15 = 7 -1x + 15 = 7 -1x = -8 x = 8 x – 5 3. Solve for the other variable (substitute) Step 3 y = x – 5 y = (8) – 5 y = 3 4. Write the ordered pair Step 4 (8, 3) 11.(2,6) 12.(8,11) 14. (13, -5) 15.(3,0) 17. (-11,-19) Pg 375

9. 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 1. Solve for 1 variable L = 60 - S 2.Substitute the variable you just solved for into the other equation AND solve. 5L + 3S = 220 5( ) + 3S = 22060 - S 300 – 5S + 3S = 220 300 – 2S= 220 – 2S = -80 S = 40 They sold 40 small snack packs

10. I can solve a system using substitution. Assignment: Pg 375: 11, 12, 14,15, 17 11.(2,6) 12.(8,11) 14. (13, -5) 15.(3,0) 17. (-11,-19)