1. Objectives: 1.Be able to solve a system of equations by using the substitution method. 2.Be able to solve a system of equations by using the elimination method. Critical Vocabulary: System of Equations, Substitution Daily Warm Up: Solve the following system of equations by graphing -x + y = 1 -2x + y = -2 Multiply By -1

2. Daily Warm Up: Solve the following system of equations by graphing -x + y = 1 -2x + y = -2 Equation 1: -x + y = 1 +x y = x + 1 Slope: 1 (Up 1 Right 1 and Down 1 Left 1) Y-intercept: (0, 1) Equation 2: -2x + y = -2 +2x y = 2x - 2 Slope: 2 (Up 2 Right 1 and Down 2 Left 1) Y-intercept: (0, -2) Solution: (3, 4)

3. y = x + 1 y = -2x - 2 Example 1: Solve the system of equations I. Substitution x + 1 = -2x - 2 Since both these equations are equal to “y” we can set them equal to each other Solve for “x” [Standard 4.2] 3x + 1 = - 2 3x = - 3 x = - 1 Check your answer using ONE of the original equations y = -2(-1) - 2 y = 2 - 2 y = 0 Solution: (-1, 0) Write your solution

4. 3x – 2 y = 12 y = x – 60 Example 2: Solve the system of equations In this case we know that “y” is equal to x-60. We can substitute x – 60 in place of “y” in the top equation 3x - 2(x - 60) = 12 - 120 -120 y = -108 - 60 y = -168 Solution: (-108, -168) 3x - 2x + 120 = 12 x + 120 = 12 x = -108 I. Substitution Solve for “x” [Standard 4.2] Check your answer using ONE of the original equations Write your solution

5. 4x + 3y = 8 x – 4y = 3 Example 3: Solve the system of equations 4(4y + 3) + 3y = 8 - 12 -12 16y + 12 +3y = 8 19y + 12 = 8 19y = -4 4x + 3y = 8 19 y = -4/19 I. Substitution x = 4y + 3 x = 4(-4/19) + 3 x = -16/19 + 3 Solution: (41/19, -4/19) x = 41/19 In this case neither equation has the “x” or “y” all by itself. We could easily get “x” all by itself in the second equation by adding 4y to both sides. No change to the first equation. Substitute 4y + 3 in place of “x” in the first equation. Solve for “y” [Standard 4.2] Check your answer using ONE of the original equations Write your solution