1. Entry Task
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2. 3-2: Solving Systems of Equations using Substitution
Target: I can solve linear systems algebraically using substitution.

3. Solving Systems of Equations using Substitution
Steps:
1. Solve one equation for one variable (y= ; x= ; a=)
2. Substitute the expression from step one into the other equation.
3. Simplify and solve the equation.
4. Substitute back into either original equation to find
the value of the other variable.
5. Check the solution in both equations of the system.

4. Example #1:
y = 4x
3x + y = -21
Step 1: Solve one equation for one variable.
y = 4x (This equation is already solved for y.)
Step 2: Substitute the expression from step one into the other equation.
3x + y = -21
3x + 4x = -21
Step 3: Simplify and solve the equation.
7x = -21
x = -3

5. y = 4x 3x + y = -21 Step 4: Substitute back into either original
equation to find the value of the other
variable.
3x + y = -21
3(-3) + y = -21
-9 + y = -21
y = -12
Solution to the system is (-3, -12).

6. y = 4x 3x + y = -21 3x + y = -21 y = 4x 3(-3) + (-12) = -21
Step 5: Check the solution in both equations.
Solution to the system is (-3,-12).
3x + y = -21
3(-3) + (-12) = -21
-9 + (-12) = -21
-21= -21
y = 4x
-12 = 4(-3)
-12 = -12

7. Example #2: x + y = 10 y = -x +10 5x - y = 2 5x -(-x +10) = 2
Step 1: Solve one equation for one variable.
x + y = 10
y = -x +10
Step 2: Substitute the expression from step one into
the other equation.
5x - y = 2
5x -(-x +10) = 2

8. 5x -(-x + 10) = 2 5x + x -10 = 2 6x -10 = 2 6x = 12 x = 2 x + y = 10
Step 3: Simplify and solve the equation.
5x -(-x + 10) = 2
5x + x -10 = 2
6x -10 = 2
6x = 12
x = 2

9. Solution to the system is (2,8).
x + y = 10
5x – y = 2
Step 4: Substitute back into either original
equation to find the value of the other
variable.
x + y = 10
2 + y = 10
y = 8
Solution to the system is (2,8).

10. 5x – y = 2 x + y =10 5(2) - (8) = 2 2 + 8 =10 10 – 8 = 2 10 =10 2 = 2
Step 5: Check the solution in both equations.
Solution to the system is (2, 8).
5x – y = 2
5(2) - (8) = 2
10 – 8 = 2
2 = 2
x + y =10
2 + 8 =10
10 =10

11. Solve by substitution:1. 2.

12. Assignment #29 pg 146 (10 pts)
p. 146 #10-21 Challenge is 54