1. L3-2 Obj: Solve a system by substitution and elimination
Warm up
Substitute 2y – 1 for x in each equation. Solve for y
1. x + 2y = 3
2. y – 2x = 8
3. 2y + 3x = -5
Substitute 2x + 5 for y in each equation and Solve for x
4. 2x – 2y = 4
5. 3x + y = 7

2. Steps for solving by substitution
Choose one equation and solve for one of the variables
Substitute the expression you found in step one into the other equation
Solve
Once you have found x or y, substitute that value into either equation to find the other variable

3. Steps for solving by elimination
Make coefficient of x or y the same (one positive and one negative)
Add the two equations
Solve for the variable
Substitute value of the variable you found in one of two original equations and solve for the other variable

4. Solve the system by substitution.
Solving Systems Algebraically
LESSON 3-2
Additional Examples
x + 3y = 12
–2x + 4y = 9
Solve the system by substitution.

5. Ex 2: x + y = x – y = 2

6. Ex 3: 4x - 3y = x + 5y = 14

7. Practice: 1. x – y = 0 x + y = 2 2. -x + 3y = -1 x - 2y = 2

8. Use the elimination method to solve the system.
Solving Systems Algebraically
LESSON 3-2
Additional Examples
3x + y = –9
–3x – 2y = 12
Use the elimination method to solve the system.

9. Solve the system by elimination. 2m + 4n = –4 3m + 5n = –3
Solving Systems Algebraically
LESSON 3-2
Additional Examples
Solve the system by elimination.
2m + 4n = –4
3m + 5n = –3

10. Solve each system by elimination.
Solving Systems Algebraically
LESSON 3-2
Additional Examples
Solve each system by elimination.
a. –3x + 5y = 6
6x – 10y = 0

11. (continued) b. –3x + 5y = 6 6x – 10y = –12
Solving Systems Algebraically
LESSON 3-2
Additional Examples
(continued)
b. –3x + 5y = 6
6x – 10y = –12
A2_0302_QC.qtb

12. Practice
Solve using elimination
1. 4x – 2y = 7
x + 2y = 3

13. 2. 3x + 7y = 15
5x + 2y = -4

14. Solving Systems Algebraically
LESSON 3-2
Additional Examples
At Renaldi’s Pizza, a soda and two slices of the pizza-of-the-day costs $ A soda and four slices of the pizza-of-the-day costs $ Find the cost of each item.

15. Break-Even Point A theater production costs $40,000 plus $2800 per performance. A sold-out performance brings in $3675. How many sold-out performances will the production need to break even?

16. It costs $5,500,000 to build a ferry boat plus $15,000 in operating costs per trip. A sold-out trip brings in $24,000. How many sold-out trips will the ferry company need to break even?
HW: L3-2 (p. 130) #4-8e 18-24e 30-40e 44-58e