1. Warm Up ( 10mins)
Algebra 2 Textbook Page 123 Check Skills You’ll Need Exercises # 2, 4, 5-7

2. Solving Systems Algebraically
Solutions
1. additive inverse of 4: –4 2. additive inverse of –x: x
3. additive inverse of 5x: –5x 4. additive inverse of 8y: –8y
5. x + 2y = 3, with x = 2y – 1:
(2y – 1) + 2y = 3
4y – 1 = 3
4y= 4
y = 1
7. 2y + 3x = –5, with x = 2y – 1:
2y + 3(2y – 1)= –5
2y + 6y – 3= –5
8y – 3= –5
8y= –2
y= –
6. y – 2x = 8, with x = 2y – 1:
y – 2(2y – 1) = 8
y – 4y + 2 = 8
–3y + 2 = 8
–3y = 6
y = –2 1 4

3. Lesson 3.2: SOLVING SYSTEMS ALGEBRAICALLY
LINEAR SYSTEMS
Lesson 3.2:
SOLVING SYSTEMS ALGEBRAICALLY

4. Objective
Students will use substitution and elimination in order to solve problems using systems of linear equations.

5. INTRODUCTORY AND DEVELOPMENTAL ACTIVITIES
Vocabulary: Find the following terms and pages 125 and write down the definitions in your notebook: •Equivalent systems

6. To solve a system of equations by substitution Step 1: Solve for one of the variables. Choose the easiest one. Step 2: Substitute the expression of the variable that you choose into the other equation. Then solve the equation. Step 3: Substitute the value of the variable found into either equation . Then solve the equation.

7. Solving Systems Algebraically
x + 3y = 12
–2x + 4y = 9
Solve the system by substitution.
Step 1:  Solve for one of the variables. Solving the first equation
for x is the easiest.
x + 3y = 12
x = –3y + 12
Step 2:  Substitute the expression for x into the other equation. Solve for y.
–2x + 4y = 9
–2(–3y + 12) + 4y = 9 Substitute for x.
6y – y = 9 Distributive Property
6y + 4y = 33
y = 3.3
Step 3:  Substitute the value of y into either equation. Solve for x.
x = –3(3.3) + 12
x = 2.1
The solution is (2.1, 3.3).

8. Solving Systems Algebraically
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.
Relate:  2 •price of a slice of pizza + price of a soda = $10.25
4 • price of a slice of pizza + price of a soda = $18.75
Define: Let p = the price of a slice of pizza.    
Lets = the price of a soda.
Write:
2p+ s = 10.25
4p+ s = 18.75
2p + s = Solve for one of the variables.
s = – 2p  

9. Solving Systems Algebraically
(continued)
4p + (10.25 – 2p) = Substitute the expression for s into the
other equation. Solve for p.
p = 4.25
2(4.25) + s = Substitute the value of p into one of
the equations. Solve for s.
s = 1.75
The price of a slice of pizza is $4.25, and the price of a soda is $1.75.

10. Pages 123-124 Complete check understanding 1-2, # 5, 14
Pages Complete check understanding 1-2, # 5, 14. Work the steps thoroughly for a better comprehension. Post # 14 on discussion board, and respond to at least two classmates.

11. To solve system of equations by elimination Step 1: Decide on which variable to eliminate Step 2: Multiply equation 1 by the coefficient of the variable that you want to eliminate in equation 2 and equation 2 by the coefficient of the variable that you want to eliminate in equation 1 Step 3: Add or subtract the two equations Step 4: Solve

12. Solving Systems Algebraically
3x + y = –9
–3x – 2y = 12
Use the elimination method to solve the system.
3x + y = –9
–3x – 2y = Two terms are additive inverses, so add.
–y = 3 Solve for y.
y = -3
3x + y = –9 Choose one of the original equations.
3x + (–3) = –9 Substitute y. Solve for x.
x = –2
The solution is (–2, –3).

13. Solving Systems Algebraically
2m + 4n = –4
3m + 5n = –3
Solve the system by elimination.
To eliminate the n terms, make them additive inverses by multiplying.
2m + 4n = –4 10m + 20n = – Multiply by 5. 1
3m + 5n = –3 –12m – 20n = Multiply by –4. 2
–2m = –8 Add.
m = 4 Solve for m.
2m + 4n = –4 Choose one of the original
equations.
2(4) + 4n = –4 Substitute for m.
8 + 4n = –4
4n = –12 Solve for n.
n = –3
The solution is (4, –3).

14. Solving Systems Algebraically
Solve each system by elimination.
a. –3x + 5y = 6
6x – 10y = 0
Multiply the first line by 2
to make the x terms
additive inverses.
–3x + 5y = 6
6x – 10y = 0
–6x + 10y = 12
0 = 12
Elimination gives an equation that is always false.
The two equations in the system represent parallel lines.
The system has no solution.

15. Solving Systems Algebraically
Solve each system by elimination.
b. –3x + 5y = 6
6x – 10y = –12
Multiply the first line by 2
to make the x terms
additive inverses.
–3x + 5y = 6
6x – 10y = –12
–6x + 10y = 12
6x + 10y = –12
0 = 0
Elimination gives an equation that is always true.
The two equations in the system represent the same line.
The system has an infinite number of solutions:
{(x, y)| y = x + } 3 5 6

16. Pages 124-125 Complete check understanding 3-5, # 32, 42
Pages Complete check understanding 3-5, # 32, 42. Work the steps thoroughly for a better comprehension. Post # 42 on the discussion board, respond to at least two classmates.