Section 3.2

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Section 3.2. Solving Systems Algebraically. Solving Systems Algebraically. ALGEBRA 2 LESSON 3-2. (For help, go to Lesson 1-1 and 1-3.). Find the additive inverse of each term. 1. 4 2. – x 3. 5 x 4. 8 y Let x = 2 y – 1. Substitute this expression for x in each equation.

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### Section 3.2

Solving Systems Algebraically

Solving Systems Algebraically

ALGEBRA 2 LESSON 3-2

(For help, go to Lesson 1-1 and 1-3.)

Find the additive inverse of each term.

1. 4 2. –x3. 5x4. 8y

Let x = 2y – 1. Substitute this expression for x in each equation.

Solve for y.

5.x + 2y = 3 6.y – 2x = 8 7. 2y + 3x = –5

Check Skills You’ll Need

3-2

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

Solving Systems Algebraically

ALGEBRA 2 LESSON 3-2

Solutions

3-2

x + 3y = 12

–2x + 4y = 9

Solving Systems Algebraically

ALGEBRA 2 LESSON 3-2

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 – 24 + 4y = 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

Quick Check

The solution is (2.1, 3.3).

3-2

2 p + s = 10.25

4 p + s = 18.75

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.

Let s = the price of a soda.

Write:

Solving Systems Algebraically

ALGEBRA 2 LESSON 3-2

At Renaldi’s Pizza, a soda and two slices of the pizza–of–the–day costs \$10.25. A soda and four slices of the pizza–of–the–day costs \$18.75. Find the cost of each item.

2p + s = 10.25 Solve for one of the variables.

s = 10.25 – 2p

3-2

Solving Systems Algebraically

ALGEBRA 2 LESSON 3-2

(continued)

4p + (10.25 – 2p) = 18.75 Substitute the expression for s into the

other equation. Solve for p.

p = 4.25

2(4.25) + s = 10.25 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.

Quick Check

3-2

3x + y = –9

–3x – 2y = 12

3x + y = –9

–3x – 2y = 12 Two terms are additive inverses, so add.

–y = 3 Solve for y.

Solving Systems Algebraically

ALGEBRA 2 LESSON 3-2

Use the elimination method to solve the system.

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).

Quick Check

3-2

2m + 4n = –4

3m + 5n = –3

2m + 4n = –4 10m + 20n = –20

1

3m + 5n = –3–12m–20n = 12

2

Multiply by 5.

2

1

Multiply by –4.

Solving Systems Algebraically

ALGEBRA 2 LESSON 3-2

Solve the system by elimination.

To eliminate the n terms, make them additive inverses by multiplying.

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

Quick Check

The solution is (4, –3).

3-2

–3x + 5y = 6

6x – 10y = 0

–6x + 10y = 12

6x – 10y = 0

Multiply the first line by 2

to make the x terms

0 = 12

Solving Systems Algebraically

ALGEBRA 2 LESSON 3-2

Solve each system by elimination.

a. –3x + 5y = 6

6x – 10y = 0

Elimination gives an equation that is always false.

The two equations in the system represent parallel lines.

The system has no solution.

3-2

3-2

–3x + 5y = 6

6x – 10y = –12

–6x + 10y = 12

6x + 10y = –12

Multiply the first line by 2

to make the x terms

0 = 0

Solving Systems Algebraically

ALGEBRA 2 LESSON 3-2

Quick Check

(continued)

b. –3x + 5y = 6

6x – 10y = –12

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.

3-2

1. Solve by substitution.

2. A bookstore took in \$167 on the sale of 5 copies of a new cookbook and

3 copies of a new novel. The next day it took in \$89 on the sales of 3 copies

of the cookbook and 1 copy of the novel. What was the price of each book?

Solve each system.

3.4. 5.

–2x + 5y = –2

x – 3y = 3

10x + 6y = 0

–7x + 2y = 31

7x + 5y = 18

–7x – 9y = 4

–3x + y = 6

6x – 2y = 25

Solving Systems Algebraically

ALGEBRA 2 LESSON 3-2

(–9, –4)

cookbook: \$25; novel: \$14

(–3, 5)

(6.5, –5.5)

no solutions

3-2