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Equations with Variables on Both SidesPowerPoint Presentation

Equations with Variables on Both Sides

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Equations with Variables on Both Sides

(For help, go to Lessons 1-7 and 2-3.)

Simplify.

1. 6x – 2x2. 2x – 6x3. 5x – 5x4. –5x + 5x

Solve each equation.

5. 4x + 3 = –5 6. –x + 7 = 12

7. 2t – 8t + 1 = 43 8. 0 = –7n + 4 – 5n

2-4

Equations with Variables on Both Sides

Solutions

1. 6x – 2x = (6 – 2)x = 4x2. 2x – 6x = (2 – 6)x = –4x

3. 5x – 5x = (5 – 5)x = 0x = 0 4. –5x + 5x = (–5 + 5)x = 0x = 0

5. 4x + 3 = –5 6. –x + 7 = 12

4x = –8 –x = 5

x = –2 x = –5

7. 2t – 8t + 1 = 43 8. 0 = –7n + 4 – 5n

–6t + 1 = 43 0 = –12n + 4

–6t = 42 12n = 4

t = –7 n =

1

3

2-4

5x – 3 – 2x = 2x + 12 – 2xSubtract 2x from each side.

3x – 3 = 12 Combine like terms.

3x – 3+ 3 = 12 + 3Add 3 to each side.

3x = 15 Simplify.

= Divide each side by 3.

3x3

153

x = 5 Simplify.

ALGEBRA 1 LESSON 2-4

Equations with Variables on Both SidesThe measure of an angle is (5x – 3)°. Its vertical angle has a measure of (2x + 12)°. Find the value of x.

5x – 3 = 2x + 12 Vertical angles are congruent.

2-4

Relate: cost of plus equipment equals skateboard and equipment

friend’s rental rental

skateboard

Define: let h = the number of hours you must skateboard

Write: 60 + 1.5 h = 5.5 h

ALGEBRA 1 LESSON 2-4

Equations with Variables on Both SidesYou can buy a skateboard for $60 from a friend and rent the safety equipment for $1.50 per hour. Or you can rent all items you need for $5.50 per hour. How many hours must you use a skateboard to justify buying your friend’s skateboard?

2-4

60 + 1.5 and equipmenth – 1.5h = 5.5h – 1.5h Subtract 1.5h from each side.

60 = 4hCombine like terms.

604

4h4

= Divide each side by 4.

15 = hSimplify.

ALGEBRA 1 LESSON 2-4

Equations with Variables on Both Sides(continued)

60 + 1.5h = 5.5h

You must use your skateboard for more than 15 hours to justify buying the skateboard.

2-4

ALGEBRA 1 LESSON 2-4 and equipment

Equations with Variables on Both SidesSolve each equation.

a. –6z + 8 = z + 10 – 7z

–6z + 8 = z + 10 – 7z

–6z + 8 = –6z + 10 Combine like terms.

–6z + 8 + 6z = –6z + 10 + 6zAdd 6z to each side.

8 = 10 Not true for any value of z!

This equation has no solution

b. 4 – 4y = –2(2y – 2)

4 – 4y = –2(2y – 2)

4 – 4y = –4y + 4Use the Distributive Property.

4 – 4y + 4y = –4y + 4 + 4yAdd 4y to each side.

4 = 4 Always true!

The equation is true for every value of y, so the equation is an identity.

2-4

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