# Equations with Variables on Both Sides ALGEBRA 1

Equations with Variables on Both Sides ALGEBRA 1 LESSON 2 -2 (For help, go to Lessons 1 -7 and 2 -3. ) Simplify. 1. 6 x – 2 x 2. 2 x – 6 x 3. 5 x – 5 x 4. – 5 x + 5 x Solve each equation. 5. 4 x + 3 = – 5 6. –x + 7 = 12 7. 2 t – 8 t + 1 = 43 8. 0 = – 7 n + 4 – 5 n 2 -2

Equations with Variables on Both Sides ALGEBRA 1 LESSON 2 -2 Solutions 1. 6 x – 2 x = (6 – 2)x = 4 x 2. 3. 5 x – 5 x = (5 – 5)x = 0 4. – 5 x + 5 x = (– 5 + 5)x = 0 5. 4 x + 3 = – 5 4 x = – 8 x = – 2 7. 2 t – 8 t + 1 = – 6 t = t= 6. 43 43 42 – 7 8. 2 -2 2 x – 6 x = (2 – 6)x = – 4 x –x + 7 = 12 –x = 5 x = – 5 0 0 12 n n = = – 7 n + 4 – 5 n – 12 n + 4 4 1 3

Equations with Variables on Both Sides ALGEBRA 1 LESSON 2 -2 5 x – 3 = 2 x + 12 5 x – 3 – 2 x = 2 x + 12 – 2 x Subtract 2 x from each side. 3 x – 3 = 12 Combine like terms. 3 x – 3 + 3 = 12 + 3 Add 3 to each side. 3 x = 15 Simplify. 3 x = 15 3 3 Divide each side by 3. x = 5 Simplify. 2 -2

Equations with Variables on Both Sides ALGEBRA 1 LESSON 2 -2 Check your answer of x = 5 5 x – 3 = 2 x + 12 5(5) – 3 = 2(5) + 12 25 – 3 = 10 + 12 22 = 22 True Equation 2 -2

Equations with Variables on Both Sides ALGEBRA 1 LESSON 2 -2 60 + 1. 5 h = 5. 5 h 60 + 1. 5 h – 1. 5 h = 5. 5 h – 1. 5 h Subtract 1. 5 h from each side. 60 = 4 h Combine like terms. 60 = 4 h 4 4 Divide each side by 4. 15 = h Simplify. 2 -2

Equations with Variables on Both Sides ALGEBRA 1 LESSON 2 -2 Check your answer of h = 15 60 + 1. 5 h = 5. 5 h 60 + 1. 5(15) = 5. 5(15) 60 + 22. 5 = 82. 5 True Equation 2 -2

Equations with Variables on Both Sides ALGEBRA 1 LESSON 2 -2 The measure of an angle is (5 x – 3)°. Its vertical angle has a measure of (2 x + 12)°. Find the value of x. 5 x – 3 = 2 x + 12 Vertical angles are congruent. 5 x – 3 – 2 x = 2 x + 12 – 2 x Subtract 2 x from each side. 3 x – 3 = 12 Combine like terms. 3 x – 3 + 3 = 12 + 3 Add 3 to each side. 3 x = 15 Simplify. 3 x = 15 3 3 Divide each side by 3. x = 5 Simplify. 2 -2

Equations with Variables on Both Sides ALGEBRA 1 LESSON 2 -2 You 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? Relate: cost of plus equipment equals skateboard and equipment friend’s rental skateboard Define: let h = the number of hours you must skateboard Write: 60 + 1. 5 h = 2 -2 5. 5 h

Equations with Variables on Both Sides ALGEBRA 1 LESSON 2 -2 60 + 1. 5 h = 5. 5 h 60 + 1. 5 h – 1. 5 h = 5. 5 h – 1. 5 h Subtract 1. 5 h from each side. 60 = 4 h Combine like terms. 60 = 4 h 4 4 Divide each side by 4. 15 = h Simplify. You must skateboard at least 11 hours. 2 -2

Equations with Variables on Both Sides ALGEBRA 1 LESSON 2 -2 Solve each equation. 1 1. 3 – 2 t = 7 t + 4 – 9 3. 3(1 – 2 x) = 4 – 5 x 2. 4 n = 2(n + 1) + 3(n – 1) 1 -1 4. You work for a delivery service. With Plan A, you can earn $5 per hour plus $. 75 per delivery. With Plan B, you can earn $7 per hour plus $. 25 per delivery. How many deliveries must you make per hour with Plan A to earn as much as with Plan B? (HINT: use the equation 5 +. 75 x = 7 +. 25 x ) 4 deliveries 2 -2

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