1. Equations with Variables on Both Sides
ALGEBRA 1 LESSON 2-2
(For help, go to Lessons 1-7 and 2-3.)
Simplify.
1. 6x – 2x 2. 2x – 6x 3. 5x – 5x 4. –5x + 5x
Solve each equation.
5. 4x + 3 = – –x + 7 = 12
7. 2t – 8t + 1 = = –7n + 4 – 5n
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2. Equations with Variables on Both Sides
ALGEBRA 1 LESSON 2-2
Solutions
1. 6x – 2x = (6 – 2)x = 4x 2. 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 = = –7n + 4 – 5n
–6t + 1 = = –12n + 4
–6t = n = 4
t = –7 n = 1 3
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3. Equations with Variables on Both Sides
ALGEBRA 1 LESSON 2-2
Solve each equation.
a. –6z + 8 = z + 10 – 7z
–6z + 8 = z + 10 – 7z
–6z + 8 = –6z + 10 Combine like terms.
–6z z = –6z z Add 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 + 4 Use the Distributive Property.
4 – 4y + 4y = –4y y Add 4y to each side.
4 = 4 Always true!
The equation is true for every value of y, so the equation is an identity.
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4. Equations with Variables on Both Sides
ALGEBRA 1 LESSON 2-2
Solve each equation.
1. 3 – 2t = 7t n = 2(n + 1) + 3(n – 1)
3. 3(1 – 2x) = 4 – 6x
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? – 1 9 1
no solution
4 deliveries
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