1. Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3
Solving Multi-Step Equations
(For help, go to Lessons 1-2 and 1-7.)
Simplify each expression.
1. 2n – 3n 2. –4 + 3b b
3. 9(w – 5) 4. –10(b – 12)
5. 3(–x + 4) 6. 5(6 – w)
Evaluate each expression.
7. 28 – a + 4a for a = x – 7x for x = –3
9. (8n + 1)3 for n = –2 10. –(17 + 3y) for y = 6
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2. Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3
Solving Multi-Step Equations
Solutions
1. 2n – 3n = (2 – 3)n = –1n = –n
2. –4 + 3b b = (3 + 5)b + (–4 + 2) = 8b – 2
3. 9(w – 5) = 9w – 9(5) = 9w – 45
4. –10(b – 12) = –10b – (–10)(12) = –10b + 120
5. 3(–x + 4) = 3(–x) + 3(4) = –3x + 12
6. 5(6 – w) = 5(6) – 5w = 30 – 5w
– a + 4a for a = 5: 28 – 5 + 4(5) = 28 – = = 43
x – 7x for x = –3: 8 + (–3) – 7(–3) = 8 + (–3) + 21 = = 26
9. (8n + 1)3 for n = –2: (8(–2) + 1)3 = (–16 + 1)3 = (–15)3 = –45
10. –(17 + 3y) for y = 6: –(17 + 3(6)) = –( ) = –35
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3. Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3
Solving Multi-Step Equations
Solve 3a a = 90
4a + 6 = 90 Combine like terms.
4a + 6 – 6 = 90 – 6 Subtract 6 from each side.
4a = 84 Simplify.
= Divide each side by 4.
a = 21 Simplify.
4a 4
84 4
3a a = 90
Check:
3(21) Substitute 21 for a.
90 = 90
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4. Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3
Solving Multi-Step Equations
You need to build a rectangular pen in your back yard for your dog. One side of the pen will be against the house. Two sides of the pen have a length of x ft and the width will be 25 ft. What is the greatest length the pen can be if you have 63 ft of fencing?
Relate: length plus 25 ft plus length equals amount of side of side of fencing
Define: Let x = length of a side adjacent to the house.
Write: x x = 63
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5. Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3
Solving Multi-Step Equations
(continued)
x x = 63
2x + 25 = 63 Combine like terms.
2x + 25 – 25 = 63 – 25 Subtract 25 from each side.
2x = 38 Simplify.
= Divide each side by 2.
2x 2
38 2
x = 19
The pen can be 19 ft long.
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6. Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3
Solving Multi-Step Equations
Solve 2(x – 3) = 8
2x – 6 = 8 Use the Distributive Property.
2x – = Add 6 to each side.
2x = 14 Simplify.
= Divide each side by 2.
2x 2
14 2
x = 7 Simplify.
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7. Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3
Solving Multi-Step Equations
Solve = 17
3x 2
x 5
Method 1: Finding common denominators
+ = 17
3x 2
x 5
x + x = 17 Rewrite the equation.
3 2
1 5
x x = 17 A common denominator of and is 10.
15 10
2 10
3 2
1 5
x = 17 Combine like terms.
17 10
( x) = (17) Multiply each each by the reciprocal of , which is
17 10
10 17
x = 10 Simplify.
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8. Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3
Solve = 17
3x 2
x 5
Method 2: Multiplying to clear fractions
+ = 17
3x 2
x 5
10( ) = 10(17) Multiply each side by 10, a common multiple of 2 and 5.
3x 2
x 5
10( ) + 10( ) = 10(17) Use the Distributive Property.
3x 2
x 5
15x + 2x = 170 Multiply.
17x = 170 Combine like terms.
= Divide each side by 17.
17x 17
170 17
x = 10 Simplify.
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9. Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3
Solving Multi-Step Equations
Solve 0.6a =
100(0.6a ) = 100(22.85) The greatest of decimal places is two places. Multiply each side by 100.
100(0.6a) + 100(18.65) = 100(22.85) Use the Distributive Property.
60a = 2285 Simplify.
60a – 1865 = 2285 – 1865 Subtract 1865 from each side.
60a = 420 Simplify.
= Divide each side by 60.
60a 60
420 60
a = 7 Simplify.
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10. Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3
Solving Multi-Step Equations
Solve each equation.
1. 4a + 3 – a = –3(x – 5) = 66
= x = 27.5 7
–17
n 3
n 4 12 57
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11. Equations with Variables on Both Sides
ALGEBRA 1 LESSON 2-4
Equations with Variables on Both Sides
(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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12. Equations with Variables on Both Sides
ALGEBRA 1 LESSON 2-4
Equations with Variables on Both Sides
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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13. Equations with Variables on Both Sides
ALGEBRA 1 LESSON 2-4
Equations with Variables on Both Sides
The 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.
5x – 3 – 2x = 2x + 12 – 2x Subtract 2x from each side.
3x – 3 = 12 Combine like terms.
3x – = Add 3 to each side.
3x = 15 Simplify.
= Divide each side by 3.
3x 3
15 3
x = 5 Simplify.
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14. Equations with Variables on Both Sides
ALGEBRA 1 LESSON 2-4
Equations with Variables on Both Sides
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 rental
skateboard
Define: let h = the number of hours you must skateboard
Write: h = h
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15. Equations with Variables on Both Sides
ALGEBRA 1 LESSON 2-4
Equations with Variables on Both Sides
(continued)
h = 5.5h
h – 1.5h = 5.5h – 1.5h Subtract 1.5h from each side.
60 = 4h Combine like terms.
60 4
4h 4
= Divide each side by 4.
15 = h Simplify.
You must use your skateboard for more than 15 hours to justify buying the skateboard.
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16. Equations with Variables on Both Sides
ALGEBRA 1 LESSON 2-4
Equations with Variables on Both Sides
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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17. Equations with Variables on Both Sides
ALGEBRA 1 LESSON 2-4
Equations with Variables on Both Sides
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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