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(For help, go to Lessons 1-2 and 1-7.) ALGEBRA 1 LESSON 2-3 Simplify each expression. 1.2n – 3n2.–4 + 3b + 2 + 5b 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 = 58.8 + x – 7x for x = –3 9.(8n + 1)3 for n = –210.–(17 + 3y) for y = 6 Solving Multi-Step Equations 2-3

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Solutions 1. 2n – 3n = (2 – 3)n = –1n = –n 2.–4 + 3b + 2 + 5b = (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 7. 28 – a + 4a for a = 5: 28 – 5 + 4(5) = 28 – 5 + 20 = 23 + 20 = 43 8. 8 + x – 7x for x = –3: 8 + (–3) – 7(–3) = 8 + (–3) + 21 = 5 + 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)) = –(17 + 18) = –35 Solving Multi-Step Equations ALGEBRA 1 LESSON 2-3 2-3

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Solve 3a + 6 + a = 90 4a + 6=90Combine like terms. 4a + 6 – 6=90 – 6Subtract 6 from each side. 4a=84Simplify. 3a + 6 + a=90 Check: 3(21) + 6 + 2190Substitute 21 for a. 63 + 6 + 2190 90=90 Solving Multi-Step Equations ALGEBRA 1 LESSON 2-3 2-3 =Divide each side by 4. a=21Simplify. 4a44a4 84 4

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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 + 25 + x = 63 Solving Multi-Step Equations ALGEBRA 1 LESSON 2-3 2-3

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x + 25 + x=63 The pen can be 19 ft long. 2x + 25=63Combine like terms. 2x + 25 – 25=63 – 25Subtract 25 from each side. 2x=38Simplify. x=19 Solving Multi-Step Equations ALGEBRA 1 LESSON 2-3 2-3 (continued) =Divide each side by 2. 2x22x2 38 2

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Solve 2(x – 3) = 8 2x – 6=8Use the Distributive Property. 2x – 6 + 6=8 + 6Add 6 to each side. 2x=14Simplify. x=7Simplify. Solving Multi-Step Equations ALGEBRA 1 LESSON 2-3 2-3 =Divide each side by 2. 2x22x2 14 2

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Solve + = 17 3x23x2 x5x5 x=10Simplify. Solving Multi-Step Equations ALGEBRA 1 LESSON 2-3 2-3 Method 1: Finding common denominators + =17 3x23x2 x5x5 x + x=17Rewrite the equation. 3232 1515 x=17Combine like terms. 17 10 ( x ) = (17)Multiply each each by the reciprocal of, which is. 17 10 10 17 17 10 10 17 x + x=17A common denominator of and is 10. 15 10 2 10 3232 1515

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Solve + = 17 3x23x2 x5x5 15x + 2x=170Multiply. 17x=170Combine like terms. x=10Simplify. Solving Multi-Step Equations ALGEBRA 1 LESSON 2-3 2-3 Method 2: Multiplying to clear fractions + =17 3x23x2 x5x5 10 ( + ) =10(17)Multiply each side by 10, a common multiple of 2 and 5. 3x23x2 x5x5 10 ( ) + 10 ( ) =10(17)Use the Distributive Property. 3x23x2 x5x5 =Divide each side by 17. 17x 17 170 17

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Solve 0.6a + 18.65 = 22.85. 100(0.6a + 18.65)=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 + 1865=2285Simplify. 60a + 1865 – 1865=2285 – 1865Subtract 1865 from each side. 60a=420Simplify. Solving Multi-Step Equations ALGEBRA 1 LESSON 2-3 2-3 =Divide each side by 60. 60a 60 420 60 a=7Simplify.

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Solve each equation. 1. 4a + 3 – a = 242. –3(x – 5) = 66 3. + = 74. 0.05x + 24.65 = 27.5 n3n3 n4n4 7–17 1257 Solving Multi-Step Equations ALGEBRA 1 LESSON 2-3 2-3

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(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 = –56.–x + 7 = 12 7.2t – 8t + 1 = 438.0 = –7n + 4 – 5n Equations with Variables on Both Sides ALGEBRA 1 LESSON 2-4 2-4

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Solutions 1.6x – 2x=(6 – 2)x = 4x2.2x – 6x= (2 – 6)x = –4x 3.5x – 5x=(5 – 5)x = 0x = 04.–5x + 5x=(–5 + 5)x = 0x = 0 5.4x + 3=–56.–x + 7=12 4x=–8–x=5 x=–2x=–5 7.2t – 8t + 1=438.0=–7n + 4 – 5n –6t + 1=430=–12n + 4 –6t=4212n=4 t=–7n= 1313 Equations with Variables on Both Sides ALGEBRA 1 LESSON 2-4 2-4

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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 + 12Vertical angles are congruent. 5x – 3 – 2x=2x + 12 – 2xSubtract 2x from each side. 3x – 3=12Combine like terms. 3x – 3 + 3=12 + 3Add 3 to each side. 3x=15Simplify. Equations with Variables on Both Sides ALGEBRA 1 LESSON 2-4 2-4 =Divide each side by 3. 3x33x3 15 3 x=5Simplify.

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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: 60 + 1.5 h = 5.5 h Equations with Variables on Both Sides ALGEBRA 1 LESSON 2-4 2-4

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60 + 1.5h=5.5h 60 + 1.5h – 1.5h=5.5h – 1.5hSubtract 1.5h from each side. 60=4hCombine like terms. You must use your skateboard for more than 15 hours to justify buying the skateboard. Equations with Variables on Both Sides ALGEBRA 1 LESSON 2-4 2-4 (continued) 60 4 4h44h4 =Divide each side by 4. 15=hSimplify.

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Solve each equation. a.–6z + 8=z + 10 – 7z –6z + 8=z + 10 – 7z –6z + 8=–6z + 10Combine like terms. –6z + 8 + 6z=–6z + 10 + 6zAdd 6z to each side. 8=10Not true for any value of z! This equation has no solution Equations with Variables on Both Sides ALGEBRA 1 LESSON 2-4 b.4 – 4y=–2(2y – 2) The equation is true for every value of y, so the equation is an identity. 4 – 4y=–2(2y – 2) 4 – 4y=–4y + 4Use the Distributive Property. 4 – 4y + 4y=–4y + 4 + 4yAdd 4y to each side. 4=4Always true! 2-4

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Solve each equation. 1.3 – 2t = 7t + 42. 4n = 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? 4 deliveries 1 no solution Equations with Variables on Both Sides ALGEBRA 1 LESSON 2-4 2-4 – 1919

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