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Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3 (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 2-3
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Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3 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 2-3
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Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3 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 2-3
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Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3 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 2-3
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Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3 (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. 2-3
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Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3 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. 2-3
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Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3 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. 2-3
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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. 2-3
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Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3 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. 2-3
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Solving Multi-Step Equations
ALGEBRA 1 LESSON 2-3 Solve each equation. 1. 4a + 3 – a = –3(x – 5) = 66 = x = 27.5 7 –17 n 3 n 4 12 57 2-3
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