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Solving Multi-Step Equations

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sdawn
17 diapositives · 233 Views · Jan 04, 2020

Learn to solve algebraic equations involving multiple steps like combining like terms and solving for variables on both sides, with detailed step-by-step solutions. Practice simplifying expressions, evaluating given expressions, and solving equations with variables on both sides. Enhance your algebra skills with this comprehensive lesson.

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Solving Multi-Step Equations
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  1. ALGEBRA 1 LESSON 2-3 Solving Multi-Step Equations (For help, go to Lessons 1-2 and 1-7.) 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 = 5 8. 8 + x – 7x for x = –3 9. (8n + 1)3 for n = –2 10. –(17 + 3y) for y = 6 2-3

  2. ALGEBRA 1 LESSON 2-3 Solving Multi-Step Equations 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 2-3

  3. 4a4 844 = Divide each side by 4. a = 21Simplify. 3a+ 6 + a = 90 Check: 3(21)+ 6 + 21 90 Substitute 21 for a. 63+ 6 + 21 90 90 = 90 ALGEBRA 1 LESSON 2-3 Solving Multi-Step Equations Solve 3a + 6 + a = 90 4a+ 6 = 90Combine like terms. 4a + 6 – 6 = 90 – 6Subtract 6 from each side. 4a = 84Simplify. 2-3

  4. 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 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? 2-3

  5. 2x+ 25 = 63 Combine like terms. 2x+ 25 – 25 = 63 – 25Subtract 25 from each side. 2x2 382 = Divide each side by 2. ALGEBRA 1 LESSON 2-3 Solving Multi-Step Equations (continued) x+ 25 + x = 63 2x = 38 Simplify. x = 19 The pen can be 19 ft long. 2-3

  6. 2x2 142 = Divide each side by 2. ALGEBRA 1 LESSON 2-3 Solving Multi-Step Equations Solve 2(x – 3) = 8 2x– 6 = 8 Use the Distributive Property. 2x– 6 + 6 = 8 + 6Add 6 to each side. 2x = 14 Simplify. x = 7 Simplify. 2-3

  7. Method 1: Finding common denominators 3x2 x5 + = 17 32 15 x+ x = 17 Rewrite the equation. 32 15 1510 210 x+ x = 17 A common denominator of and is 10. 1710 x = 17 Combine like terms. (x) = (17) Multiply each each by the reciprocal of , which is . 1017 1710 1017 1710 1017 ALGEBRA 1 LESSON 2-3 Solving Multi-Step Equations Solve + = 17 3x2 x5 x = 10 Simplify. 2-3

  8. Method 2: Multiplying to clear fractions 3x2 x5 + = 17 10(+ ) = 10(17) Multiply each side by 10, a common multiple of 2 and 5. 3x2 x5 10() + 10( ) = 10(17) Use the Distributive Property. 3x2 x5 17x17 17017 = Divide each side by 17. ALGEBRA 1 LESSON 2-3 Solving Multi-Step Equations Solve + = 17 3x2 x5 15x + 2x = 170 Multiply. 17x = 170 Combine like terms. x = 10 Simplify. 2-3

  9. 60a60 42060 = Divide each side by 60. a = 7 Simplify. ALGEBRA 1 LESSON 2-3 Solving Multi-Step Equations 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 = 2285 Simplify. 60a + 1865 – 1865 = 2285 – 1865Subtract 1865 from each side. 60a = 420 Simplify. 2-3

  10. ALGEBRA 1 LESSON 2-3 Solving Multi-Step Equations Solve each equation. 1. 4a + 3 – a = 24 2. –3(x – 5) = 66 3. + = 7 4. 0.05x + 24.65 = 27.5 7 –17 n3 n4 12 57 2-3

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

  12. ALGEBRA 1 LESSON 2-4 Equations with Variables on Both Sides Solutions 1. 6x – 2x = (6 – 2)x = 4x2. 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 = 43 8. 0 = –7n + 4 – 5n –6t + 1 = 43 0 = –12n + 4 –6t = 42 12n = 4 t = –7 n = 1 3 2-4

  13. 5x – 3 – 2x = 2x + 12 – 2xSubtract 2x from each side. 3x – 3 = 12 Combine like terms. 3x – 3+ 3 = 12 + 3Add 3 to each side. 3x = 15 Simplify. = Divide each side by 3. 3x3 153 x = 5 Simplify. 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. 2-4

  14. 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 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? 2-4

  15. 60 + 1.5h – 1.5h = 5.5h – 1.5h Subtract 1.5h from each side. 60 = 4hCombine like terms. 604 4h4 = Divide each side by 4. 15 = hSimplify. ALGEBRA 1 LESSON 2-4 Equations with Variables on Both Sides (continued) 60 + 1.5h = 5.5h You must use your skateboard for more than 15 hours to justify buying the skateboard. 2-4

  16. 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 + 8 + 6z = –6z + 10 + 6zAdd 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 + 4Use the Distributive Property. 4 – 4y + 4y = –4y + 4 + 4yAdd 4y to each side. 4 = 4 Always true! The equation is true for every value of y, so the equation is an identity. 2-4

  17. 1 9 – ALGEBRA 1 LESSON 2-4 Equations with Variables on Both Sides Solve each equation. 1. 3 – 2t = 7t + 4 2. 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? 1 no solution 4 deliveries 2-4