(8) Lesson 2.5 - Solve Multi-Step Equations

(8) Lesson 2.5 - Solve Multi-Step Equations

1. Course 3, Lesson 2-5 Solve each equation. Check your solution. 1. 8 b – 12 = 5 b 2. 5 c + 24 = c 3. 3 x + 2 = 2 x – 3 4. 4 n – 3 = 2 n + 7 5. Todd is trying to decide between two jobs. Job A pays $400 per week plus a 20% commission on everything sold. Job B pays $500 per week plus a 15% commission on everything sold. How much would Todd have to sell each week for both jobs to pay the same? Write an equation and solve.

2. Course 3, Lesson 2-5 ANSWERS 1. 4 2 . − 6 3. − 5 4 . 5 5. 400 + 0.20 x = 500 + 0.15 x ; $ 2,000

3. WHAT is equivalence? Expressions and Equations Course 3, Lesson 2-5

4. 8.EE.7 Solve linear equations in one variable. 8.EE.7a Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. Show which of these possibilities is the case by successively transforming the given equation into simpler forms, until an equivalent equation of the form x = a , a = a , or a = b results (where a and b are different numbers). 8.EE.7b Solve linear equations with rational numbers coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms . Course 3, Lesson 2-5 Common Core State Standards © Copyright 2010. National Governors Association Center for Best Practices and Council of Chief State School Officers. All rights reserved. Expressions and Equations

5. Mathematical Practices 1 Make sense of problems and persevere in solving them. 2 Reason abstractly and quantitatively. 3 Construct viable arguments and critique the reasoning of others. 4 Model with mathematics. Course 3, Lesson 2-5 Common Core State Standards © Copyright 2010. National Governors Association Center for Best Practices and Council of Chief State School Officers. All rights reserved. Expressions and Equations

6. To solve multi-step equations, equations with no solutions, equations with an infinite number of solutions Course 3, Lesson 2-5 Expressions and Equations

7. Symbol null set Ø empty set { } identity Course 3, Lesson 2-5 Expressions and Equations

8. Number of Solutions Course 3, Lesson 2-5 Expressions and Equations Null Set One Solution Identify Words no solution one solution infinitely many solutions Symbols a = b x = a a = a Example 3 x + 4 = 3 x 2 x = 20 4 x + 2 = 4 x + 2 4 = 0 x = 10 2 = 2 Since 4 ≠ 0, Since 2 = 2, the there is no solution is all solution. numbers.

9. Need Another Example? Step-by-Step Example 1. Solve 15(20 + d ) = 420. 1 2 3 4 15(20 + d ) = 420 Write the equation. Subtraction Property of Equality Simplify. 300 + 15 d = 420 Distributive Property Simplify. 15 d = 120 Division Property of Equality d = 8 – 300 = – 300

10. Answer Need Another Example? Solve –3(4 p – 6) = 54. –3

11. 1 Need Another Example? 2 3 4 5 6 Step-by-Step Example 2. Solve 6( x – 3) + 10 = 2(3 x – 4). 6( x – 3) + 10 = 2(3 x – 4) Write the equation. Collect like terms. Simplify. 6 x – 18 + 10 = 6 x – 8 Distributive Property 6 x – 8 = 6 x – 8 Addition Property of Equality 6 x = 6 x Simplify. Division Property of Equality x = x The statement x = x is always true. The equation is an identity and the solution set is all numbers. Write the original equation. Check 6( x – 3) + 10 = 2(3 x – 4) Substitute any value for x . 6( 5 – 3) + 10 = 2[3 (5) – 4] Simplify. 6(2) + 10 = 2(15 – 4) 22 = 22 ? ? + 8 = + 8

12. Answer Need Another Example? Solve 3(4 x + 8) = 2(6 x + 12). identity; all numbers

13. 1 Need Another Example? 2 3 4 5 Step-by-Step Example 3. Solve 8(4 – 2 x ) = 4(3 – 5 x ) + 4 x . 8(4 – 2 x ) = 4(3 – 5 x ) + 4 x Write the equation. Collect like terms. Simplify. 32 – 16 x = 12 – 20 x + 4 x Distributive Property 32 – 16 x = 12 – 16 x Addition Property of Equality 32 = 12 The statement 32 = 12 is never true. The equation has no solution and the solution set is . Write the equation. Check 8(4 – 2 x ) = 4(3 – 5 x ) + 4 x Substitute any value for x . 8[4 – 2 (2) ] = 4[3 – 5 (2) ] + 4 (2) Simplify. 0 ≠ –20 ? 8(0) = 4(–7) + 8 ? Since 0 ≠ –20, the equation has no solution. + 16 x = + 16 x

14. Answer Need Another Example? Solve 4(5 x + 3) – 6 x = 7(2 x + 3). null set; no solution

15. 1 Need Another Example? 2 3 4 5 6 Step-by-Step Example 4. At the fair, Hunter bought 3 snacks and 10 ride tickets. Each ride ticket costs $1.50 less than a snack. If he spent a total of $24.00, what was the cost of each snack? Write an equation to represent the problem. Write the equation. Simplify. 3 s + 10( s – 1.5) = 24 Addition Property of Equality 13 s = 39 3 s + 10 s – 15 = 24 Distributive Property 13 s – 15 = 24 Collect like terms. Division Property of Equality Simplify. s = 3 So, the cost of each snack was $3. + 15 = + 15

16. Answer Need Another Example? The length of Philip’s stride when walking is 4 inches greater than the length of Anne’s stride. If it takes Philip 5 steps and Anne 6 steps to walk the same distance, what is the length of Anne’s stride? 20 in.

17. How did what you learned today help you answer the WHAT is equivalence? Course 3, Lesson 2-5 Expressions and Equations

18. How did what you learned today help you answer the WHAT is equivalence? Course 3, Lesson 2-5 Expressions and Equations Sample answers: When the expressions on each side of the equals sign are the same, the equation is an identity and the solution is all real numbers. When the final step in solving an equation produces expressions that are not the same, the solution to the equation is the null set.

19. Describe how the previous lesson on solving equations with variables on each side helped you with today’s lesson on solving multi-step equations. Ratios and Proportional Relationships Expressions and Equations Course 3, Lesson 2-5