Linear Equations and Inequalities - XYZ Custom Plus

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146Chapter 2 Linear Equations and Inequalities5. Solve 4 − 2(3y + 1) = −16.Example 5Solve 7 − 3(2y + 1) = 16.Solution We begin by multiplying −3 times the sum of 2y and 1:Step 1a: 7 − 6y − 3 = 16 Distributive propertyStep 1c: −6y + 4 = 16 Simplify the left sideStep 2:−6y + 4 + (−4) = 16 + (−4) Add −4 to both sides{ −6y = 12Step 3:{ − 1 _6 ( −6y) = − 1 _6 (12) Multiply both sides by −1 _6 y = −2Step 4: Replacing y with −2 in the original equation yields a true statement.There are two things to notice about the example that follows: first, the distributiveproperty is used to remove parentheses that are preceded by a negativesign, and, second, the addition property and the multiplication property are notshown in as much detail as in the previous examples.6. Solve.4(3x − 2) − (6x − 5) = 6 − (3x + 1)Example 6Solve 3(2x − 5) − (2x − 4) = 6 − (4x + 5).Solution When we apply the distributive property to remove the groupingsymbols and separate terms, we have to be careful with the signs. Remember, wecan think of −(2x − 4) as −1(2x − 4), so that−(2x − 4) = −1(2x − 4) = −2x + 4It is not uncommon for students to make a mistake with this type of simplificationand write the result as −2x − 4, which is incorrect. Here is the complete solutionto our equation:3(2x − 5) − (2x − 4) = 6 − (4x + 5) Original equation6x − 15 − 2x + 4 = 6 − 4x − 5 Distributive property4x − 11 = −4x + 1 Simplify each side8x − 11 = 1Add 4x to each side8x = 12Add 11 to each sidex = _128 Multiply each side by 1 _8 x = _3 Reduce to lowest terms2The solution, _ 3 , checks when replacing x in the original equation.2AnswersGetting Ready for ClassAfter reading through the preceding section, respond in your ownwords and in complete sentences.1. What is the first step in solving a linear equation containingparentheses?2. What is the last step in solving a linear equation?3. Explain in words how you would solve the equation 2x − 3 = 8.4. If an equation contains decimals, what can you do to eliminate thedecimals?5. 3 6. 8 _9
2.4 Problem Set147Problem Set 2.4A Solve each of the following equations using the four steps shown in this section. [Examples 1–2]1. 2(x + 3) = 1232. 3(x − 2) = 643. 6(x − 1) = −18−24. 4(x + 5) = 16−15. 2(4a + 1) = −6−16. 3(2a − 4) = 1247. 14 = 2(5x − 3)28. −25 = 5(3x + 4)−39. −2(3y + 5) = 14−410. −3(2y − 4) = −6311. −5(2a + 4) = 0−212. −3(3a − 6) = 0213. 1 = _1 ( 4 x + 2)2014. 1 = _1 ( 6 x + 3)3015. 3(t − 4) + 5 = −4116. 5(t − 1) + 6 = −9−2A Solve each equation. [Examples 2–6]17. 4(2y + 1) − 7 = 118. 6(3y + 2) − 8 = −2_ 1 2 _− 1 3 19. _1 2 ( x − 3) = 1 _( x + 1)4720. _1 3 ( x − 4) = 1 _( x − 6)21021. −0.7(2x − 7) = 0.3(11 − 4x)822. −0.3(2x − 5) = 0.7(3 − x)623. −2(3y + 1) = 3(1 − 6y) − 924. −5(4y − 3) = 2(1 − 8y) + 11_− 1 3 _ 1 2 25. _3 4 ( 8 x − 4) + 3 = 2 _( 5 x + 10) − 1526. _5 6 ( 6 x + 12) + 1 = 2 _( 9 x − 3) + 53_ 3 4 827. 0.06x + 0.08(100 − x) = 6.57528. 0.05x + 0.07(100 − x) = 6.240Selected exercises available online at www.webassign.net/brookscole
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