116 Chapter 2 Linear Equations and Inequalities in One Variable Check: 5x - 3 - 2x = 8x - 1 5a- 2 5 b - 3 - 2a- 2 5 b = 8a- 2 5 b - 1 - 10 5 - 3 + 4 5 = - 16 5 - 1 - 10 5 - 15 5 + 4 5 =- 16 5 - 5 5 - 21 5 =- 21 5 Original equation Replace x with - 2 5 . Multiply. Convert each number to a fraction with the common denominator of 5. Add. Since x=- 2 5 makes the equation true, the solution set is e- 2 5 f . 2d. 8(2x - 1) = 10x + 7 16x - 8 = 10x + 7 Apply the distributive property. 16x - 8 - 10x = 10x + 7 - 10x Subtract 10x from each side. 6x - 8 =7 Simplify. 6x - 8 + 8 = 7 + 8 Add 8 to each side. 6x = 15 Simplify. 6x 6 = 15 6 Divide each side by 6. x = 5 2 Simplify. Recall 15 6 = 5 2 . Check: 8(2x - 1) = 10x + 7 Original equation 8 c2a5 2 b-1d = 10a5 2 b + 7 Replace x with 5 2 . 8(5 - 1) = 25 + 7 Multiply. 8(4) = 32 Subtract. 32 = 32 Simplify. Since x = 5 2 makes the equation true, the solution set is e 5 2 f . Student Check 2 Solve each equation. a. -3x + 2=-7 b. 2y - 6=-9y + 1 c. 7x + 4 + 5x = 14x + 3 d. 6(x - 2) = 9x + 8 Applications As we have already noted, expressing relationships mathematically is a key component to successfully solving word problems. While there may be some problems that we can solve by reasoning, it is important that we be able to justify how we came up with our answer and why we know it is valid. The ability to speak, think, and write mathematically equips us with the tools needed to provide such justification. The more we practice the skills at this level, the easier it is for us to extend them to more difficult concepts. ���������������������� Solving Word Problems Step 1: Read the problem and determine what is unknown. Assign a variable for the unknown quantity. Objective 3 ▶ Solve application problems.
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