Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Appendix A - Equations and Inequalities - Exercises - Page 416: 12

Answer

{$\frac{5}{12}$}

Work Step by Step

Step 1: $\frac{7}{4}+\frac{1}{5}x-\frac{3}{2}=\frac{4}{5}x$ Step 2: Multiplying both sides by 20, the LCD of the fractions: $20(\frac{7}{4}+\frac{1}{5}x-\frac{3}{2})=20(\frac{4}{5}x)$ Step 3: $20(\frac{7}{4})+20(\frac{1}{5}x)-20(\frac{3}{2})=20(\frac{4}{5}x)$ Step 4: $35+4x-30=16x$ Step 5: $4x+5=16x$ Step 6: Subtracting $16x$ from both sides, $4x+5-16x=16x-16x$ Step 7: $-12x+5=0$ Step 8: Subtracting $5$ from both sides, $-12x+5-5=0-5$ Step 9: $-12x=-5$ Step 10: Dividing both sides by $-12$, $\frac{-12x}{-12}=\frac{-5}{-12}$ Step 11: $x=\frac{5}{12}$ Check: Step 1: Substituting the answer in the initial equation, $\frac{7}{4}+\frac{1}{5}(\frac{5}{12})-\frac{3}{2}=\frac{4}{5}(\frac{5}{12})$ Step 2: $\frac{7}{4}+(\frac{1}{12})-\frac{3}{2}=(\frac{4}{12})$ Step 3: $\frac{7(3)+1-6(3)}{12}=\frac{1}{3}$ Step 4: $\frac{21+1-18}{12}=\frac{1}{3}$ Step 5: $\frac{22-18}{12}=\frac{1}{3}$ Step 6: $\frac{4}{12}=\frac{1}{3}$ Step 7: $\frac{1}{3}=\frac{1}{3}$ Since the solution checks, the solution set is {$\frac{5}{12}$}
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