Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 1 - Section 1.5 - Problem Solving and Using Formulas - Exercise Set - Page 69: 95

Answer

$C=\frac{-VL+SN}{N-L}$

Work Step by Step

$V=C - \frac{C-S}{L}N$ In solving for $C$, it must be isolated on one side. Step 1. Add $\frac{C-S}{L}N$ on both sides: $V+\frac{C-S}{L}N=C - \frac{C-S}{L}N +\frac{C-S}{L}N$ $V+\frac{C-S}{L}N=C$ Step 2. Subtract $V$ on both sides. $V-V+\frac{C-S}{L}N=C-V$ $\frac{C-S}{L}N=C-V$ Step 3. Multiply the whole equation by $L$. $(\frac{C-S}{L}N)(L)=(C-V)L$ $(C-S)N =(C-V)L$ Step 4. Expand the equations. $(C-S)N$ = $CN-SN$ $(C-V)L$ = $CL-VL$ Step 5. Rewrite the equation from step 3. $CN-SN$= $CL-VL$ Step 6. Subtract $CL$ from both sides. $CN-SN-CL$= $CL-CL-VL$ $CN-SN-CL$= $-VL$ Step 7. Add $SN$ to both sides. $CN-SN+SN-CL$= $-VL+SN$ $CN-CL=-VL+SN$ Step 8. Use the distributive property to convert the two occurrences of $C$ into one. $C(N-L)=-VL+SN$ Step 9. Divide the whole equation by $N-L$. $\frac{C(N-L)}{N-L}=\frac{-VL+SN}{N-L}$ $C=\frac{-VL+SN}{N-L}$
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