Intermediate Algebra for College Students (7th Edition)

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

Chapter 4 - Cumulative Review Exercises - Page 311: 14

Answer

The rooms with kitchen facilities $=46$. The rooms without kitchen factilities $=14$.

Work Step by Step

Step 1:- Assume unknown quantities as variables. Let the rooms with kitchen facilities $=x$. Let the rooms without kitchen factilities $=y$. Step 2:- Write system of equations. The given values are Total number of rooms $=60$ Total nightly revenue is $= \$5260$ In the equation form $\Rightarrow x+y=60$ ...... (1) $\Rightarrow 90x+80y=5260$ ...... (2) Step 3:- Solve the system of equations. Multiply the equation (1) by $-80$. $\Rightarrow -80x-80y=-4800$ ...... (3) Add equation (2) and (3). $\Rightarrow 90x+80y-80x-80y=5260-4800$ Simplify. $\Rightarrow 10x=460$ Divide both sides by $10$. $\Rightarrow \frac{10x}{10}=\frac{460}{10}$ Simplify. $\Rightarrow x=46$ Plug the value of $y$ into equation (1). $\Rightarrow 46+y=60$ Subtract $46$ from both sides. $\Rightarrow 46+y-46=60-46$ Simplify. $\Rightarrow y=14$ Step 4:- Check the answers. Substitute the values of $x$ and $y$ into equation (2). $\Rightarrow 90(46)+80(14)=5260$ $\Rightarrow 4140+1120=5260$ $\Rightarrow 5260=5260$. True.
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