Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 3 Linear Systems and Matrices - 3.1 Solve Linear Systems by Graphing - 3.1 Exercises - Problem Solving - Page 157: 37

Answer

$11\text{ days}$

Work Step by Step

Let's note: $x$=the number of days $y$=the total cost We build the system of equations: $$\begin{cases} y=x+121\\ y=12x. \end{cases}$$ In order to find the value of $x$ so that the costs are equal, we graph the lines $y=x+121$ and $y=12x$ (see graph). The intersection point appears to be $(11,132)$. We check the solution algebraically: $$\begin{align*} y&=x+121\\ 132&\stackrel{?}{=}11+121\\ 132&=132\checkmark\\\\ y&=12x\\ 132&\stackrel{?}{=}12(11)\\ 132&=132\checkmark. \end{align*}$$ The number of days after which the costs are equal is $x=11$. If the daily cost for Option B increases to $12+a$, where $a>0$, the system of equations becomes: $$\begin{cases} y=x+121\\ y=(12+a)x. \end{cases}$$
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