Intermediate Algebra for College Students (7th Edition)

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

Chapter 3 - Cumulative Review Exercises - Page 251: 17

Answer

The cost of one pad $=\$0.80$. The cost of one pen $=\$0.20$.

Work Step by Step

Step 1:- Assume unknown quantities as variables. Let the cost of one pad $=x$. Let the cost of one pen $=y$. Step 2:- Write system of equations. The given values are Cost of $2$ pads and $19$ pens equals $\$5.40$. and cost of $7$ pads and $4$ pens equals $\$6.40$. In the equation form. $\Rightarrow 2x+19y=5.40$ ...... (1) $\Rightarrow 7x+4y=6.40$ ...... (2) Step 3:- Solve the system of equations. Multiply the equation (1) by $7$ and equation (2) by $-2$. $\Rightarrow 14x+133y= 37.80$ ...... (3) $\Rightarrow -14x-8y=-12.80$ ...... (4) Add equation (3) and (4). $\Rightarrow 14x+133y-14x-8y=37.80-12.80$ Simplify. $\Rightarrow 125y=25$ Divide both sides by $125$. $\Rightarrow \frac{125y}{125}=\frac{25}{125}$ Simplify. $\Rightarrow y=0.20$ Plug the value of $y$ into equation (1). $\Rightarrow 2x+19(0.20)=5.40$ Simplify. $\Rightarrow 2x+3.80=5.40$ Isolate $x$. $\Rightarrow x=\frac{5.40-3.80}{2}$ Simplify. $\Rightarrow x=0.80$. Step 4:- Check the answers. Substitute the values of $x$ and $y$ into equation (2). $\Rightarrow 7(0.80)+4(0.20)=6.40$ $\Rightarrow 5.60+0.80=6.40$ $\Rightarrow 6.40=6.40$. True.
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