Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 2 - Review Exercises - Page 209: 4

Answer

$x\text{-intercept: } \left( \dfrac{28}{5},0 \right) \\y\text{-intercept: } (0,4)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the definition of intercepts to find the $x$- and $y$-intercepts of the given equation, $ 5x+7y=28 .$ Then plot the points corresponding to the intercepts and connect these with a line to get the graph. $\bf{\text{Solution Details:}}$ The $x$-intercept is the value of $x$ when $y=0.$ Substituting $y=0$ in the given equation, then \begin{array}{l}\require{cancel} 5x+7y=28 \\\\ 5x+7(0)=28 \\\\ 5x+0=28 \\\\ 5x=28 \\\\ x=\dfrac{28}{5} .\end{array} Hence, the $x$-intercept is $ \left( \dfrac{28}{5},0 \right) .$ The $y$-intercept is the value of $y$ when $x=0.$ Substituting $x=0$ in the given equation, then \begin{array}{l}\require{cancel} 5(0)+7y=28 \\\\ 0+7y=28 \\\\ 7y=28 \\\\ y=\dfrac{28}{7} \\\\ y=4 .\end{array} Hence, the $y$-intercept is $ (0,4) .$ Connecting the following intercepts, \begin{array}{l}\require{cancel} x\text{-intercept: } \left( \dfrac{28}{5},0 \right) \\y\text{-intercept: } (0,4) ,\end{array} gives the graph of the given equation.
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