Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 1 - 1.4 - Quadratic Equations and Applications - 1.4 Exercises - Page 111: 106


$x_1=7.5$ m, $y_1=\dfrac{70}{3}$ m $x_2=17.5$ m, $y_2=10$ m

Work Step by Step

We can write the system of equations: $\begin{cases} 4x+3y=100\\ (2x)y=350 \end{cases}$ Express $y$ in terms of $x$ from the first equation and use it in the second: $3y=100-4x$ $y=\dfrac{100-4x}{3}$ $2x\cdot \dfrac{100-4x}{3}=350$ $\dfrac{2x(100-4x)}{3}=350$ $2x(100-4x)=3(350)$ $x(100-4x)=3(175)$ $100x-4x^2=525$ $4x^2-100x+525=0$ Solve the equation using the quadratic formula: $x=\dfrac{-(-100)\pm\sqrt{(-100)^2-4(4)(525)}}{2(4)}=\dfrac{100\pm\sqrt{1600}}{8}=\dfrac{100\pm 40}{8}$ $x_1=\dfrac{100-40}{8}=\dfrac{60}{8}=7.5$ $x_2=\dfrac{100+40}{8}=\dfrac{140}{8}=17.5$ Determine $y$: $x_1=7.5\Rightarrow y_1=\dfrac{100-4(7.5)}{3}=\dfrac{70}{3}$ $x_2=17.5\Rightarrow y_2=\dfrac{100-4(17.5)}{3}=\dfrac{30}{3}=10$ There are two solutions: $x_1=7.5$ m, $y_1=\dfrac{70}{3}$ m $x_2=17.5$ m, $y_2=10$ m
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.