## Algebra and Trigonometry 10th Edition

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