Answer
$x_1=7.5$ 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
