Answer
$\left\{\left(-\dfrac{5}{3},1\right)\right\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
2x+4y=\dfrac{2}{3}\\
3x-5y=-10
\end{cases}$
Use the elimination method. Multiply the first equation by 5, multiply the second equation by 4, and add them to eliminate $y$ and determine $x$:
$\begin{cases}
5(2x+4y)=5\left(\dfrac{2}{3}\right)\\
4(3x-5y)=4(-10)
\end{cases}$
$\begin{cases}
10x+20y=\dfrac{10}{3}\\
12x-20y=-40
\end{cases}$
$10x+20y+12x-20y=\dfrac{10}{3}-40$
$22x=-\dfrac{110}{3}$
$x=-\dfrac{110}{66}$
$x=-\dfrac{5}{3}$
Determine $y$ using the second equation:
$3x-5y=-10$
$3\left(-\dfrac{5}{3}\right)-5y=-10$
$-5-5y=-10$
$5y=5$
$y=1$
The solution set of the system is:
$\left\{\left(-\dfrac{5}{3},1\right)\right\}$
