Answer
The solution to this system of equations is $\left(-\dfrac{4}{7}, \dfrac{1}{14}\right)$.
Work Step by Step
First, we rewrite the equations so that the variables are on one side while the constant is on the other:
$-3x + 4y = 2$
$ 5x - 2y =-3$
We see that if we multiply the second equation by $2$, then the $y$ terms will have opposite numerical coefficients. If we add these two equations together, we can eliminate the variable $y$ and just deal with one variable instead of two. Let us multiply the second equation by $2$ to obtain the equivalent system:
$-3x + 4y = 2$
$10x - 4y = -6$
Add the two equations together to come up with one single equation:
$(-3x+4y)+(10x-4y)=2+(-6)\\
7x = -4$
Divide each side by $7$ to solve for $x$:
$x = -\dfrac{4}{7}$
Now that we have the value for $x$, we can plug it into one of the equations to solve for $y$.
Let us plug in the value for $x$ into the first equation:
$2-4y-3x\\
2 = 4y - 3\left(-\frac{4}{7}\right)$
$2 = 4y + \frac{12}{7}$
Subtract $\dfrac{12}{7}$ from both sides of the equation to isolate constants to the right side of the equation:
$2 - \dfrac{12}{7}=4y$
Convert $2$ into a fraction with $7$ as its denominator:
$\dfrac{14}{7} - \dfrac{12}{7}=4y$
$ \dfrac{2}{7}=4y$
Multiply $\dfrac{1}{4}$ to both sides of the equation:
$ \dfrac{1}{4} \cdot \dfrac{2}{7}=\dfrac{1}{4} \cdot 4y\\
\dfrac{2}{28}=y$
Divide both numerator and denominator by $2$ to simplify the fraction:
$\dfrac{1}{14}=y$
The solution to this system of equations is $\left(-\dfrac{4}{7}, \dfrac{1}{14}\right)$.
