#### Answer

The solution is $(0 , 5)$.

#### Work Step by Step

For this system of equations, we can use the addition method to solve for $x$ and $y$.
We first need to transform the equations so that the coefficients of one of the variables differs only in sign. We see that if we multiply the first equation by $3$ and the second equation by $2$, we get coefficients of $6$ and $-6$ for the $y$ term:
$$3(3x + 2y) = 3(10)$$
$$2(4x - 3y) = 2(-15)$$
Distribute first:
$$9x + 6y = 30$$
$$8x - 6y = -30$$
Now we add the two equations together to get the following solution. The $y$ term cancels out:
$$x = 0$$
We plug in $0$ for $x$ into the first equation to solve for $y$:
$$3(0) + 2y = 10$$
Multiply first:
$$2y = 10$$
Divide both sides by $2$ to solve for $y$:
$$y = 5$$
The solution is $(0 , 5)$.