#### Answer

The solution to this system of equations is $(\frac{3}{4}, -\frac{9}{2})$.

#### Work Step by Step

Let's solve one of the equations for one of the variables, and then use that equation to substitute into the other equation. Let's solve the second equation for $y$:
$2x + y = -3$
Subtract $2x$ from each side of the equation to solve for $y$:
$y = -2x - 3$
We can use this equation to substitute for $y$ in the first equation. Then we can solve for $x$ first:
$2x - 3 = (-2x - 3) + 3$
Use distribution to get rid of the parentheses:
$2x - 3 = -2x - 3 + 3$
Add $2x$ to each side to combine $x$ terms:
$4x - 3 = -3 + 3$
Simplify by adding:
$4x - 3 = 0$
Add $3$ to each side of the equation to isolate constants on the right side of the equation:
$4x = 3$
Divide each side by $4$ to solve for $x$:
$x = \frac{3}{4}$
$2(\frac{3}{4}) + y = -3$
Multiply to simplify:
$\frac{6}{4} + y = -3$
$\frac{3}{2} + y = -3$
Subtract $\frac{3}{2}$ from each side of the equation to solve for $y$:
$y = -3 - \frac{3}{2}$
Change $3$ into an equivalent fraction that has $2$ as its denominator so we can subtract the constants:
$y = -\frac{6}{2} - \frac{3}{2}$
Subtract the fractions to solve for $y$:
$y = -\frac{9}{2}$
The solution to this system of equations is $(\frac{3}{4}, -\frac{9}{2})$.