## Geometry: Common Core (15th Edition)

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