# How do you solve the system of equations -\frac { 3} { 4} x + y = - 1 and - 3x + \frac { 1} { 2} y = 10?

$x = - 4$, and $y = - 1 \frac{3}{16}$
I'm going to use a method called substitution. First, add $\frac{3}{4} x$ to both sides of the first equation. This leaves you with $y = \frac{3}{4} x - 1$. Substitute this value of $y$ into the second equation to get $- 3 x + \frac{1}{2} \left(\frac{3}{4} x - 1\right) = 10$. Simplify this equation to get $- 3 x + \frac{3}{8} x - \frac{1}{2} = 10$. Combine like terms and simplify to get $- \frac{21}{8} x = \frac{21}{2}$. Divide on both sides by $- 21$ to get $\frac{1}{8} x = - \frac{1}{2}$. Multiply by 8 on both sides to get $x = - 4$. Substitute this value of $x$ into our equation of $y = \frac{3}{4} x - 1$ to get $y = - \frac{3}{16} - 1 = - 1 \frac{3}{16}$.