Answer
The solution to the system is the ordered pair $(-\frac{15}{7},-\frac{5}{7})$.
Work Step by Step
Equation 1: $-x-\frac{y}{2}=\frac{5}{2}$
Equation 2: $\frac{x}{6}-\frac{y}{2}=0$
Multiply equation 1 by $-1$:
$$[-x-\frac{y}{2}=\frac{5}{2}]\cdot-1$$ $$x+\frac{y}{2}=-\frac{5}{2}$$
Add this equation to equation 2:
$$x+\frac{y}{2}=-\frac{5}{2}$$ $$+$$ $$\frac{x}{6}-\frac{y}{2}=0$$ $$=$$ $$\frac{7x}{6}=-\frac{5}{2}$$
Multiply both sides by $6$:
$$(\frac{7x}{6})\cdot6=(-\frac{5}{2})\cdot6$$ $$7x=-15$$
Divide both sides by $7$:
$$\frac{7x}{7}=\frac{-15}{7}$$ $$x=-\frac{15}{7}$$
Substitute this value of $x$ to equation 1:
$$-x-\frac{y}{2}=\frac{5}{2}$$ $$-(-\frac{15}{7})-\frac{y}{2}=\frac{5}{2}$$ $$\frac{15}{7}-\frac{y}{2}=\frac{5}{2}$$
Subtract $\frac{15}{7}$ from both sides:
$$\frac{15}{7}-\frac{15}{7}-\frac{y}{2}=\frac{5}{2}-\frac{15}{7}$$ $$-\frac{y}{2}=\frac{5}{2}-\frac{15}{7}$$
Multiply both sides by $14$:
$$(-\frac{y}{2})\cdot14=(\frac{5}{2}-\frac{15}{7})\cdot14$$ $$-7y=35-30$$ $$-7y=5$$
Divide both sides by $-7$:
$$\frac{-7y}{-7}=\frac{5}{-7}$$ $$y=-\frac{5}{7}$$
Therefore, the solution to the system is the ordered pair $(-\frac{15}{7},-\frac{5}{7})$.