Answer
$(x,y)= (-6,-7)$ (consistent independent system)
Work Step by Step
Given $$\begin{cases}
\frac{2}{3}x+\frac{1}{2}y &= -\frac{15}{2}\\
\frac{3}{4}x-y&= \frac{5}{2}\\
\end{cases}$$ Lets first get rid of the fraction in the first equation to make things easier.
$$\begin{aligned}
(2)\cdot(3)\cdot\left(\frac{2}{3}x+\frac{1}{2}y\right) &= -\frac{15}{2}\cdot(2)\cdot(3)\\
4x+3y&=-45\\
\end{aligned}$$ Now solve for $y$ in the second equation: $$\begin{aligned}
\frac{3}{4}x-y&= \frac{5}{2}\\
-y&= -\frac{3}{4}x+\frac{5}{2}\\
y&=\frac{3}{4}x-\frac{5}{2}\\
&=0.75x-2.5\\
\end{aligned}$$ Substitute the result into the above equation.
$$\begin{aligned}
4x+3\cdot\left(0.75x-2.5 \right)&=-45\\
4x+2.25x-7.5&=-45\\
6.25x=7.5-45\\
x&=-\frac{37.5}{6.25}\\
x&= -6
\end{aligned}$$ and $$\begin{aligned}
y&=0.75\cdot(-6)-2.5\\
&= -7.
\end{aligned}$$ Check $$\begin{aligned}
\frac{3}{4}\cdot(-6)-(-7)&\stackrel{?}{=} \frac{5}{2}\\
-\frac{9}{2}+7&\stackrel{?}{=} \frac{5}{2}\\
\frac{14-9}{2}&\stackrel{?}{=}\frac{5}{2}\\
\frac{5}{2}&= \frac{5}{2}\checkmark\\
\frac{2}{3}(-6)+\frac{1}{2}(-7) &\stackrel{?}{=} -\frac{15}{2}\\
-\frac{15}{2}&=-\frac{15}{2}\checkmark.
\end{aligned}$$ The solution set is $(x,y)= (-6,-7)$ . The system is consistent and independent because it has a single solution.
