Answer:
[tex](8,-12)[/tex]
Step-by-step explanation:
The two equations are:
[tex]\frac{1}{2}x+\frac{1}{3}y=0\\\frac{1}{4}x-\frac{1}{2}y=8[/tex]
We can solve the first equation for [tex]x[/tex] and then substitute into second equation to find the value of [tex]y[/tex].
[tex]\frac{1}{2}x+\frac{1}{3}y=0\\\frac{1}{2}x=-\frac{1}{3}y\\x=\frac{-\frac{1}{3}y}{\frac{1}{2}}\\x=-\frac{2}{3}y[/tex]
Now,
[tex]\frac{1}{4}x-\frac{1}{2}y=8\\\frac{1}{4}(-\frac{2}{3}y)-\frac{1}{2}y=8\\-\frac{2}{12}y-\frac{1}{2}y=8\\-\frac{2}{3}y=8\\y=\frac{8}{-\frac{2}{3}}\\y=-12[/tex]
Substituting [tex]y=-12[/tex] into the "solved for [tex]x[/tex]" version of first equation, we get the value of [tex]x[/tex]. So,
[tex]x=-\frac{2}{3}y\\x=-\frac{2}{3}(-12)\\x=8[/tex]
Hence the ordered pair is [tex](8,-12)[/tex] and this is the solution to the system of equations shown.
