Answer:
[tex](4.5,-3)\textrm{ and }(0,6)[/tex]
Step-by-step explanation:
Given:
[tex]x-6=-\frac{1}{6}y^{2}[/tex]
[tex]2x+y=6[/tex]
Express [tex]x[/tex] in terms of [tex]y[/tex] and then plug in first equation to solve for [tex]y[/tex].
[tex]2x+y=6\\2x=6-y\\x=\frac{6-y}{2}[/tex]
Now,
[tex]x-6=-\frac{1}{6}y^{2}\\\frac{6-y}{2}-6=-\frac{1}{6}y^{2}\\\frac{1}{6}y^{2}+\frac{6-y}{2}-6=0[/tex].
Multiply 6 on both sides.
[tex]6(\frac{1}{6}y^{2}+\frac{6-y}{2}-6)=6(0)\\y^{2}+3(6-y)-36=0\\y^{2}+18-3y-36=0\\y^{2}-3y-18=0\\(y-6)(y+3)=0\\y=6\textrm{ or }y=-3[/tex]
Now, for [tex]y=-3,x=\frac{6-(-3)}{2}=\frac{6+3}{2}=\frac{9}{2}=4.5[/tex]
For, [tex]y=6, x=\frac{6-6}{2}=0[/tex].
Therefore, the solutions for the system of equations are [tex](4.5,-3)\textrm{ and }(0,6)[/tex]
