Answer:
To solve the system of equations,
[tex]\begin{gathered} 5x-y=0 \\ \frac{y^2}{90}-\frac{x^2}{36}=1 \end{gathered}[/tex]Solving 1st equation we get,
[tex]y=5x[/tex][tex]\frac{y^2}{90}-\frac{x^2}{36}=1[/tex]Substitute y=5x in the above equation, we get
[tex]\frac{(5x)^2}{90}-\frac{x^2}{36}=1[/tex][tex]\frac{25x^2}{90}-\frac{x^2}{36}=1[/tex][tex]\frac{5x^2}{18}-\frac{x^2}{36}=1[/tex][tex]\frac{10x^2-x^2}{36}=1[/tex][tex]\frac{9x^2}{36}=1[/tex][tex]\frac{x^2}{4}=1[/tex][tex]x^2=4[/tex][tex]x=\pm2[/tex]when x=2, we get y=5x=5(2)=10
when x=-2, we get y=5x=5(-2)=-10
There are two solution for the given system.
[tex](2,10),(2,-10)[/tex]Answer is: x=2,y=10 and x=2,y=-10
