Answer:
[tex]x=\pm\sqrt{3}[/tex] and they are actual solutions
Step-by-step explanation:
we have
[tex]\frac{x^2}{2x-6}=\frac{9}{6x-18}[/tex]
Factor the denominators both sides
[tex]\frac{x^2}{2(x-3)}=\frac{9}{6(x-3)}[/tex]
Simplify
[tex]\frac{x^2}{2}=\frac{9}{6}[/tex]
[tex]x^2=\frac{18}{6}[/tex]
[tex]x=\pm\sqrt{3}[/tex]
Verify
1) For [tex]x=\sqrt{3}[/tex]
[tex]\frac{\sqrt{3}^2}{2(\sqrt{3}-3)}=\frac{9}{6(\sqrt{3}-3)}[/tex]
[tex]\frac{3}{2(\sqrt{3}-3)}=\frac{9}{6(\sqrt{3}-3)}[/tex]
[tex]18=18[/tex] ---> is true
therefore
[tex]x=\sqrt{3}[/tex] ----> is an actual solution
2) For [tex]x=-\sqrt{3}[/tex]
[tex]\frac{-\sqrt{3}^2}{2(-\sqrt{3}-3)}=\frac{9}{6(-\sqrt{3}-3)}[/tex]
[tex]\frac{3}{2(-\sqrt{3}-3)}=\frac{9}{6(-\sqrt{3}-3)}[/tex]
[tex]18=18[/tex] ---> is true
therefore
[tex]x=-\sqrt{3}[/tex] ----> is an actual solution
therefore
