Answer:
[tex]x = 0[/tex] and [tex]x = \textrm{negative one fifth}[/tex]
Step-by-step explanation:
Given:
The rational equation to solve is given as:
[tex]\frac{x}{6}=\frac{x^2}{x-1}
[/tex]
We use the cross multiplication rule and multiply the numerators one side with the denominators of the other side. Therefore,
[tex]x\times (x-1)=6\times x^2\\x^2-x=6x^2\\\textrm{Bringing all variables to the right side, we get:}\\6x^2-x^2+x=0\\5x^2+x=0\\x(5x+1)=0\\x=0\ or\ 5x+1=0\\x=0\ or\ x=-\frac{1}{5}
[/tex]
Now, for [tex]x=0[/tex], the rational equation is equal to 0. So, [tex]x=0[/tex] is a solution.
Also, [tex]x=-\frac{1}{5}[/tex] is also a solution as the equation's denominator is defined at that point.
So, no extraneous solution.
