Question

If $y\ne 4a$, and $x=\frac{y+a^2}{y-4a}$, find the value of $y$ in terms of $a$ and $x$.

• A. $\frac{4a-4a^{2}x}{x+1}$
• B. $\frac{a^2-4ax}{x+1}$
• C. $\frac{a^2+4ax}{x+1}$
• D. $\frac{a^2+4ax}{x-1}$
• E. $\frac{a^2-4ax}{x-1}$

Answer 1

The correct option is D $\frac{a^2+4ax}{x-1}$

Given, $x=\frac{y+a^2}{y-4a}$

$x(y-4a)=y+a^2\Rightarrow xy-4ax=y+a^2\Rightarrow xy-y=a^2+4ax\Rightarrow y\ast (x-1)=a^2+4ax\Rightarrow y=\frac{a^2+4ax}{x-1}$