Question

If $x+y=2z$, then the value of $\frac{x}{x-z}+\frac{z}{y-z}$ is:

• A. $1$
• B. $3$
• C. $\frac{1}{2}$
• D. $2$

Answer 1

Answer: (A)

$1$

Given: $x+y=2z$

$\frac{x}{x-z}+\frac{z}{y-z}$

$=\frac{x(y-z)+z(x-z)}{(x-z)(y-z)}$

$=\frac{xy-xz+zx-z^2}{xy-xz-zy+z^2}$

$=\frac{xy-z^2}{xy-z(x+y)+z^2}$

$=\frac{xy-z^2}{xy-z\left(2z\right)+z^2}$

$=\frac{xy-z^2}{xy-z^2}=1$

$\therefore \frac{x}{x-z}+\frac{z}{y-z}=1$.