Solve $\dfrac{5}{(2-x)^2}+\dfrac{1}{x-2}=4$. Answer x = ...

1. Use the fact that the squared denominator is unchanged and state the restriction.

Since $(2-x)^2=(x-2)^2$, the equation becomes
$\frac{5}{(x-2)^2}+\frac{1}{x-2}=4,$
where $x\ne 2$.

2. Clear the fractional denominators to form a quadratic equation.

Multiply throughout by $(x-2)^2$:
$5+(x-2)=4(x-2)^2.$
Let $y=x-2$. Then
$5+y=4y^2.$
So
$4y^2-y-5=0.$

3. Solve the quadratic equation.

Factorise:
$4y^2-y-5=0$
$(4y-5)(y+1)=0.$
Hence
$y=\frac{5}{4}\quad\text{or}\quad y=-1.$

4. Substitute back to find $x$ and check against the restriction.

Since $y=x-2$,
$x-2=\frac{5}{4}\quad\text{or}\quad x-2=-1.$
Thus
$x=\frac{13}{4}\quad\text{or}\quad x=1.$
Both values are not equal to $2$, so both are valid.