#1**+3 **

I think this isn't your question:

\(\frac{z}{z}-5+\frac{5}{z}+5=\frac{50}{z^2}-25\)

I think this is your question:

\(\frac{z}{z-5}+\frac{5}{z+5}=\frac{50}{z^2-25}\)

Here, take note of what z values cause a zero in the denominator.

z ≠ 5, z ≠ -5

Get a common denominator on the left side

\(\frac{z(z+5)}{(z-5)(z+5)}+\frac{5(z-5)}{(z-5)(z+5)}=\frac{50}{z^2-25}\)

Factor the denominator on the right side.

\(\frac{z(z+5)}{(z-5)(z+5)}+\frac{5(z-5)}{(z-5)(z+5)}=\frac{50}{(z-5)(z+5)}\)

Multiply through by (z-5)(z+5).

\(z(z+5)+5(z-5)=50\)

Distribute and combine like terms.

\(z^2+5z+5z-25=50 \\ z^2+10z-75=0\)

Factor and set each factor = 0.

\(z^2+10z-75=0 \\ (z+15)(z-5)=0\)

z+15 = 0 or z-5 = 0

z = -15 or z = 5

HOWEVER...in the original problem, z = 5 causes a zero in the denominator.

So, z cannot = 5.

The solution set is just: { -15 }

hectictar
Apr 10, 2017