+9466 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 }
