Question

Solve the fractions for z:
z/z-3 + z/z^2-9 = z+4/z+3

Answer 1

\[\frac{z}{z-3}+\frac{z}{z^2-9}=\frac{z+4}{z+3}\]



\[\frac{z}{z-3}+\frac{z}{(z-3)(z+3)}=\frac{z+4}{z+3}\]



\[z(z+3)+z=(z+4)(z-3)\]



\[z^2+3z+z=z^2+z-12\]



\[z^2+3z+z-z^2-z+12=0\]



\[3z+12=0\]



\[3z=-12\]



\[z=\frac{-12}{3}\]



\[z=-4\]
I'll let you perform the check.

Answer 2

z = -4 is correct