what is the solution of $\frac{1}{c - 3}-\frac{1}{c}=\frac{3}{c(c - 3)}$?\no $c = 0$ and $c = 3$\no all real…

Answer:

all real numbers, except c = 0 and c = 3

Explanation:

Step1: Find common denominator

The common denominator of $\frac{1}{c - 3}$, $\frac{1}{c}$ and $\frac{3}{c(c - 3)}$ is $c(c - 3)$.

Step2: Rewrite fractions

$\frac{1}{c - 3}\times\frac{c}{c}-\frac{1}{c}\times\frac{c - 3}{c - 3}=\frac{3}{c(c - 3)}$, which gives $\frac{c}{c(c - 3)}-\frac{c - 3}{c(c - 3)}=\frac{3}{c(c - 3)}$.

Step3: Combine fractions on left - hand side

$\frac{c-(c - 3)}{c(c - 3)}=\frac{3}{c(c - 3)}$, simplifies to $\frac{c - c+3}{c(c - 3)}=\frac{3}{c(c - 3)}$, then $\frac{3}{c(c - 3)}=\frac{3}{c(c - 3)}$. This is an identity, but we must exclude values that make the original denominators zero. The denominators $c-3 = 0$ (i.e., $c = 3$) and $c=0$ would make the original rational - expressions undefined. So the solution is all real numbers except $c = 0$ and $c = 3$.