what is the solution of $\frac{1}{c - 3}-\frac{1}{c}=\frac{3}{c(c - 3)}$?\n$c = 0$ and $c = 3$\nall real…

what is the solution of $\frac{1}{c - 3}-\frac{1}{c}=\frac{3}{c(c - 3)}$?\n$c = 0$ and $c = 3$\nall real numbers\nall real numbers, except $c\neq0$ and $c\neq3$\nno solution

what is the solution of $\frac{1}{c - 3}-\frac{1}{c}=\frac{3}{c(c - 3)}$?\n$c = 0$ and $c = 3$\nall real numbers\nall real numbers, except $c\neq0$ and $c\neq3$\nno solution

Answer

Explanation:

Step1: Find common denominator

The left - hand side has denominators $c - 3$ and $c$. The common denominator is $c(c - 3)$. Rewrite the left - hand side: $\frac{c}{c(c - 3)}-\frac{c - 3}{c(c - 3)}=\frac{3}{c(c - 3)}$.

Step2: Combine fractions on left - hand side

$\frac{c-(c - 3)}{c(c - 3)}=\frac{3}{c(c - 3)}$, which simplifies to $\frac{c - c+3}{c(c - 3)}=\frac{3}{c(c - 3)}$, and further to $\frac{3}{c(c - 3)}=\frac{3}{c(c - 3)}$. However, we must consider the domain of the original rational equation. The denominators $c-3$, $c$ and $c(c - 3)$ cannot be zero. So $c\neq0$ and $c\neq3$.

Answer:

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