Question

What is the solution of StartFraction 1 Over c minus 3 EndFraction minus StartFraction 1 Over c EndFraction = StartFraction 3 Over c (c minus 3) EndFraction?
c = 0 and c = 3
all real numbers
all real numbers, except c ¹ 0 and c ¹ 3
no solution

Answer 1

Answer:

• All real numbers except c = 0 and c = 3.

Explanation:

The equation is:

   $\dfrac{1}{c-3}-\dfrac{1}{c}=\dfrac{3}{c(c-3)}$

Since c - 3, c, and c(c - 3) are he denominators, none of them can be equal to zero:

• c ≠ 0

• c - 3 ≠ 0 ⇒ c ≠ 3

• c (c - 3) ≠ 0 ⇒ c ≠ 0 and c ≠ 3.

Now you can multiply both sides of the equation by the common denominator: c (c - 3):

       $c-(c-3)=3\\\\c-c+3=3\\\\0=3-3\\\\0=0$

That means the equality is valid for all real numbers for which it is defined, which is all real numbers except c = 0 and c = 3.

Answer 2

The solution of the given equation can be all real numbers, except c = 0 and c = 3.

Given the following equation:

• $\frac{1}{c\;-\;3} -\frac{1}{c} =\frac{3}{c(c\;-\;3)}$

Note: The denominators cannot be equal to zero (0) because a division by zero (0) is undefined.

Next, we would multiply both sides of the given by the lowest common multiple (LCM).

The lowest common multiple (LCM) is c(c - 3).

$c(c\;-\;3) \times (\frac{1}{c\;-\;3} -\frac{1}{c}) =\frac{3}{c(c\;-\;3)} \times c(c\;-\;3)\\\\c - [c\;-\;3]=3\\\\c-c+3=3\\\\0=3-3\\\\0=0$

Therefore, the solution of the given equation can be all real numbers, except c = 0 and c = 3.

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