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What is the solution of [tex]\frac{1}{c-3} - \frac{1}{c} = \frac{3}{c(c-3)}[/tex]?

A. [tex]c = 0[/tex] and [tex]c = 3[/tex]

B. all real numbers

C. all real numbers, except [tex]c = 0[/tex] and [tex]c = 3[/tex]

D. no solution


Sagot :

To solve the equation [tex]\( \frac{1}{c-3} - \frac{1}{c} = \frac{3}{c(c-3)} \)[/tex], let's proceed step by step:

1. Identify the common denominator:
The common denominator of the fractions on both sides of the equation is [tex]\( c(c-3) \)[/tex].

2. Express each term with the common denominator:
[tex]\[ \frac{1}{c-3} = \frac{c}{c(c-3)}, \quad \frac{1}{c} = \frac{c-3}{c(c-3)} \][/tex]

3. Rewriting the equation with a common denominator:
[tex]\[ \frac{c}{c(c-3)} - \frac{c-3}{c(c-3)} = \frac{3}{c(c-3)} \][/tex]

4. Combine the fractions on the left side:
[tex]\[ \frac{c - (c-3)}{c(c-3)} = \frac{3}{c(c-3)} \][/tex]

5. Simplify the numerator on the left:
[tex]\[ \frac{c - c + 3}{c(c-3)} = \frac{3}{c(c-3)} \][/tex]
[tex]\[ \frac{3}{c(c-3)} = \frac{3}{c(c-3)} \][/tex]

6. Since both sides of the equation are equal:
This identity confirms that the equation holds true generally for [tex]\( c \)[/tex] except where the denominators are undefined, i.e., where [tex]\( c \)[/tex] causes division by zero.

7. Determine the values that make the denominators zero:
[tex]\[ c(c-3) \neq 0 \][/tex]
[tex]\[ c \neq 0 \quad \text{and} \quad c \neq 3 \][/tex]

8. Conclude that the equation has no solutions at [tex]\( c=0 \)[/tex] and [tex]\( c=3 \)[/tex]:

Since the integers [tex]\( c = 0 \)[/tex] and [tex]\( c = 3 \)[/tex] would cause division by zero, these values are not valid solutions.

Therefore, the solution to the equation [tex]\( \frac{1}{c-3} - \frac{1}{c} = \frac{3}{c(c-3)} \)[/tex] is:

[tex]\[ \text{No solution} \][/tex]