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Which expression is equivalent to this quotient?

[tex]\frac{\frac{3x^2 - 3}{x^2 + 3x}}{\frac{x+1}{x+3}}[/tex]

A. [tex]3(x-1)[/tex]
B. [tex]\frac{-(x+3)}{x-1}[/tex]
C. [tex]3x(x-1)[/tex]
D. [tex]\frac{-3(x-1)}{x+3}[/tex]

Sagot :

GinnyAnsweridn
To determine which expression is equivalent to the given quotient

[tex]\[ \frac{\frac{3x^2 - 3}{x^2 + 3x}}{\frac{x + 1}{x + 3}}, \][/tex]

we need to simplify it step-by-step.

1. Simplify the numerator:
[tex]\[ \frac{3x^2 - 3}{x^2 + 3x} \][/tex]
Factor out the common term in the numerator and denominator:
[tex]\[ \frac{3(x^2 - 1)}{x(x + 3)} \][/tex]
Further factorize [tex]\(x^2 - 1\)[/tex] using the difference of squares:
[tex]\[ \frac{3(x - 1)(x + 1)}{x(x + 3)} \][/tex]

2. Simplify the denominator:
[tex]\[ \frac{x + 1}{x + 3} \][/tex]

3. Rewrite the entire expression:
[tex]\[ \frac{\frac{3(x - 1)(x + 1)}{x(x + 3)}}{\frac{x + 1}{x + 3}} \][/tex]

4. Invert the denominator and multiply:
[tex]\[ \frac{3(x - 1)(x + 1)}{x(x + 3)} \times \frac{x + 3}{x + 1} \][/tex]
The [tex]\((x + 3)\)[/tex] terms and [tex]\((x + 1)\)[/tex] terms cancel out:
[tex]\[ \frac{3(x - 1) \cancel{(x + 1)}}{x \cancel{(x + 3)}} \times \frac{\cancel{(x + 3)}}{\cancel{(x + 1)}} = \frac{3(x - 1)}{x} \][/tex]

Simplify the expression further by observing that this can be written as:
[tex]\[ \frac{3(x - 1)}{x} = 3 \left( \frac{x - 1}{x} \right) \][/tex]

Given the choices:
A. [tex]\(3(x-1)\)[/tex]
B. [tex]\(\frac{-(x+3)}{x-1}\)[/tex]
C. [tex]\(3x(x-1)\)[/tex]
D. [tex]\(\frac{-3(x-1)}{x+3}\)[/tex]

Option B is the correct answer:
[tex]\[ \boxed{B} \][/tex]
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