For all real numbers x such that x \neq -3, x \neq 0, and x \neq 3, which of the following… — GRE General Test

For all real numbers

x

such that

x \neq -3

x \neq 0

, and

x \neq 3

, which of the following expressions is equivalent to

\frac{x^{-2} - 9^{-1}}{x^{-1} + 3^{-1}} \cdot \frac{3x^2 + 9x}{x - 3}?

-x - 3

The numerator of the first factor is a difference of squares

(x^{-1} - 3^{-1})(x^{-1} + 3^{-1})

. Dividing by

x^{-1} + 3^{-1}

leaves

x^{-1} - 3^{-1} = \frac{1}{x} - \frac{1}{3} = \frac{3 - x}{3x}

. Factoring

3x

out of the numerator of the second expression gives

3x(x + 3)

. Multiplying these terms yields

\frac{3 - x}{3x} \cdot \frac{3x(x + 3)}{x - 3}

. Since

3 - x = -(x - 3)

, the binomials

(3 - x)

and

(x - 3)

cancel to

-1

, and

3x

cancels out completely, resulting in

-(x + 3) = -x - 3

Factor the numerator of the first expression as a difference of squares.

x^{-2} - 9^{-1} = (x^{-1})^2 - (3^{-1})^2 = (x^{-1} - 3^{-1})(x^{-1} + 3^{-1})

Expressing negative exponents as squares allows cancellation with the denominator.

Divide by

(x^{-1} + 3^{-1})

and convert negative exponents into a single rational term.

\frac{(x^{-1} - 3^{-1})(x^{-1} + 3^{-1})}{x^{-1} + 3^{-1}} = x^{-1} - 3^{-1} = \frac{1}{x} - \frac{1}{3} = \frac{3 - x}{3x}

Finding a common denominator simplifies the first factor.

Factor the numerator of the second expression

3x^2 + 9x

3x^2 + 9x = 3x(x + 3)

Factoring out the greatest common factor

3x

enables further simplification.

Multiply the simplified expressions and reduce.

\left(\frac{3 - x}{3x}\right) \cdot \left(\frac{3x(x + 3)}{x - 3}\right) = \frac{-(x - 3)}{3x} \cdot \frac{3x(x + 3)}{x - 3} = -(x + 3) = -x - 3

Canceling

3x

and noting that

(3 - x)/(x - 3) = -1

leaves the linear expression

-x - 3

Simplifying algebraic expressions containing negative exponents, rational fractions, and opposite-sign binomial factors

Question