For all real values of x and y where the expression is defined, simplify the algebraic fra… — ACT

For all real values of $x$ and $y$ where the expression is defined, simplify the algebraic fraction:

\frac{(x + y)^3 x^{-2} y}{(x+y) (x y^{-1})^{-2}}

Which expression is equivalent to this fraction?

The expression

\frac{(x+y)^2}{y}

The correct expression is derived by first simplifying the denominator

(xy^{-1})^{-2}

x^{-2}y^2

. We then group and simplify like terms:

\frac{(x+y)^3}{x+y} = (x+y)^2

\frac{x^{-2}}{x^{-2}} = 1

, and

\frac{y}{y^2} = \frac{1}{y}

. Multiplying these yields the simplified expression.

Simplify the term

(x y^{-1})^{-2}

in the denominator

x^{-2} y^2

Apply the power of a product rule:

(ab)^n = a^n b^n

, which gives

x^{-2} (y^{-1})^{-2}

. Then, use the power of a power rule:

(y^{-1})^{-2} = y^{(-1) \cdot (-2)} = y^2

Rewrite the original fraction with the simplified denominator

\frac{(x + y)^3 x^{-2} y}{(x+y) x^{-2} y^2}

Substitute the simplified term back into the expression to align similar bases.

Simplify the fraction by dividing terms with like bases using the quotient rule

\frac{(x+y)^2}{y}

Divide each component:

\frac{(x+y)^3}{x+y} = (x+y)^{3-1} = (x+y)^2

\frac{x^{-2}}{x^{-2}} = x^{-2 - (-2)} = x^0 = 1

, and

\frac{y}{y^2} = y^{1-2} = y^{-1} = \frac{1}{y}

. Multiplying these results gives

\frac{(x+y)^2}{y}

Properties of exponents in algebraic expressions, including power of a product, power of a power, and quotient rules.

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