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Exercise

$\frac{x^2-1}{y^2+5y+4}\cdot\frac{y^2+2y-8}{x^2-2x}$

Step-by-step Solution

1

Multiplying fractions $\frac{x^2-1}{y^2+5y+4} \times \frac{y^2+2y-8}{x^2-2x}$

$\frac{\left(x^2-1\right)\left(y^2+2y-8\right)}{\left(y^2+5y+4\right)\left(x^2-2x\right)}$

Final answer to the exercise

$\frac{\left(x^2-1\right)\left(y^2+2y-8\right)}{\left(y^2+5y+4\right)\left(x^2-2x\right)}$

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Main Topic: Equivalent Expressions

Two algebraic expressions are equivalent if they stay equal to each other, for any values their variables can take.

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