Question

Operations with rational expressions

Answer 1

$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}=-\frac{w+7}{9}$

Solution:

Given expression:

$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}$

To solve the given expression:

First simplify: $\frac{50-2 w^{2}}{3 w^{2}+9 w-30}$

$\frac{50-2 w^{2}}{3 w^{2}+9 w-30}=-\frac{2(w+5)(w-5)}{3(w-2)(w+5)}$

Cancel the common factor (w + 5).

                      $=-\frac{2(w-5)}{3(w-2)}$

Now substitute this in the given expression.

$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}=-\frac{2(w-5)}{3(w-2)} \cdot \frac{w^{2}+5 w-14}{6 w-30}$

Multiply the fractions $\frac{a}{b} \cdot \frac{c}{d}=\frac{a \cdot c}{b \cdot d}$

                               $=-\frac{2(w-5)\left(w^{2}+5 w-14\right)}{3(w-2)(6 w-30)}$

Factor the denominator $3(w-2)(6 w-30) =18(w-2)(w-5)$

                               $=-\frac{2(w-5)\left(w^{2}+5 w-14\right)}{18(w-2)(w-5)}$                        

Cancel the common factor 2(w – 5).

                               $=-\frac{w^{2}+5 w-14}{9(w-2)}$

Factor the numerator $w^{2}+5 w-14=(w-2)(w+7)$

                               $=-\frac{(w-2)(w+7)}{9(w-2)}$

Cancel the common factor (w – 2).

                               $=-\frac{w+7}{9}$

$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}=-\frac{w+7}{9}$