Mathematics: Operations with rational expressions

The answer for #1 and 2 1)B 5.007x10 -32)B 103,150...

Operations with rational expressions

kat122402

20.01.2022

$$\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}$

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