Question

PLEASE HHELP!!! Bring the fraction:

Answer 1

Answer:

$\frac{b}{7a^2c} = \frac{5abc^2}{35a^3c^3}$

$\frac{a}{a - 4} = \frac{-a^2 - 4a}{16 -a^2}$

Step-by-step explanation:

Given

$\frac{b}{7a^2c}$

Express the denominator as $35a^3c^3$

To do this, we divide$35a^3c^3$ by the denominator

$\frac{35a^3c^3}{7a^2c} = 5ac^2$

So, the required fraction is:

$\frac{b}{7a^2c} * \frac{5ac^2}{5ac^2}$

$\frac{5abc^2}{35a^3c^3}$

Hence:

$\frac{b}{7a^2c} = \frac{5abc^2}{35a^3c^3}$

Given

$\frac{a}{a - 4}$

Express the denominator as $16 - a^2$

Multiply the fraction a+4/a+4

So, we have:

$\frac{a}{a - 4} * \frac{(a + 4)}{(a + 4)}$

Apply difference of two squares to the denominator

$\frac{a^2 + 4a}{a^2 - 16}$

Take the additive inverse of the numerator and denominator

$\frac{-(a^2 + 4a)}{-(a^2 - 16)}$

$\frac{-a^2 - 4a}{16 -a^2}$

Hence:

$\frac{a}{a - 4} = \frac{-a^2 - 4a}{16 -a^2}$