Answer:
[tex]\frac{b}{7a^2c} = \frac{5abc^2}{35a^3c^3}[/tex]
[tex]\frac{a}{a - 4} = \frac{-a^2 - 4a}{16 -a^2}[/tex]
Step-by-step explanation:
Given
[tex]\frac{b}{7a^2c}[/tex]
Express the denominator as [tex]35a^3c^3[/tex]
To do this, we divide[tex]35a^3c^3[/tex] by the denominator
[tex]\frac{35a^3c^3}{7a^2c} = 5ac^2[/tex]
So, the required fraction is:
[tex]\frac{b}{7a^2c} * \frac{5ac^2}{5ac^2}[/tex]
[tex]\frac{5abc^2}{35a^3c^3}[/tex]
Hence:
[tex]\frac{b}{7a^2c} = \frac{5abc^2}{35a^3c^3}[/tex]
Given
[tex]\frac{a}{a - 4}[/tex]
Express the denominator as [tex]16 - a^2[/tex]
Multiply the fraction a+4/a+4
So, we have:
[tex]\frac{a}{a - 4} * \frac{(a + 4)}{(a + 4)}[/tex]
Apply difference of two squares to the denominator
[tex]\frac{a^2 + 4a}{a^2 - 16}[/tex]
Take the additive inverse of the numerator and denominator
[tex]\frac{-(a^2 + 4a)}{-(a^2 - 16)}[/tex]
[tex]\frac{-a^2 - 4a}{16 -a^2}[/tex]
Hence:
[tex]\frac{a}{a - 4} = \frac{-a^2 - 4a}{16 -a^2}[/tex]
