Question

Find any domain restrictions on the given rational equation

Answer 1

Answer:

x ≠ -4, -1, 3

Step-by-step explanation:

$\frac{x}{x+4}+ \frac{12}{x^2} +5x+4 = \frac{8x}{5x-15}$

$12/(x^2+5x+4) = 8x/(5x-15)$

$\frac{12}{((x+1)(x+4))} = \frac{8x}{(5(x-3))}$

Answer 2

$\\ \tt\hookrightarrow \dfrac{x}{x+4}+\dfrac{12}{x^2+5x+4}=\dfrac{8x}{5x-15}$

$\\ \tt\hookrightarrow \dfrac{x}{x+4}+\dfrac{12}{(x+1)(x+4)}=\dfrac{8x}{5(x-3)}$

• For x=-4 First fraction becomes undefined
• For x=-1 second fraction becomes undefined
• For x=3 third fraction also becomes undefined

So Domain=(-4,-1,3)