What is ((8+4x))/((4x+x^2))<= 2/x+3/((4+x)) ?

The solution to ((8+4x))/((4x+x^2)) -4

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((8+4x))/((4x+x^2))<= 2/x+3/((4+x))

\frac{( 8 + 4 x )}{( 4 x + x ^{2} )} \leq \frac{2}{x} + \frac{3}{( 4 + x )}

x > - 4

Interval Notation

(- 4, \infty)

Solution steps

\frac{8 + 4 x}{4 x + x ^{2}} \leq \frac{2}{x} + \frac{3}{4 + x}

Rewrite in standard form

\frac{1}{x + 4} \geq 0

\frac{1}{a} \geq 0

then

a > 0

x + 4 > 0

Move

4

to the right side

x > - 4

\frac{x ^{2} + ( 4 x ^{2} )}{( ( x - 2 ) ^{2} )} \leq 5

(3 x + 1)^{2} + 4 x < 9 x^{2} + x - 4

((x^{2} - 5 x + 6)^{2})^{0} ((x^{2} - 11 x + 24)^{3})^{1} > 0

\frac{( 2 ( x - 4 ) )}{( ( x - 1 ) ( x - 7 ) )} \geq \frac{1}{( x - 2 )}

- 3 m^{2} + 2 m + 1 \leq 0

What is ((8+4x))/((4x+x^2))<= 2/x+3/((4+x)) ?
The solution to ((8+4x))/((4x+x^2))<= 2/x+3/((4+x)) is x>-4