(4/x+2)+(2/2x+4)=16

Solution for (4/x+2)+(2/2x+4)=16 equation:

(4/x+2)+(2/2x+4)=16

We move all terms to the left:

(4/x+2)+(2/2x+4)-(16)=0


Domain of the equation: x+2)!=0

x∈R



Domain of the equation: 2x+4)!=0

x∈R


We get rid of parentheses

4/x+2/2x+2+4-16=0

We calculate fractions


8x/2x^2+2x/2x^2+2+4-16=0

We add all the numbers together, and all the variables

8x/2x^2+2x/2x^2-10=0

We multiply all the terms by the denominator


8x+2x-10*2x^2=0

We add all the numbers together, and all the variables

10x-10*2x^2=0

Wy multiply elements

-20x^2+10x=0

a = -20; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·(-20)·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*-20}=\frac{-20}{-40}
=1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*-20}=\frac{0}{-40}
=0
$