1/(x+1)+4=9/x+1

Solution for 1/(x+1)+4=9/x+1 equation:

1/(x+1)+4=9/x+1

We move all terms to the left:

1/(x+1)+4-(9/x+1)=0


Domain of the equation: (x+1)!=0

We move all terms containing x to the left, all other terms to the right

x!=-1

x∈R



Domain of the equation: x+1)!=0

x∈R


We get rid of parentheses

1/(x+1)-9/x-1+4=0

We calculate fractions


x/(x^2+x)+(-9x-9)/(x^2+x)-1+4=0

We add all the numbers together, and all the variables

x/(x^2+x)+(-9x-9)/(x^2+x)+3=0

We multiply all the terms by the denominator


x+(-9x-9)+3*(x^2+x)=0

We multiply parentheses

3x^2+x+(-9x-9)+3x=0

We get rid of parentheses

3x^2+x-9x+3x-9=0

We add all the numbers together, and all the variables

3x^2-5x-9=0

a = 3; b = -5; c = -9;
Δ = b2-4ac
Δ = -52-4·3·(-9)
Δ = 133
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{133}}{2*3}=\frac{5-\sqrt{133}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{133}}{2*3}=\frac{5+\sqrt{133}}{6}
$