3y-5/y+1=4-y+3/y-1

Solution for 3y-5/y+1=4-y+3/y-1 equation:

3y-5/y+1=4-y+3/y-1

We move all terms to the left:

3y-5/y+1-(4-y+3/y-1)=0


Domain of the equation: y!=0

y∈R



Domain of the equation: y-1)!=0

y∈R


We add all the numbers together, and all the variables

3y-5/y-(-1y+3/y+3)+1=0

We get rid of parentheses

3y-5/y+1y-3/y-3+1=0

We multiply all the terms by the denominator


3y*y+1y*y-3*y+1*y-5-3=0

We add all the numbers together, and all the variables

-2y+3y*y+1y*y-8=0

Wy multiply elements

3y^2+y^2-2y-8=0

We add all the numbers together, and all the variables

4y^2-2y-8=0

a = 4; b = -2; c = -8;
Δ = b2-4ac
Δ = -22-4·4·(-8)
Δ = 132
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{33}}{2*4}=\frac{2-2\sqrt{33}}{8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{33}}{2*4}=\frac{2+2\sqrt{33}}{8}
$