4/y-3=5/2y+3

Solution for 4/y-3=5/2y+3 equation:

4/y-3=5/2y+3

We move all terms to the left:

4/y-3-(5/2y+3)=0


Domain of the equation: y!=0

y∈R



Domain of the equation: 2y+3)!=0

y∈R


We get rid of parentheses

4/y-5/2y-3-3=0

We calculate fractions


8y/2y^2+(-5y)/2y^2-3-3=0

We add all the numbers together, and all the variables

8y/2y^2+(-5y)/2y^2-6=0

We multiply all the terms by the denominator


8y+(-5y)-6*2y^2=0

Wy multiply elements

-12y^2+8y+(-5y)=0

We get rid of parentheses

-12y^2+8y-5y=0

We add all the numbers together, and all the variables

-12y^2+3y=0

a = -12; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-12)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-12}=\frac{-6}{-24}
=1/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-12}=\frac{0}{-24}
=0
$