2/3y-5=1/2y-3

Solution for 2/3y-5=1/2y-3 equation:

2/3y-5=1/2y-3

We move all terms to the left:

2/3y-5-(1/2y-3)=0


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R



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

y∈R


We get rid of parentheses

2/3y-1/2y+3-5=0

We calculate fractions


4y/6y^2+(-3y)/6y^2+3-5=0

We add all the numbers together, and all the variables

4y/6y^2+(-3y)/6y^2-2=0

We multiply all the terms by the denominator


4y+(-3y)-2*6y^2=0

Wy multiply elements

-12y^2+4y+(-3y)=0

We get rid of parentheses

-12y^2+4y-3y=0

We add all the numbers together, and all the variables

-12y^2+y=0

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

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