1/3y-4=1/2y-3

Solution for 1/3y-4=1/2y-3 equation:

1/3y-4=1/2y-3

We move all terms to the left:

1/3y-4-(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

1/3y-1/2y+3-4=0

We calculate fractions


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

We add all the numbers together, and all the variables

2y/6y^2+(-3y)/6y^2-1=0

We multiply all the terms by the denominator


2y+(-3y)-1*6y^2=0

Wy multiply elements

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

We get rid of parentheses

-6y^2+2y-3y=0

We add all the numbers together, and all the variables

-6y^2-1y=0

a = -6; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-6)·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*-6}=\frac{0}{-12}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-6}=\frac{2}{-12}
=-1/6
$