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

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

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

We move all terms to the left:

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


Domain of the equation: 2y!=0

y!=0/2

y!=0

y∈R



Domain of the equation: 3y+4)!=0

y∈R


We get rid of parentheses

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

We calculate fractions


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

We add all the numbers together, and all the variables

3y/6y^2+(-4y)/6y^2-7=0

We multiply all the terms by the denominator


3y+(-4y)-7*6y^2=0

Wy multiply elements

-42y^2+3y+(-4y)=0

We get rid of parentheses

-42y^2+3y-4y=0

We add all the numbers together, and all the variables

-42y^2-1y=0

a = -42; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-42)·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*-42}=\frac{0}{-84}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-42}=\frac{2}{-84}
=-1/42
$