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
$