1/3y-4=1/4y-5

Solution for 1/3y-4=1/4y-5 equation:

1/3y-4=1/4y-5

We move all terms to the left:

1/3y-4-(1/4y-5)=0


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R



Domain of the equation: 4y-5)!=0

y∈R


We get rid of parentheses

1/3y-1/4y+5-4=0

We calculate fractions


4y/12y^2+(-3y)/12y^2+5-4=0

We add all the numbers together, and all the variables

4y/12y^2+(-3y)/12y^2+1=0

We multiply all the terms by the denominator


4y+(-3y)+1*12y^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
$