-1/3y-6=-17+1/4y

Solution for -1/3y-6=-17+1/4y equation:

-1/3y-6=-17+1/4y

We move all terms to the left:

-1/3y-6-(-17+1/4y)=0


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R



Domain of the equation: 4y)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

-1/3y-(1/4y-17)-6=0

We get rid of parentheses

-1/3y-1/4y+17-6=0

We calculate fractions


(-4y)/12y^2+(-3y)/12y^2+17-6=0

We add all the numbers together, and all the variables

(-4y)/12y^2+(-3y)/12y^2+11=0

We multiply all the terms by the denominator


(-4y)+(-3y)+11*12y^2=0

Wy multiply elements

132y^2+(-4y)+(-3y)=0

We get rid of parentheses

132y^2-4y-3y=0

We add all the numbers together, and all the variables

132y^2-7y=0

a = 132; b = -7; c = 0;
Δ = b2-4ac
Δ = -72-4·132·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7}{2*132}=\frac{0}{264}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7}{2*132}=\frac{14}{264}
=7/132
$