32-8/3y=3/4y+5

Solution for 32-8/3y=3/4y+5 equation:

32-8/3y=3/4y+5

We move all terms to the left:

32-8/3y-(3/4y+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

-8/3y-3/4y-5+32=0

We calculate fractions


(-32y)/12y^2+(-9y)/12y^2-5+32=0

We add all the numbers together, and all the variables

(-32y)/12y^2+(-9y)/12y^2+27=0

We multiply all the terms by the denominator


(-32y)+(-9y)+27*12y^2=0

Wy multiply elements

324y^2+(-32y)+(-9y)=0

We get rid of parentheses

324y^2-32y-9y=0

We add all the numbers together, and all the variables

324y^2-41y=0

a = 324; b = -41; c = 0;
Δ = b2-4ac
Δ = -412-4·324·0
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-41}{2*324}=\frac{0}{648}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+41}{2*324}=\frac{82}{648}
=41/324
$