5/3y-7/3=5/4y-5

Solution for 5/3y-7/3=5/4y-5 equation:

5/3y-7/3=5/4y-5

We move all terms to the left:

5/3y-7/3-(5/4y-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

5/3y-5/4y+5-7/3=0

We calculate fractions


20y/108y^2+(-135y)/108y^2+(-28y)/108y^2+5=0

We multiply all the terms by the denominator


20y+(-135y)+(-28y)+5*108y^2=0

Wy multiply elements

540y^2+20y+(-135y)+(-28y)=0

We get rid of parentheses

540y^2+20y-135y-28y=0

We add all the numbers together, and all the variables

540y^2-143y=0

a = 540; b = -143; c = 0;
Δ = b2-4ac
Δ = -1432-4·540·0
Δ = 20449
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{20449}=143$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-143)-143}{2*540}=\frac{0}{1080}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-143)+143}{2*540}=\frac{286}{1080}
=143/540
$