-4/3y-3/2=5/4y-5

Solution for -4/3y-3/2=5/4y-5 equation:

-4/3y-3/2=5/4y-5

We move all terms to the left:

-4/3y-3/2-(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

-4/3y-5/4y+5-3/2=0

We calculate fractions


(-144y^2)/48y^2+(-64y)/48y^2+(-60y)/48y^2+5=0

We multiply all the terms by the denominator


(-144y^2)+(-64y)+(-60y)+5*48y^2=0

Wy multiply elements

(-144y^2)+240y^2+(-64y)+(-60y)=0

We get rid of parentheses

-144y^2+240y^2-64y-60y=0

We add all the numbers together, and all the variables

96y^2-124y=0

a = 96; b = -124; c = 0;
Δ = b2-4ac
Δ = -1242-4·96·0
Δ = 15376
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{15376}=124$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-124)-124}{2*96}=\frac{0}{192}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-124)+124}{2*96}=\frac{248}{192}
=1+7/24
$