3/4y+1=2/5y-3

Solution for 3/4y+1=2/5y-3 equation:

3/4y+1=2/5y-3

We move all terms to the left:

3/4y+1-(2/5y-3)=0


Domain of the equation: 4y!=0

y!=0/4

y!=0

y∈R



Domain of the equation: 5y-3)!=0

y∈R


We get rid of parentheses

3/4y-2/5y+3+1=0

We calculate fractions


15y/20y^2+(-8y)/20y^2+3+1=0

We add all the numbers together, and all the variables

15y/20y^2+(-8y)/20y^2+4=0

We multiply all the terms by the denominator


15y+(-8y)+4*20y^2=0

Wy multiply elements

80y^2+15y+(-8y)=0

We get rid of parentheses

80y^2+15y-8y=0

We add all the numbers together, and all the variables

80y^2+7y=0

a = 80; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·80·0
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*80}=\frac{-14}{160}
=-7/80
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*80}=\frac{0}{160}
=0
$