5/8y+7=3/4y-3

Solution for 5/8y+7=3/4y-3 equation:

5/8y+7=3/4y-3

We move all terms to the left:

5/8y+7-(3/4y-3)=0


Domain of the equation: 8y!=0

y!=0/8

y!=0

y∈R



Domain of the equation: 4y-3)!=0

y∈R


We get rid of parentheses

5/8y-3/4y+3+7=0

We calculate fractions


20y/32y^2+(-24y)/32y^2+3+7=0

We add all the numbers together, and all the variables

20y/32y^2+(-24y)/32y^2+10=0

We multiply all the terms by the denominator


20y+(-24y)+10*32y^2=0

Wy multiply elements

320y^2+20y+(-24y)=0

We get rid of parentheses

320y^2+20y-24y=0

We add all the numbers together, and all the variables

320y^2-4y=0

a = 320; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·320·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*320}=\frac{0}{640}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*320}=\frac{8}{640}
=1/80
$