3/4y-4=7/8y+1

Solution for 3/4y-4=7/8y+1 equation:

3/4y-4=7/8y+1

We move all terms to the left:

3/4y-4-(7/8y+1)=0


Domain of the equation: 4y!=0

y!=0/4

y!=0

y∈R



Domain of the equation: 8y+1)!=0

y∈R


We get rid of parentheses

3/4y-7/8y-1-4=0

We calculate fractions


24y/32y^2+(-28y)/32y^2-1-4=0

We add all the numbers together, and all the variables

24y/32y^2+(-28y)/32y^2-5=0

We multiply all the terms by the denominator


24y+(-28y)-5*32y^2=0

Wy multiply elements

-160y^2+24y+(-28y)=0

We get rid of parentheses

-160y^2+24y-28y=0

We add all the numbers together, and all the variables

-160y^2-4y=0

a = -160; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·(-160)·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*-160}=\frac{0}{-320}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*-160}=\frac{8}{-320}
=-1/40
$