3/8y-5=-1+1/4y

Solution for 3/8y-5=-1+1/4y equation:

3/8y-5=-1+1/4y

We move all terms to the left:

3/8y-5-(-1+1/4y)=0


Domain of the equation: 8y!=0

y!=0/8

y!=0

y∈R



Domain of the equation: 4y)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

3/8y-(1/4y-1)-5=0

We get rid of parentheses

3/8y-1/4y+1-5=0

We calculate fractions


12y/32y^2+(-8y)/32y^2+1-5=0

We add all the numbers together, and all the variables

12y/32y^2+(-8y)/32y^2-4=0

We multiply all the terms by the denominator


12y+(-8y)-4*32y^2=0

Wy multiply elements

-128y^2+12y+(-8y)=0

We get rid of parentheses

-128y^2+12y-8y=0

We add all the numbers together, and all the variables

-128y^2+4y=0

a = -128; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-128)·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*-128}=\frac{-8}{-256}
=1/32
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-128}=\frac{0}{-256}
=0
$