1/4y-6=1/8y+4

Solution for 1/4y-6=1/8y+4 equation:

1/4y-6=1/8y+4

We move all terms to the left:

1/4y-6-(1/8y+4)=0


Domain of the equation: 4y!=0

y!=0/4

y!=0

y∈R



Domain of the equation: 8y+4)!=0

y∈R


We get rid of parentheses

1/4y-1/8y-4-6=0

We calculate fractions


8y/32y^2+(-4y)/32y^2-4-6=0

We add all the numbers together, and all the variables

8y/32y^2+(-4y)/32y^2-10=0

We multiply all the terms by the denominator


8y+(-4y)-10*32y^2=0

Wy multiply elements

-320y^2+8y+(-4y)=0

We get rid of parentheses

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