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

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

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

We move all terms to the left:

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

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

We get rid of parentheses

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

We calculate fractions


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

We multiply all the terms by the denominator


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

Wy multiply elements

192y^2+4y+(-8y)=0

We get rid of parentheses

192y^2+4y-8y=0

We add all the numbers together, and all the variables

192y^2-4y=0

a = 192; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·192·0
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*192}=\frac{0}{384}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*192}=\frac{8}{384}
=1/48
$