1/4y+2=1/8y+3

Solution for 1/4y+2=1/8y+3 equation:

1/4y+2=1/8y+3

We move all terms to the left:

1/4y+2-(1/8y+3)=0


Domain of the equation: 4y!=0

y!=0/4

y!=0

y∈R



Domain of the equation: 8y+3)!=0

y∈R


We get rid of parentheses

1/4y-1/8y-3+2=0

We calculate fractions


8y/32y^2+(-4y)/32y^2-3+2=0

We add all the numbers together, and all the variables

8y/32y^2+(-4y)/32y^2-1=0

We multiply all the terms by the denominator


8y+(-4y)-1*32y^2=0

Wy multiply elements

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

We get rid of parentheses

-32y^2+8y-4y=0

We add all the numbers together, and all the variables

-32y^2+4y=0

a = -32; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-32)·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*-32}=\frac{-8}{-64}
=1/8
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-32}=\frac{0}{-64}
=0
$