1/4y+10=1/8y

Solution for 1/4y+10=1/8y equation:

1/4y+10=1/8y

We move all terms to the left:

1/4y+10-(1/8y)=0


Domain of the equation: 4y!=0

y!=0/4

y!=0

y∈R



Domain of the equation: 8y)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

1/4y-(+1/8y)+10=0

We get rid of parentheses

1/4y-1/8y+10=0

We calculate fractions


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
$