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

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

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

We move all terms to the left:

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


Domain of the equation: 4y!=0

y!=0/4

y!=0

y∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply all the terms by the denominator


-y*4y-8*4y+8*4y+3+1=0

We add all the numbers together, and all the variables

-y*4y-8*4y+8*4y+4=0

Wy multiply elements

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

We add all the numbers together, and all the variables

-4y^2+4=0

a = -4; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·(-4)·4
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*-4}=\frac{-8}{-8}
=1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*-4}=\frac{8}{-8}
=-1
$