y/8y+35=3y

Solution for y/8y+35=3y equation:

y/8y+35=3y

We move all terms to the left:

y/8y+35-(3y)=0


Domain of the equation: 8y!=0

y!=0/8

y!=0

y∈R


We add all the numbers together, and all the variables

-3y+y/8y+35=0

We multiply all the terms by the denominator


-3y*8y+y+35*8y=0

We add all the numbers together, and all the variables

y-3y*8y+35*8y=0

Wy multiply elements

-24y^2+y+280y=0

We add all the numbers together, and all the variables

-24y^2+281y=0

a = -24; b = 281; c = 0;
Δ = b2-4ac
Δ = 2812-4·(-24)·0
Δ = 78961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{78961}=281$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(281)-281}{2*-24}=\frac{-562}{-48}
=11+17/24
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(281)+281}{2*-24}=\frac{0}{-48}
=0
$