7/6y+6=8+3/8y

Solution for 7/6y+6=8+3/8y equation:

7/6y+6=8+3/8y

We move all terms to the left:

7/6y+6-(8+3/8y)=0


Domain of the equation: 6y!=0

y!=0/6

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

7/6y-(3/8y+8)+6=0

We get rid of parentheses

7/6y-3/8y-8+6=0

We calculate fractions


56y/48y^2+(-18y)/48y^2-8+6=0

We add all the numbers together, and all the variables

56y/48y^2+(-18y)/48y^2-2=0

We multiply all the terms by the denominator


56y+(-18y)-2*48y^2=0

Wy multiply elements

-96y^2+56y+(-18y)=0

We get rid of parentheses

-96y^2+56y-18y=0

We add all the numbers together, and all the variables

-96y^2+38y=0

a = -96; b = 38; c = 0;
Δ = b2-4ac
Δ = 382-4·(-96)·0
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-38}{2*-96}=\frac{-76}{-192}
=19/48
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+38}{2*-96}=\frac{0}{-192}
=0
$