7/16y+6=8+3/8y

Solution for 7/16y+6=8+3/8y equation:

7/16y+6=8+3/8y

We move all terms to the left:

7/16y+6-(8+3/8y)=0


Domain of the equation: 16y!=0

y!=0/16

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

7/16y-(3/8y+8)+6=0

We get rid of parentheses

7/16y-3/8y-8+6=0

We calculate fractions


56y/128y^2+(-48y)/128y^2-8+6=0

We add all the numbers together, and all the variables

56y/128y^2+(-48y)/128y^2-2=0

We multiply all the terms by the denominator


56y+(-48y)-2*128y^2=0

Wy multiply elements

-256y^2+56y+(-48y)=0

We get rid of parentheses

-256y^2+56y-48y=0

We add all the numbers together, and all the variables

-256y^2+8y=0

a = -256; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·(-256)·0
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{-(8)-8}{2*-256}=\frac{-16}{-512}
=1/32
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*-256}=\frac{0}{-512}
=0
$