1/y+13=3/16y

Solution for 1/y+13=3/16y equation:

1/y+13=3/16y

We move all terms to the left:

1/y+13-(3/16y)=0


Domain of the equation: y!=0

y∈R



Domain of the equation: 16y)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

1/y-(+3/16y)+13=0

We get rid of parentheses

1/y-3/16y+13=0

We calculate fractions


16y/16y^2+(-3y)/16y^2+13=0

We multiply all the terms by the denominator


16y+(-3y)+13*16y^2=0

Wy multiply elements

208y^2+16y+(-3y)=0

We get rid of parentheses

208y^2+16y-3y=0

We add all the numbers together, and all the variables

208y^2+13y=0

a = 208; b = 13; c = 0;
Δ = b2-4ac
Δ = 132-4·208·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-13}{2*208}=\frac{-26}{416}
=-1/16
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+13}{2*208}=\frac{0}{416}
=0
$