1/3y-4=1/7y

Solution for 1/3y-4=1/7y equation:

1/3y-4=1/7y

We move all terms to the left:

1/3y-4-(1/7y)=0


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R



Domain of the equation: 7y)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

1/3y-(+1/7y)-4=0

We get rid of parentheses

1/3y-1/7y-4=0

We calculate fractions


7y/21y^2+(-3y)/21y^2-4=0

We multiply all the terms by the denominator


7y+(-3y)-4*21y^2=0

Wy multiply elements

-84y^2+7y+(-3y)=0

We get rid of parentheses

-84y^2+7y-3y=0

We add all the numbers together, and all the variables

-84y^2+4y=0

a = -84; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-84)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-84}=\frac{-8}{-168}
=1/21
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-84}=\frac{0}{-168}
=0
$