1/4y-5=1/7y

Solution for 1/4y-5=1/7y equation:

1/4y-5=1/7y

We move all terms to the left:

1/4y-5-(1/7y)=0


Domain of the equation: 4y!=0

y!=0/4

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

1/4y-(+1/7y)-5=0

We get rid of parentheses

1/4y-1/7y-5=0

We calculate fractions


7y/28y^2+(-4y)/28y^2-5=0

We multiply all the terms by the denominator


7y+(-4y)-5*28y^2=0

Wy multiply elements

-140y^2+7y+(-4y)=0

We get rid of parentheses

-140y^2+7y-4y=0

We add all the numbers together, and all the variables

-140y^2+3y=0

a = -140; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-140)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-140}=\frac{-6}{-280}
=3/140
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-140}=\frac{0}{-280}
=0
$