1/2y-4=1/5y+7

Solution for 1/2y-4=1/5y+7 equation:

1/2y-4=1/5y+7

We move all terms to the left:

1/2y-4-(1/5y+7)=0


Domain of the equation: 2y!=0

y!=0/2

y!=0

y∈R



Domain of the equation: 5y+7)!=0

y∈R


We get rid of parentheses

1/2y-1/5y-7-4=0

We calculate fractions


5y/10y^2+(-2y)/10y^2-7-4=0

We add all the numbers together, and all the variables

5y/10y^2+(-2y)/10y^2-11=0

We multiply all the terms by the denominator


5y+(-2y)-11*10y^2=0

Wy multiply elements

-110y^2+5y+(-2y)=0

We get rid of parentheses

-110y^2+5y-2y=0

We add all the numbers together, and all the variables

-110y^2+3y=0

a = -110; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-110)·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*-110}=\frac{-6}{-220}
=3/110
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-110}=\frac{0}{-220}
=0
$