5/8y-1=1+7/10y

Solution for 5/8y-1=1+7/10y equation:

5/8y-1=1+7/10y

We move all terms to the left:

5/8y-1-(1+7/10y)=0


Domain of the equation: 8y!=0

y!=0/8

y!=0

y∈R



Domain of the equation: 10y)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

5/8y-(7/10y+1)-1=0

We get rid of parentheses

5/8y-7/10y-1-1=0

We calculate fractions


50y/80y^2+(-56y)/80y^2-1-1=0

We add all the numbers together, and all the variables

50y/80y^2+(-56y)/80y^2-2=0

We multiply all the terms by the denominator


50y+(-56y)-2*80y^2=0

Wy multiply elements

-160y^2+50y+(-56y)=0

We get rid of parentheses

-160y^2+50y-56y=0

We add all the numbers together, and all the variables

-160y^2-6y=0

a = -160; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·(-160)·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*-160}=\frac{0}{-320}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*-160}=\frac{12}{-320}
=-3/80
$