(1/5y-1)=2/3y+17

Solution for (1/5y-1)=2/3y+17 equation:

(1/5y-1)=2/3y+17

We move all terms to the left:

(1/5y-1)-(2/3y+17)=0


Domain of the equation: 5y-1)!=0

y∈R



Domain of the equation: 3y+17)!=0

y∈R


We get rid of parentheses

1/5y-2/3y-1-17=0

We calculate fractions


3y/15y^2+(-10y)/15y^2-1-17=0

We add all the numbers together, and all the variables

3y/15y^2+(-10y)/15y^2-18=0

We multiply all the terms by the denominator


3y+(-10y)-18*15y^2=0

Wy multiply elements

-270y^2+3y+(-10y)=0

We get rid of parentheses

-270y^2+3y-10y=0

We add all the numbers together, and all the variables

-270y^2-7y=0

a = -270; b = -7; c = 0;
Δ = b2-4ac
Δ = -72-4·(-270)·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7}{2*-270}=\frac{0}{-540}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7}{2*-270}=\frac{14}{-540}
=-7/270
$