1/3y+7=3y-10

Solution for 1/3y+7=3y-10 equation:

1/3y+7=3y-10

We move all terms to the left:

1/3y+7-(3y-10)=0


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R


We get rid of parentheses

1/3y-3y+10+7=0

We multiply all the terms by the denominator


-3y*3y+10*3y+7*3y+1=0

Wy multiply elements

-9y^2+30y+21y+1=0

We add all the numbers together, and all the variables

-9y^2+51y+1=0

a = -9; b = 51; c = +1;
Δ = b2-4ac
Δ = 512-4·(-9)·1
Δ = 2637
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2637}=\sqrt{9*293}=\sqrt{9}*\sqrt{293}=3\sqrt{293}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(51)-3\sqrt{293}}{2*-9}=\frac{-51-3\sqrt{293}}{-18}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(51)+3\sqrt{293}}{2*-9}=\frac{-51+3\sqrt{293}}{-18}
$