2/3y+10=-1/2y+8

Solution for 2/3y+10=-1/2y+8 equation:

2/3y+10=-1/2y+8

We move all terms to the left:

2/3y+10-(-1/2y+8)=0


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R



Domain of the equation: 2y+8)!=0

y∈R


We get rid of parentheses

2/3y+1/2y-8+10=0

We calculate fractions


4y/6y^2+3y/6y^2-8+10=0

We add all the numbers together, and all the variables

4y/6y^2+3y/6y^2+2=0

We multiply all the terms by the denominator


4y+3y+2*6y^2=0

We add all the numbers together, and all the variables

7y+2*6y^2=0

Wy multiply elements

12y^2+7y=0

a = 12; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·12·0
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*12}=\frac{-14}{24}
=-7/12
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*12}=\frac{0}{24}
=0
$