2.3y+3.2-y=2.1+1/3y+1.1

Solution for 2.3y+3.2-y=2.1+1/3y+1.1 equation:

2.3y+3.2-y=2.1+1/3y+1.1

We move all terms to the left:

2.3y+3.2-y-(2.1+1/3y+1.1)=0


Domain of the equation: 3y+1.1)!=0

y∈R


We add all the numbers together, and all the variables

2.3y-y-(1/3y+3.2)+3.2=0

We add all the numbers together, and all the variables

1.3y-(1/3y+3.2)+3.2=0

We get rid of parentheses

1.3y-1/3y-3.2+3.2=0

We multiply all the terms by the denominator


(1.3y)*3y-(3.2)*3y+(3.2)*3y-1=0

We add all the numbers together, and all the variables

(+1.3y)*3y-(3.2)*3y+(3.2)*3y-1=0

We multiply parentheses

3y^2-9.6y+9.6y-1=0

We add all the numbers together, and all the variables

3y^2-1=0

a = 3; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·3·(-1)
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{3}}{2*3}=\frac{0-2\sqrt{3}}{6}
=-\frac{2\sqrt{3}}{6}
=-\frac{\sqrt{3}}{3}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{3}}{2*3}=\frac{0+2\sqrt{3}}{6}
=\frac{2\sqrt{3}}{6}
=\frac{\sqrt{3}}{3}
$