1/2(2x+1)=1/3(3+2x)

Solution for 1/2(2x+1)=1/3(3+2x) equation:

1/2(2x+1)=1/3(3+2x)

We move all terms to the left:

1/2(2x+1)-(1/3(3+2x))=0


Domain of the equation: 2(2x+1)!=0

x∈R



Domain of the equation: 3(3+2x))!=0

x∈R


We add all the numbers together, and all the variables

1/2(2x+1)-(1/3(2x+3))=0

We calculate fractions


(3x2/(2(2x+1)*3(2x+3)))+(-2x2/(2(2x+1)*3(2x+3)))=0


We calculate terms in parentheses: +(3x2/(2(2x+1)*3(2x+3))), so:

3x2/(2(2x+1)*3(2x+3))

We multiply all the terms by the denominator


3x2

We add all the numbers together, and all the variables

3x^2

Back to the equation:

+(3x^2)



We calculate terms in parentheses: +(-2x2/(2(2x+1)*3(2x+3))), so:

-2x2/(2(2x+1)*3(2x+3))

We multiply all the terms by the denominator


-2x2

We add all the numbers together, and all the variables

-2x^2

Back to the equation:

+(-2x^2)


We add all the numbers together, and all the variables

3x^2+(-2x^2)=0

We get rid of parentheses

3x^2-2x^2=0

We add all the numbers together, and all the variables

x^2=0

a = 1; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·1·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{0}{2}=0$