1/2(2h-1)=1/3(2h+1)

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

1/2(2h-1)=1/3(2h+1)

We move all terms to the left:

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


Domain of the equation: 2(2h-1)!=0

h∈R



Domain of the equation: 3(2h+1))!=0

h∈R


We calculate fractions


(3h2/(2(2h-1)*3(2h+1)))+(-2h2/(2(2h-1)*3(2h+1)))=0


We calculate terms in parentheses: +(3h2/(2(2h-1)*3(2h+1))), so:

3h2/(2(2h-1)*3(2h+1))

We multiply all the terms by the denominator


3h2

We add all the numbers together, and all the variables

3h^2

Back to the equation:

+(3h^2)



We calculate terms in parentheses: +(-2h2/(2(2h-1)*3(2h+1))), so:

-2h2/(2(2h-1)*3(2h+1))

We multiply all the terms by the denominator


-2h2

We add all the numbers together, and all the variables

-2h^2

Back to the equation:

+(-2h^2)


We add all the numbers together, and all the variables

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

We get rid of parentheses

3h^2-2h^2=0

We add all the numbers together, and all the variables

h^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:
$h=\frac{-b}{2a}=\frac{0}{2}=0$