2/3(2k-5)=1/2(2k-5)

Solution for 2/3(2k-5)=1/2(2k-5) equation:

2/3(2k-5)=1/2(2k-5)

We move all terms to the left:

2/3(2k-5)-(1/2(2k-5))=0


Domain of the equation: 3(2k-5)!=0

k∈R



Domain of the equation: 2(2k-5))!=0

k∈R


We calculate fractions


(4k2/(3(2k-5)*2(2k-5)))+(-3k2/(3(2k-5)*2(2k-5)))=0


We calculate terms in parentheses: +(4k2/(3(2k-5)*2(2k-5))), so:

4k2/(3(2k-5)*2(2k-5))

We multiply all the terms by the denominator


4k2

We add all the numbers together, and all the variables

4k^2

Back to the equation:

+(4k^2)



We calculate terms in parentheses: +(-3k2/(3(2k-5)*2(2k-5))), so:

-3k2/(3(2k-5)*2(2k-5))

We multiply all the terms by the denominator


-3k2

We add all the numbers together, and all the variables

-3k^2

Back to the equation:

+(-3k^2)


We add all the numbers together, and all the variables

4k^2+(-3k^2)=0

We get rid of parentheses

4k^2-3k^2=0

We add all the numbers together, and all the variables

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