2/3(k+2)=3/4(2k-1)

Solution for 2/3(k+2)=3/4(2k-1) equation:

2/3(k+2)=3/4(2k-1)

We move all terms to the left:

2/3(k+2)-(3/4(2k-1))=0


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

k∈R



Domain of the equation: 4(2k-1))!=0

k∈R


We calculate fractions


(8k2/(3(k+2)*4(2k-1)))+(-9kk/(3(k+2)*4(2k-1)))=0


We calculate terms in parentheses: +(8k2/(3(k+2)*4(2k-1))), so:

8k2/(3(k+2)*4(2k-1))

We multiply all the terms by the denominator


8k2

We add all the numbers together, and all the variables

8k^2

Back to the equation:

+(8k^2)



We calculate terms in parentheses: +(-9kk/(3(k+2)*4(2k-1))), so:

-9kk/(3(k+2)*4(2k-1))

We multiply all the terms by the denominator


-9kk

Back to the equation:

+(-9kk)


We get rid of parentheses

8k^2-9kk=0