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

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

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

We move all terms to the left:

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


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

k∈R



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

k∈R


We calculate fractions


(2k2/(3(3k+5)*2(2k-5)))+(-9k3/(3(3k+5)*2(2k-5)))=0


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

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

We multiply all the terms by the denominator


2k2

We add all the numbers together, and all the variables

2k^2

Back to the equation:

+(2k^2)



We calculate terms in parentheses: +(-9k3/(3(3k+5)*2(2k-5))), so:

-9k3/(3(3k+5)*2(2k-5))

We multiply all the terms by the denominator


-9k3

We add all the numbers together, and all the variables

-9k^3

We do not support ekpression: k^3