2(1+3k)=4+3(2k-5)

Solution for 2(1+3k)=4+3(2k-5) equation:

2(1+3k)=4+3(2k-5)

We move all terms to the left:

2(1+3k)-(4+3(2k-5))=0

We add all the numbers together, and all the variables

2(3k+1)-(4+3(2k-5))=0

We multiply parentheses

6k-(4+3(2k-5))+2=0


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

4+3(2k-5)

determiningTheFunctionDomain
3(2k-5)+4

We multiply parentheses

6k-15+4

We add all the numbers together, and all the variables

6k-11

Back to the equation:

-(6k-11)


We get rid of parentheses

6k-6k+11+2=0

We add all the numbers together, and all the variables

13!=0

There is no solution for this equation