3(2z+1)+4(z+3)=5(2z+1)+4(3z+1)

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

3(2z+1)+4(z+3)=5(2z+1)+4(3z+1)

We move all terms to the left:

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

We multiply parentheses

6z+4z-(5(2z+1)+4(3z+1))+3+12=0


We calculate terms in parentheses: -(5(2z+1)+4(3z+1)), so:

5(2z+1)+4(3z+1)

We multiply parentheses

10z+12z+5+4

We add all the numbers together, and all the variables

22z+9

Back to the equation:

-(22z+9)


We add all the numbers together, and all the variables

10z-(22z+9)+15=0

We get rid of parentheses

10z-22z-9+15=0

We add all the numbers together, and all the variables

-12z+6=0

We move all terms containing z to the left, all other terms to the right

-12z=-6

z=-6/-12

z=1/2