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

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



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

We move all terms to the left:

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

We multiply parentheses

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



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

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

We multiply parentheses

(2z-1)+12z-4

We get rid of parentheses

2z+12z-1-4

We add all the numbers together, and all the variables

14z-5

Back to the equation:

-(14z-5)



We add all the numbers together, and all the variables

10z-(14z-5)+9=0

We get rid of parentheses

10z-14z+5+9=0

We add all the numbers together, and all the variables

-4z+14=0

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

-4z=-14

z=-14/-4

z=3+1/2

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