14(z+4)-4(z-2)=6(z-1)+3(z-2)

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Solution for 14(z+4)-4(z-2)=6(z-1)+3(z-2) equation: 



14(z+4)-4(z-2)=6(z-1)+3(z-2)

We move all terms to the left:

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

We multiply parentheses

14z-4z-(6(z-1)+3(z-2))+56+8=0



We calculate terms in parentheses: -(6(z-1)+3(z-2)), so:

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

We multiply parentheses

6z+3z-6-6

We add all the numbers together, and all the variables

9z-12

Back to the equation:

-(9z-12)



We add all the numbers together, and all the variables

10z-(9z-12)+64=0

We get rid of parentheses

10z-9z+12+64=0

We add all the numbers together, and all the variables

z+76=0

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

z=-76

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