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

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



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

We move all terms to the left:

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

We multiply parentheses

14z-3z-(4(z-1)+6(z-2))+28+9=0



We calculate terms in parentheses: -(4(z-1)+6(z-2)), so:

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

We multiply parentheses

4z+6z-4-12

We add all the numbers together, and all the variables

10z-16

Back to the equation:

-(10z-16)



We add all the numbers together, and all the variables

11z-(10z-16)+37=0

We get rid of parentheses

11z-10z+16+37=0

We add all the numbers together, and all the variables

z+53=0

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

z=-53

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