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

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Solution for 10(z+4)-3(z-3)=4(z-4)+2(z-4) equation: 



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

We move all terms to the left:

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

We multiply parentheses

10z-3z-(4(z-4)+2(z-4))+40+9=0



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

4(z-4)+2(z-4)

We multiply parentheses

4z+2z-16-8

We add all the numbers together, and all the variables

6z-24

Back to the equation:

-(6z-24)



We add all the numbers together, and all the variables

7z-(6z-24)+49=0

We get rid of parentheses

7z-6z+24+49=0

We add all the numbers together, and all the variables

z+73=0

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

z=-73

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