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

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



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

We move all terms to the left:

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

We multiply parentheses

10z-4z-(2(z-2)+3(z-4))+20+8=0



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

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

We multiply parentheses

2z+3z-4-12

We add all the numbers together, and all the variables

5z-16

Back to the equation:

-(5z-16)



We add all the numbers together, and all the variables

6z-(5z-16)+28=0

We get rid of parentheses

6z-5z+16+28=0

We add all the numbers together, and all the variables

z+44=0

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

z=-44

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