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

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Solution for 10(z+2)-4(z-1)=3(z-2)+2(z-2) equation: 



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

We move all terms to the left:

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

We multiply parentheses

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



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

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

We multiply parentheses

3z+2z-6-4

We add all the numbers together, and all the variables

5z-10

Back to the equation:

-(5z-10)



We add all the numbers together, and all the variables

6z-(5z-10)+24=0

We get rid of parentheses

6z-5z+10+24=0

We add all the numbers together, and all the variables

z+34=0

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

z=-34

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