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

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



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

We move all terms to the left:

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

We multiply parentheses

10z-4z-(2(z-2)+3(z-2))+20+12=0



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

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

We multiply parentheses

2z+3z-4-6

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)+32=0

We get rid of parentheses

6z-5z+10+32=0

We add all the numbers together, and all the variables

z+42=0

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

z=-42

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