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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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



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

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

We multiply parentheses

4z+2z-8-4

We add all the numbers together, and all the variables

6z-12

Back to the equation:

-(6z-12)



We add all the numbers together, and all the variables

10z-(6z-12)+43=0

We get rid of parentheses

10z-6z+12+43=0

We add all the numbers together, and all the variables

4z+55=0

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

4z=-55

z=-55/4

z=-13+3/4

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