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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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



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

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

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

We add all the numbers together, and all the variables

2(z-3)-6

We multiply parentheses

2z-6-6

We add all the numbers together, and all the variables

2z-12

Back to the equation:

-(2z-12)



We add all the numbers together, and all the variables

6z-(2z-12)+18=0

We get rid of parentheses

6z-2z+12+18=0

We add all the numbers together, and all the variables

4z+30=0

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

4z=-30

z=-30/4

z=-7+1/2

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