(4z-5)+(2z+7)=(3z-4)+2(z+3)

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Solution for (4z-5)+(2z+7)=(3z-4)+2(z+3) equation: 



(4z-5)+(2z+7)=(3z-4)+2(z+3)

We move all terms to the left:

(4z-5)+(2z+7)-((3z-4)+2(z+3))=0

We get rid of parentheses

4z+2z-((3z-4)+2(z+3))-5+7=0



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

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

We multiply parentheses

(3z-4)+2z+6

We get rid of parentheses

3z+2z-4+6

We add all the numbers together, and all the variables

5z+2

Back to the equation:

-(5z+2)



We add all the numbers together, and all the variables

6z-(5z+2)+2=0

We get rid of parentheses

6z-5z-2+2=0

We add all the numbers together, and all the variables

z=0

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