(7z+40)=(3z+112)

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Solution for (7z+40)=(3z+112) equation: 



(7z+40)=(3z+112)

We move all terms to the left:

(7z+40)-((3z+112))=0

We get rid of parentheses

7z-((3z+112))+40=0



We calculate terms in parentheses: -((3z+112)), so:

(3z+112)

We get rid of parentheses

3z+112

Back to the equation:

-(3z+112)



We get rid of parentheses

7z-3z-112+40=0

We add all the numbers together, and all the variables

4z-72=0

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

4z=72

z=72/4

z=18

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