150=(3z+2)+(4z+3)+(3z+2)+(4z+3)

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Solution for 150=(3z+2)+(4z+3)+(3z+2)+(4z+3) equation: 



150=(3z+2)+(4z+3)+(3z+2)+(4z+3)

We move all terms to the left:

150-((3z+2)+(4z+3)+(3z+2)+(4z+3))=0



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

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

We get rid of parentheses

3z+4z+3z+4z+2+3+2+3

We add all the numbers together, and all the variables

14z+10

Back to the equation:

-(14z+10)



We get rid of parentheses

-14z-10+150=0

We add all the numbers together, and all the variables

-14z+140=0

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

-14z=-140

z=-140/-14

z=+10

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