((2z-36)+z)/2=z+15

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Solution for ((2z-36)+z)/2=z+15 equation: 



((2z-36)+z)/2=z+15

We move all terms to the left:

((2z-36)+z)/2-(z+15)=0

We get rid of parentheses

((2z-36)+z)/2-z-15=0

We multiply all the terms by the denominator


((2z-36)+z)-z*2-15*2=0



We calculate terms in parentheses: +((2z-36)+z), so:

(2z-36)+z

We add all the numbers together, and all the variables

z+(2z-36)

We get rid of parentheses

z+2z-36

We add all the numbers together, and all the variables

3z-36

Back to the equation:

+(3z-36)



We add all the numbers together, and all the variables

(3z-36)-z*2-30=0

Wy multiply elements

(3z-36)-2z-30=0

We get rid of parentheses

3z-2z-36-30=0

We add all the numbers together, and all the variables

z-66=0

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

z=66

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