15-(3z+1)=4(z+1)+3z

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Solution for 15-(3z+1)=4(z+1)+3z equation: 



15-(3z+1)=4(z+1)+3z

We move all terms to the left:

15-(3z+1)-(4(z+1)+3z)=0

We get rid of parentheses

-3z-(4(z+1)+3z)-1+15=0



We calculate terms in parentheses: -(4(z+1)+3z), so:

4(z+1)+3z

We add all the numbers together, and all the variables

3z+4(z+1)

We multiply parentheses

3z+4z+4

We add all the numbers together, and all the variables

7z+4

Back to the equation:

-(7z+4)



We add all the numbers together, and all the variables

-3z-(7z+4)+14=0

We get rid of parentheses

-3z-7z-4+14=0

We add all the numbers together, and all the variables

-10z+10=0

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

-10z=-10

z=-10/-10

z=1

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