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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