5(z-4)=2(3-z)+5

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Solution for 5(z-4)=2(3-z)+5 equation:



5(z-4)=2(3-z)+5
We move all terms to the left:
5(z-4)-(2(3-z)+5)=0
We add all the numbers together, and all the variables
5(z-4)-(2(-1z+3)+5)=0
We multiply parentheses
5z-(2(-1z+3)+5)-20=0
We calculate terms in parentheses: -(2(-1z+3)+5), so:
2(-1z+3)+5
We multiply parentheses
-2z+6+5
We add all the numbers together, and all the variables
-2z+11
Back to the equation:
-(-2z+11)
We get rid of parentheses
5z+2z-11-20=0
We add all the numbers together, and all the variables
7z-31=0
We move all terms containing z to the left, all other terms to the right
7z=31
z=31/7
z=4+3/7

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