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1/3(1z+4)-6=2/3(5-1z)
We move all terms to the left:
1/3(1z+4)-6-(2/3(5-1z))=0
Domain of the equation: 3(1z+4)!=0
z∈R
Domain of the equation: 3(5-1z))!=0We add all the numbers together, and all the variables
z∈R
1/3(z+4)-(2/3(-1z+5))-6=0
We calculate fractions
(3z(-)/(3(z+4)*3(-1z+5)))+(-6zz/(3(z+4)*3(-1z+5)))-6=0
We calculate terms in parentheses: +(3z(-)/(3(z+4)*3(-1z+5))), so:
3z(-)/(3(z+4)*3(-1z+5))
We add all the numbers together, and all the variables
3z0/(3(z+4)*3(-1z+5))
We multiply all the terms by the denominator
3z0
We add all the numbers together, and all the variables
3z
Back to the equation:
+(3z)
We calculate terms in parentheses: +(-6zz/(3(z+4)*3(-1z+5))), so:We get rid of parentheses
-6zz/(3(z+4)*3(-1z+5))
We multiply all the terms by the denominator
-6zz
Back to the equation:
+(-6zz)
3z-6zz-6=0
We move all terms containing z to the left, all other terms to the right
3z-6zz=6
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