1/2(z+4)-6=2/3(5-z)

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Solution for 1/2(z+4)-6=2/3(5-z) equation:



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

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