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(1/2x)-7=(1/3)(x-12)
We move all terms to the left:
(1/2x)-7-((1/3)(x-12))=0
Domain of the equation: 2x)!=0
x!=0/1
x!=0
x∈R
Domain of the equation: 3)(x-12))!=0We add all the numbers together, and all the variables
x∈R
(+1/2x)-((+1/3)(x-12))-7=0
We get rid of parentheses
1/2x-((+1/3)(x-12))-7=0
We multiply parentheses ..
-((+x^2+1/3*-12))+1/2x-7=0
We calculate fractions
(-((+x^2+1*2x)/(-12))*2x+3*)+()/(-12))*2x+3*)-7=0
We calculate terms in parentheses: +(-((+x^2+1*2x)/(-12))*2x+3*), so:
-((+x^2+1*2x)/(-12))*2x+3*
We add all the numbers together, and all the variables
-((+x^2+1*2x)/(-12))*2x
We multiply all the terms by the denominator
-((+x^2+1*2x)
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(+x^2+1*2x
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(+x^2+1*2x
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