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1/2(10x+4)-20=-1/3(18x-9)
We move all terms to the left:
1/2(10x+4)-20-(-1/3(18x-9))=0
Domain of the equation: 2(10x+4)!=0
x∈R
Domain of the equation: 3(18x-9))!=0We calculate fractions
x∈R
(3x1/(2(10x+4)*3(18x-9)))+(-(-2x1)/(2(10x+4)*3(18x-9)))-20=0
We calculate terms in parentheses: +(3x1/(2(10x+4)*3(18x-9))), so:
3x1/(2(10x+4)*3(18x-9))
We multiply all the terms by the denominator
3x1
We add all the numbers together, and all the variables
3x
Back to the equation:
+(3x)
We calculate terms in parentheses: +(-(-2x1)/(2(10x+4)*3(18x-9))), so:We add all the numbers together, and all the variables
-(-2x1)/(2(10x+4)*3(18x-9))
We add all the numbers together, and all the variables
-(-2x)/(2(10x+4)*3(18x-9))
We multiply all the terms by the denominator
-(-2x)
We get rid of parentheses
2x
Back to the equation:
+(2x)
5x-20=0
We move all terms containing x to the left, all other terms to the right
5x=20
x=20/5
x=4
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