4(x-3)/6=2(x+1)/7

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Solution for 4(x-3)/6=2(x+1)/7 equation:



4(x-3)/6=2(x+1)/7
We move all terms to the left:
4(x-3)/6-(2(x+1)/7)=0
We calculate fractions
4x/()+(-(2(x+1)*6)/()=0
We calculate terms in parentheses: +(-(2(x+1)*6)/(), so:
-(2(x+1)*6)/(
We multiply all the terms by the denominator
-(2(x+1)*6)
We calculate terms in parentheses: -(2(x+1)*6), so:
2(x+1)*6
We multiply parentheses
12x+12
Back to the equation:
-(12x+12)
We get rid of parentheses
-12x-12
Back to the equation:
+(-12x-12)
We get rid of parentheses
4x/()-12x-12=0
We multiply all the terms by the denominator
4x-12x*()-12*()=0
We add all the numbers together, and all the variables
4x-12x*()=0

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