P Panama New member Joined Nov 29, 2008 Messages 29 Nov 29, 2008 #1 Without actually solving the equation list all possible numbers that would be rejected if they appeared as potential solutions 1/(2x) + 1/(3x) = x/(3) I think the answer is 0. Is this correct?
Without actually solving the equation list all possible numbers that would be rejected if they appeared as potential solutions 1/(2x) + 1/(3x) = x/(3) I think the answer is 0. Is this correct?
D Deleted member 4993 Guest Nov 29, 2008 #2 Panama said: Without actually solving the equation list all possible numbers that would be rejected if they appeared as potential solutions 1/(2x) + 1/(3x) = x/(3) I think the answer is 0. Is this correct? <<<< Correct Click to expand... As x increases - the left-hand-side(LHS) decreases but the right-hand-side(RHS) increases. At x=3, quick mental math tells you that RHS is way bigger. So |x|<3 is an approximate bound on the value of x.
Panama said: Without actually solving the equation list all possible numbers that would be rejected if they appeared as potential solutions 1/(2x) + 1/(3x) = x/(3) I think the answer is 0. Is this correct? <<<< Correct Click to expand... As x increases - the left-hand-side(LHS) decreases but the right-hand-side(RHS) increases. At x=3, quick mental math tells you that RHS is way bigger. So |x|<3 is an approximate bound on the value of x.
