(2/x)-(2/(2x+1))=1/(x+1)
Could you please show work? I've been having trouble with this. The answer should be x=-2/3 but I can't seem to get to it.
Could you please show work? I've been having trouble with this. The answer should be x=-2/3 but I can't seem to get to it.
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Using the usual trick for adding fractions, a/b - c/d = (ad)/(bd) - (bc)/(bd) = (ad - bc)/(bd), the left hand side is
2/x - 2/(2x+1) = [2(2x+1) - 2(x)]/[x(2x+1)] = [4x + 2 - 2x]/[x(2x+1)] = [2x + 2]/[x(2x+1)]
So [2x + 2]/[x(2x+1)] = 1/(x+1). Multiplying both sides of this by x(x+1)(2x+1) we deduce (after some cancellation) that
(2x+2)(x+1) = x(2x+1).
The left hand side is 2x^2 + 2x + 2x + 2 = 2x^2 + 4x + 2 and the right hand side is 2x^2 + x. So 2x^2 + 4x + 2 = 2x^2 + x. Cancelling the 2x^2 we see that 4x + 2 = x and hence that 4x - x = -2 and hence that 3x = -2 and x = -2/3.
2/x - 2/(2x+1) = [2(2x+1) - 2(x)]/[x(2x+1)] = [4x + 2 - 2x]/[x(2x+1)] = [2x + 2]/[x(2x+1)]
So [2x + 2]/[x(2x+1)] = 1/(x+1). Multiplying both sides of this by x(x+1)(2x+1) we deduce (after some cancellation) that
(2x+2)(x+1) = x(2x+1).
The left hand side is 2x^2 + 2x + 2x + 2 = 2x^2 + 4x + 2 and the right hand side is 2x^2 + x. So 2x^2 + 4x + 2 = 2x^2 + x. Cancelling the 2x^2 we see that 4x + 2 = x and hence that 4x - x = -2 and hence that 3x = -2 and x = -2/3.