OG - pg. 339 - #8

[margaretmyoder]

I would have posted this question in the Official Guide forum... but I can't write a new post in that forum. Apologies.

OG pg 339 #8: (1-x) / (x-1) = (1/x)
Quantity A = x
Quantity B = -.5

This is how I tried to solve the problem:

[ (1-x) / (x-1) = (1/x) ] * x <----- I multiplied both sides of the equation by x
(x-x^2) / (x-1) = 1
[ (x-x^2) / (x-1) = 1 ] * (x-1) <----- I multiplied both sides by (x-1) to get rid of the denominator on the left side of the equation
x - x^2 = x - 1
[ x - x^2 = x - 1 ] <----- I subtracted x from each side
-x^2 = -1
[ -x^2 = -1 ] * (-1) <----- I multiplied each side by -1
x^2 = 1
x = 1 or -1 <----- I solved for x by taking its square root

My original answer was D because 1 is greater than -.5 and -1 is less than -.5. However, the answer is B. Is this because x can't equal 1 because that would make the fraction in the original equation (1-x)/(x-1) undefined? Therefore, x is equal to -1 and -1 is less than -.5. Is that correct?

[tommywallach]

Yep! You nailed it. You did everything right, but you ALWAYS have to be on the lookout for "undefined" situations when you see fractions with variables.

-t