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Precalculus Problem

  1. Jun 17, 2004 #1
    Problem: 1 - x-3/x-2 = 2x-3/x-2

    OK I know the answer in the back of our text book is: No Solution


    I work out the problem to get, x=2

    Here is how I got that.
    [x-2) - (x-3)]/x-2 = 2x-3/x-2

    Now I know that if 2 is put back in for x undefined values occur.

    1 - (undefined value) = (undefined value)

    So what I don't understand is how is x=2 a solution if it doesn't work in the original equation?

    And if it is not an actual solution than how come it came up as one?

    2nd Problem:
    [Earth Science]
    In 1984, the Soviets led the world in drilling the deepest hole in the Earth's crust--more than 12 kilometers deep. They found that below 3 kilometers the temperature T increased 2.5 (degrees) C for each additional 100 meters of depth.

    (A) If the temperature at 3 kilometers is 30 (degrees) C and x is the depth of the hole in kilometers, write an equation using x that will give the temperature T in the hole at any distance beyond 3 kilometers.

    Ok I figured that: T = 2.5[(x-3)/100] + 30

    But the answer is different...
    [Answer: T = 30 + 25(x-3)]

    But...if you put in a number, x in meters, in my equation, and put in a number x in km, in the 'back of the books equation' you get the same answer.

    T(4000 m) = 55 (degrees) C
    T(4 km) = 55 (degrees) C

    So...not sure which is 'right' per say, some help would be nice...


    -Tony Zalles
    Last edited: Jun 17, 2004
  2. jcsd
  3. Jun 17, 2004 #2


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    Look at each step you did, and consider when they can be reversed...

    Try putting the units on all of the numbers involved, then compare the two solutions to see if they're the same.
  4. Jun 17, 2004 #3

    Well looking at each step I took everything can be reversed except...the step I took between:

    [(x-2) - (x-3)]/x-2 = 2x-3/x-2



    Where I got rid of 'x-2' in the denominator on both sides....so I guess the point you're trying to drive at is that by getting rid of that part of the equation than I effectively get rid of the part that would otherwise make, x=2, not a solution, therefore coming to the conclusion in the back of the book that there is "No Solution".

    Is that right?

    Given that in the word problem they give their distance values in kilometers, I'm assuming that therefore they want an equation expressing T using x (the distance) expressed in kilometers. Which would make 'the back of the book equation' the correct answer.

    Was that what you were trying to point out to me?

    By the way thanks for the reply.
  5. Jun 17, 2004 #4


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    Furthermore, whenever you multiply both sides of an equation by something, you introduce potentially false "solutions" corresponding to when the thing you multiplied is zero. For instance, suppose we start with the nice, simple equation x = 3.
    Now, let's do some arithmetic:

    x = 3
    x (x - 2) = 3 (x - 2)
    x^2 - 2x = 3x - 6
    x ^ 2 - 5x + 6 = 0
    x = 2 or x = 3

    We see that multiplying by (x - 2) introduced a new "solution", x = 2.

    It is often wise to avoid multiplying (or dividing!) by things that are sometimes zero. To be entirely safe, you should either:

    (a) Try factoring instead
    (b) Split the problem into two cases; one where the thing always zero, and one where the thing is never zero (and thus it is safe to multiply or divide)
  6. Jun 17, 2004 #5
    Thanks that makes a lot of sense. :)

    (a) I understand,

    But would you mind providing an example where you would do (b) if you can think of one...I kind of get it but an example would reassure me that I really understand what you're explaining.

    BTW thanks again for the quick reply.
  7. Jun 17, 2004 #6


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    Suppose I want to solve the equation:

    x (x - 2) = 3 (x - 2)

    I split it into two cases: x = 2 and x != 2.

    In the case where x = 2, the equation is simply 0 = 0, so x = 2 is a solution.

    When x != 2, we can confidently divide by (x - 2) to get x = 3. Since all our steps are reversible, and 3 != 2, we see that 3 is a solution.
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