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Two quick questions regarding simplification

  1. Sep 23, 2011 #1
    Could someone please explain how

    [tex]\frac{x^2}{x^2+2}=1-\frac{2}{x^2+2}[/tex]

    and how

    [tex]xlnx-x=xln(x-1)[/tex]

    Also, is there a list of types of simplifications like these two that I could study? I'm in Calc II and my own trouble comes from simple things like this that I should already know.
     
  2. jcsd
  3. Sep 23, 2011 #2

    Mark44

    Staff: Mentor

    Start on the right side, and get a common denominator.
    Are you sure you copied this correctly, because as you have it, it's not an identity.

    xlnx - x = x(lnx - 1), by factoring x from both terms on the left side of the equation.

    I'm not sure that there's a list of types of simplifications. It's probably better to go back and review the techniques of algebra, in your textbook, if you still have it, or online at a site such as Khan Academy.
     
  4. Sep 23, 2011 #3

    symbolipoint

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    You can multiply the left side and the right side by x2+2.
     
  5. Sep 23, 2011 #4
    Let's not make things unnecessarily complicated. He only needs to deal with one side of this equation.
     
  6. Sep 23, 2011 #5

    Mentallic

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    It basically takes advantage of this rule for fractions

    [tex]\frac{a}{a+b}=\frac{a+b-b}{a+b}=\frac{a+b}{a+b}+\frac{-b}{a+b}=1-\frac{b}{a+b}[/tex]
     
  7. Sep 23, 2011 #6

    symbolipoint

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    Yes, I see what you mean:

    From equation shown, [tex]\frac{x^2}{x^2+2}=1-\frac{2}{x^2+2}[/tex]

    then,
    [tex]\[
    \begin{array}{l}
    \left( {\frac{{x^2 }}{{x^2 + 2}}} \right)\left( {x^2 + 2} \right) = \left( {1 - \frac{2}{{x^2 + 2}}} \right)\left( {x^2 + 2} \right) \\
    x^2 = (x^2 + 2) - 2 \\
    x^2 = x^2 \\
    \end{array}
    \]
    [/tex]
     
  8. Sep 24, 2011 #7

    Mark44

    Staff: Mentor

    The trouble with this approach is that you are tacitly assuming that the two quantities of the first equation are equal when you multiply both sides by x2 + 2. The bottom line here shows that x2 = x2, which is true, but not useful. A complete proof would go on to show that since each operation applied is reversible (one-to-one, then each equation above is equivalent.

    An easier and more straightforward way to do this problem is to show that the right side of the original equation can be manipulated to produce the left side.
     
  9. Sep 24, 2011 #8
    To simplify your solution:

    [tex]\frac{x^2}{x^2+2}=1-\frac{2}{x^2+2}[/tex]

    [tex]\frac{x^2}{x^2+2}=(1*\frac{x^2+2}{x^2+2})-\frac{2}{x^2+2}[/tex]

    Multiplying the numerator and denominator by the same values does not change the quotient of a fraction.

    Now we simplify and the conclusion is obvious:

    [tex]\frac{x^2}{x^2+2}=\frac{x^2}{x^2+2}[/tex]
     
  10. Sep 24, 2011 #9

    symbolipoint

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    Now it's clearer. The poster wanted to know HOW. I missed seeing that and attempted to solve an equation, while what was really needed was "verify the identity".
     
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