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3 unknown variables- I'm lost one this one right now

  1. Feb 25, 2004 #1
    Can someone explain to me how to do this?

    If a(x-2)+4(5x+b)=23x-30 find the values of a and b



    Thank you
     
  2. jcsd
  3. Feb 25, 2004 #2
    Expand out the left side:

    a(x-2)+4(5x+b)
    ax - 2a + 20x + 4b
    (a + 20)x + (4b - 2a)

    This gives us the equation:

    (a + 20)x + (4b - 2a) = 23x - 30

    Now, the coefficients on the x terms on each side and the constant terms on each side must be equal, so this single equation is actual a pair of equations. Can you see them?

    cookiemonster
     
  4. Feb 26, 2004 #3

    HallsofIvy

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    IF dmbeluke's TITLE was correct, then this problem is impossible. If a, b, and x are all unknown numbers then we have a single equation in 3 unkowns and there are an infinite number of possible solutions.

    However,cookiemonster is assuming (very likely correctly) what dmbeluke didn't say: that x is a variable and the equation must be true for all values of x. If two polynomials are equal for all x, then corresponding coefficients must be equal.

    Another way to do this: since this is true for all x, choose any two convenient values of x to get two equations for a and b.

    Since a(x-2)+4(5x+b)=23x-30 for all x, in particular, if we take x= 2, we get a(2-2)+ 4(5(2)+ b= 23(2)- 30
    4(10+b)= 46-30 or 40+ 4b= 16.
    (I chose x= 2, of course, precisely because of that "a(x-2)")

    After you have found b, take x= 0 to get a simple equation for a.
     
    Last edited: Feb 27, 2004
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