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Trying to teach myself equations

  1. Mar 19, 2006 #1
    I'm trying to teach myself equations and I'm stuck on equations with just letters. I can easily solve for x with: [tex] 34 - 10x = 6x + 2 [/tex]

    x=2

    I'm having trouble with:
    [tex] ax = bx + c [/tex]
     
    Last edited: Mar 19, 2006
  2. jcsd
  3. Mar 19, 2006 #2

    Integral

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    How did you arrive at the conclusion x=2 in your example?
     
  4. Mar 19, 2006 #3
    I subtracted 2 from each side using the "golden rule of equations"
    .

    That gave [tex] 32 - 10x = 6x [/tex]

    then [tex] 32 - 10x + 10x =6x + 10x [/tex]

    This LaTeX is giving me some trouble here, bare with me.
    thanks for your help!
     
    Last edited: Mar 20, 2006
  5. Mar 19, 2006 #4
    I think you have the concept down perfectly. Now can you apply what you did with that equation to the other one

    ax = bx + c ?
     
  6. Mar 19, 2006 #5
    then [tex]\frac{32} {16}[/tex] [tex]\frac{16x} {16}[/tex] x=2
     
  7. Mar 19, 2006 #6

    chroot

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    Move ax and bx to the same side of the equation by subtracting bx from both sides:

    ax - bx = c

    Factor out the x on the left:

    (a - b) x = c

    Divide both sides by (a-b):

    x = c / (a-b)

    You already know how to do algebra! The fact that they're letters instead of numbers changes nothing.

    - Warren
     
  8. Mar 20, 2006 #7
    I know where you're coming from though with the letters and all. However, in pre calc we use ALOT of problems with letters instead of numbers to find other formulas.
     
  9. Mar 20, 2006 #8
    Thanks for the help!

    When chroot factored out the [tex]x[/tex] on the left, what rule/ property/operations did he use to acheive that (and when to use parenthesis ( ).. )? My book never showed me examples or really explained why or how to do that.(atleast from what I read)

    All I know is:
    Postualates of equality/operations


    Also, I know there can be several ways to work out an equation, but where do you people usually start to break down these problems when solving for x.

    I usually do it the wrong way before I get it the right way. I know there are no shortcuts to learning but I'm just wondering if I overlooked something here.
     
  10. Mar 20, 2006 #9

    chroot

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    I just used the distributive property of multiplication.

    (a-b)*x = ax - bx.

    - Warren
     
  11. Mar 20, 2006 #10
    distributive property of multiplication. Ha, thanks
     
  12. Mar 20, 2006 #11

    chroot

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    Got it?

    - Warren
     
  13. Mar 21, 2006 #12

    I need to search around for more practice equations until it comes natural to me.

    My question is what does the / stand for in the answer ..... divide?
     
  14. Mar 21, 2006 #13
    b + ax = c
    b - b + ax = c - b
    ax = (c - b)

    What are we doing to both sides to isolate the x on that last line? I got the answer only because I saw a pattern, but I'm unsure how I got it. What let me isolate x by moving a to the other side of the problem?

    x = (c - b)/a
     
    Last edited: Mar 21, 2006
  15. Mar 21, 2006 #14

    Hootenanny

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    Think of it like this; You divided both sides by [itex]a[/itex]. You could add an extra step between the penultimate and final line if you like;

    [tex]ax = (c-b)[/tex]

    [tex]\frac{ax}{a} = \frac{(c-b)}{a} \Rightarrow \frac{\not{a}x}{\not{a}} = \frac{(c-b)}{a} [/tex]

    [tex]x = \frac{(c-b)}{a}[/tex]

    You can do this because you did the same thing to both sides, i.e. divided both sides by a. You can do anything to can equation as long as you do it to both sides of the equals sign. :smile:
     
  16. Mar 21, 2006 #15
    Yes , thanks for showing me that, I understand them now.
    However, when I look at this problem I want to try using the d first on both sides...... Is there something you see in the equation that tells you to use b first. Because while using d I came to this

    ax - b = cx + d

    ax- b - d = cx + d - d

    ax - (b - d) = cx

    ax - ax - (b - d) = cx - ax


    [tex]
    \frac{-(b - d)}{c - a} [/tex][tex]
    \frac{cx - ax}{c - a}
    [/tex]

    -(b - d)/(c-a) = x ............. what does the -(b - d) part equal? Is it (b + d)?
     
    Last edited: Mar 21, 2006
  17. Mar 21, 2006 #16
    Then I tried the b first:

    ax - b = cx + d

    ax - b + b = cx + d + b

    ax = cx + (d + b)

    ax - cx = cx - cx + (d + b)

    [tex]
    \frac{ax - cx}{a - c}[/tex][tex]
    \frac{(d+b)}{a-c}
    [/tex]

    x= (d+b)/(a-c)
     
  18. Mar 21, 2006 #17
    That step is incorrect:
    -b-d =/= -(b-d)
     
  19. Mar 21, 2006 #18

    Hootenanny

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    A simpler way of doing to would be to move the cx over to the left first. Then the b over to the right. Then factorise;

    [tex]ax - b = cx + d[/tex]

    Add b to both sides and subtract cx from both sides

    [tex]ax - cx = d + b[/tex]

    Factorise x;

    [tex]x(a - c) = d + b[/tex]

    Divide both sides by the bracket;

    [tex]x = \frac{d+b}{a-c}[/tex]

    To answer your earlier question;
    No [itex]-(b-d) = d-b[/itex]. You have made an error in your first attempt.

    This step

    [tex] \frac{ax - cx}{a - c}= \frac{(d+b)}{a-c}[/tex]
    [tex]x= \frac{(d+b)}{a-c}[/tex]

    Is not correct because

    [tex]\frac{ax - cx}{a - c} \not{=} x[/tex]

    [Edit] Libertine got there before me :smile:[/Edit]
     
  20. Mar 22, 2006 #19
    Ok, I've been doing them fairly good now using th distributive property of multiplication chroot showed me. I did these correct I presume?

    10 + 5x = 100 - 4x

    (4 + 5)x = 100 - 10

    [tex] \frac{9x}{9}= \frac{90}{9}[/tex]

    x=10


    --------------------------------------
    10 + 5x = 100 - 4x

    10 - 10 + 5x = 100 - 10 - 4x

    5x = 90 - 4x

    5x + 4x = 90

    [tex] \frac{9x}{9}= \frac{90}{9}[/tex]

    x=10


    Thanks for taking the time with me.


    cheers everyone
     
  21. Mar 22, 2006 #20

    chroot

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    Looks like you've got it, Vincent Vega.

    I assume you understand the objections raised by Hootenanny and Libertine in the last few posts?

    - Warren
     
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