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Very Easy Algebra 1 Help

  1. Aug 21, 2009 #1
    Does this...

    (AX + B)(C - D)

    when you multiply

    the first numbers... AX and C

    mean the same thing as

    (AX)C -D(AX) +BC -BD

    or this

    AXC - DAX + BC - BD

    which one is it???
  2. jcsd
  3. Aug 21, 2009 #2
    Please I need to know
  4. Aug 21, 2009 #3
    Both are valid. The brackets do not change anything at this point.

  5. Aug 21, 2009 #4
    ok well the reason why I'm asking because in a physics problem it does so...

    like which one is it??

    like with out the brackets really changes it

    if you put the brackets on
  6. Aug 21, 2009 #5
    Could you possibly explain what A,X,B,C,D are? Are they numbers, matrices or what? Any concrete problem?

  7. Aug 21, 2009 #6
    ya sure...

    - (180 s)(a t3)

    this was line before

    -[Vo t3 + 2^-1 a t3^2] = 180 s (a t3) - a t3^2 + Vo (180 s) - Vo t3 - 1,100 m + Xo

    line before that

    -[Vo t3 + 2^-1 a t3^2] = (a t3 + Vo) (180 s - t3) - 1,100 m + Xo

    so there would be no ( ) around a t3???

    like aren't you doing this...

    (a t3)(180 s)

    and here's the line before that line

    1,100 m - [Xo + Vo t3 + 2^-1 a t3^2] = (a t3 + Vo) (180 s - t3)
  8. Aug 21, 2009 #7
    -[Vo t3 + 2^-1 a t3^2] = 180 s (a t3) - a t3^2 + Vo (180 s) - Vo t3 - 1,100 m + Xo

    see like in that line

    would you get this

    (180 s)a + (180 s)t3

    or just (180 s)a t3
  9. Aug 21, 2009 #8
    were s is seconds

    and the 3 is just a subscript
  10. Aug 21, 2009 #9
    please someone!!!
  11. Aug 21, 2009 #10


    User Avatar
    Homework Helper
    Education Advisor
    Gold Member

    You really need to understand commutative and associative properties of equality for Addition and Multiplication. Also you must understand the distributive property. This is year-1 Algebra stuff which you must learn to know formally and intuitively. Much of your progress relies on these properties and a few others.
  12. Aug 21, 2009 #11
    so which one is it???
  13. Aug 21, 2009 #12
    oh come on
  14. Aug 21, 2009 #13
    You're still not getting it ...

    A(BC) = ABC

    The parentheses make no difference so long as there are no sums in them.

    A(B+C), on the other hand, would give you AB + BC.

    Can you see where you were wrong above now? In post #7 one option is definitely correct and the other is definitely incorrect.
  15. Aug 21, 2009 #14
    ok yes just got anohter question...

    Ok well I got down to this and don't know what to do please someone help me...

    -2^-1 t3^2 - 180 s t3 + t3^2 = a^-1 (Vo (180 s) - 1,100 m + Xo)

    how do I solve this for t3

    the 3 is a subscript to t
    180 s is 180 seconds and is considered one term
    Vo is one term
    Xo is one term
  16. Aug 21, 2009 #15
    Everything in this equation is a constant except t3, correct?

    If so, then you've two terms with t3^2, one with t3 to the first power, and one term with no t3 (on the right side), i.e. a constant.

    That makes this a quadratic equation. Do you remember how to solve them?

    I'm guessing that you're not all that comfortable with algebra, given your original question, but maybe you are (?). If not, though, you need to brush up on it.
  17. Aug 21, 2009 #16
    I rember how to solve them but not when it's in this format
  18. Aug 21, 2009 #17
    y = ax^2 + bx + C

    I know how to solve for x... I think... it's been a while though please help...

    sorry lol
  19. Aug 21, 2009 #18
    Okay ... take what I'm about to type and have it tatooed on the back of one of your hands.

    if you have ax^2 + bx + c = 0 (no y, BTW - we're dealing with only one variable here),

    then the solution for x is given by the quadratic formula:

    x = (-b +/-sqrt(b^2 - 4ac))/2a ("sqrt" = square root)

    You must learn this now and never again forget it!! ;-)

    So ... you need to rearrange your equation for t3 so that it's in the form above, which you can do since you have terms with t3^2, t3 and constants, just like above for x.
  20. Aug 21, 2009 #19
    Ok I graphed into a graphic calculator

    y1 = -2 X^2 - 180 x + X^2

    y2 = .2^-1 (5.494*180-1100)

    and got this

    (3.111, -555.4)

    I assume I need the 3.111 seconds for my answer
    but sense this is physics not algebra how do I rearange for y3???
  21. Aug 21, 2009 #20
    ok but what does this have to do with the problem

    .2^-1 (5.494*180-1100)

    the equation on the right side of the equal sign

    I just ignore it....???
  22. Aug 21, 2009 #21
    Whoa ... I have no idea where you're going with that ... what's y??
  23. Aug 21, 2009 #22
    x = (-b +/-sqrt(b^2 - 4ac))/2a ("sqrt" = square root)

    so is the 2a part...

    the a is -2^-1 so does it become part of numerator and I get this???
    x = a(-b +/-sqrt(b^2 - 4ac))/2 ("sqrt" = square root)
  24. Aug 21, 2009 #23
    Of course not! That's part of the equation. You need to have a zero on the right side to make it look like ax^2 + bx + c = 0, so what to you have to do to get zero on the right side?
  25. Aug 21, 2009 #24
    i have no idea but the 2a part... read #22 please
    also why do I just ignore the other part of the equation that seems rather strange
  26. Aug 21, 2009 #25
    No, first you have to get the equation into the standard form so that you can clearly identify a, b, and c.

    a is not simply -2^-1. Gather all your t3^2 terms together.
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