Results 1 to 11 of 11
Like Tree3Thanks
  • 1 Post By mathguy25
  • 1 Post By Soroban
  • 1 Post By HallsofIvy

Math Help - Basic algebra problems

  1. #1
    Newbie
    Joined
    Apr 2013
    From
    Australia
    Posts
    19

    Basic algebra problems

    I have a couple of questions in this weeks work that I'm unsure how to approach. The first question is as follows -

    "Use algebra to express Basic algebra problems-equationa.jpg in the form Basic algebra problems-equationb.jpg ." I have made a few attempts at trying to complete this question, but as yet have been unsuccessful.

    Another question I'm having difficulty with is "Find all linear functions f(x) = ax + b such that f(x) = f-1 (x), where f(x) = (x - 3)2(4 - x)(2x + 1)". I'm not entirely sure what this question is asking.

    Any help would be greatly appreciated.
    Thank you.
    Last edited by deSitter; April 5th 2013 at 03:45 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Apr 2013
    From
    Green Bay
    Posts
    68
    Thanks
    16

    Re: Basic algebra problems

    Note that a/(x-b) + b = a/(x - b) + b(x-b)/(x-b) = (a + b(x - b)) / (x - b). Then (a + bx - b^2) / (x - b).

    Cross multiply.

    (1 - 2x)(x - b) = (3x - 2)(a + bx - b^2)
    x - b - 2x^2 + 2bx = 3ax + 3bx^2 - 3b^2 x + 2b^2 - 2a - 2bx
    -2x^2 + (2b + 1)x - b = 3bx^2 + (3a - 3b^2 - 2b)x + 2b^2 - 2a

    Equate coefficients

    (*) -2 = 3b
    (**) 2b + 1 = 3a - 3b^2 - 2b
    (***) 2b^2 - 2a = -b

    (*) => b = -2/3. Plug this into (**) and (***).

    (**) => 2(-2/3) + 1 = 3a - 3(-2/3)^2 - 2(-2/3)
    => -4/3 + 1 = 3a - 3(4/9) + 4/3
    => -1/3 = 3a - 4/3 + 4/3
    => -1/3 = 3a
    => a = -1/9

    (***) => 2(-2/3)^2 - 2a = -(-2/3)
    => 2(4/9) - 2a = 4/3
    => 8/9 - 2a = 12/9
    => -2a = 12/9 - 8/9
    => -2a = 4/9
    => a = -2/9

    We see that a = -1/9 and a = -2/9 which is impossible. There must be some typo.
    Last edited by mathguy25; April 6th 2013 at 06:17 AM.
    Thanks from deSitter
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,686
    Thanks
    617

    Re: Basic algebra problems

    Hello, deSitter!

    There is a typo in the problem . . .


    \text{Use algebra to express }\,F \:=\:\frac{1-2x}{3x-2}\,\text{ in the form }\,\frac{a}{x-b} + b

    Long division:
    . . \begin{array}{cccccc}&&&&\text{-}\frac{2}{3} \\ && -- & -- & -- \\ 3x-2 & ) & \text{-}2x &+& 1 \\ && \text{-}2x &+& \frac{4}{3} \\ && -- & -- & -- \\ &&&& \text{-}\frac{1}{3} \end{array}


    We have: . F \:=\:\frac{\text{-}\frac{1}{3}}{3x-2} - \frac{2}{3}


    Divide numerator and denominator by 3:

    . . . . F \:=\:\frac{\text{-}\frac{1}{9}}{x - \frac{2}{3}} - \frac{2}{3}
    Look at what we have: . \frac{\overbrace{\text{-}\tfrac{1}{9}}^a}{x - \underbrace{\left(\tfrac{2}{3}\right)}_{b?}} + \underbrace{\left(\text{-}\tfrac{2}{3}\right)}_{b?}
    Thanks from deSitter
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Apr 2013
    From
    Australia
    Posts
    19

    Re: Basic algebra problems

    Thanks guys, you have both been very helpful. Any thoughts on the second problem? I'm still unsure what the question is asking. I'm sure it's something simple but the wording has me confused.

    Find all linear functions f(x) = ax + b such that f(x) = f -1 (x), where f(x) = (x - 3)2(4 - x)(2x + 1)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Apr 2013
    From
    Australia
    Posts
    19

    Re: Basic algebra problems

    I think I have made an error with the last question. It should be -

    "Find all linear functions f(x) = ax + b such that f(x) = f -1 (x)"

    (x-3)2(4-x)(2x+1) was from a previous question and I has assumed it carried over, but this is probably incorrect. I had supposed that f(x) = ax + b was just the form of the linear functions I was to attempt to find. Does anyone have any clues I where I should start with the above question?

    PS. yes there was a typo in the denominator, it should have been 3x + 2. Thank you again for you assistance.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,412
    Thanks
    1328

    Re: Basic algebra problems

    Yes, the added "f(x) = (x - 3)2(4 - x)(2x + 1)" makes no sense because that is NOT a linear function.

    You are correct that a linear function is of the form y= ax+ b. Now can you find the inverse of that?

    Since inverting y= f(x) gives x= f-1(y), start by swapping x and y: x= ay+ b and solve for y. Then use f-1(x)= f(x) to solve for a and b.
    (You cannot solve for a and b as specific numbers because there are an infinite number of such functions.)
    Thanks from deSitter
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Apr 2013
    From
    Green Bay
    Posts
    68
    Thanks
    16

    Re: Basic algebra problems

    Remember, given f(x) we can determine f^(-1) (x) by the following steps
    1. Set y = f(x).
    2. Interchange x and y.
    3. Solve for new y.
    4. Set new y = f^(-1) (x)

    Using these steps will yied f^(-1) (x) = x/a - b/a

    Set f(x) = f^(-1) (x).
    Then ax + b = x/a - b/a

    Now equate coefficients. Then a = 1/a and b = -a/b. Then a^2 = 1 and b^2 = -a. Since a^2 = 1, we see that either a = 1 or a = -1. If a = 1, then -a = -1. Then b^2 = -1 which yields no real solution for b. Thus, a = -1. Then b^2 = -(-1) = 1. Then b = 1 or b = -1.

    Solution: f(x) = -x + 1 and f(x) = -x - 1.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Apr 2013
    From
    Green Bay
    Posts
    68
    Thanks
    16

    Re: Basic algebra problems

    EDIT: Cross multiply should then be (1 - 2x)(x - b) = (3x + 2)(a + bx - b^2)

    x - b - 2x^2 + 2bx = 3ax + 3bx^2 - 3b^2 x + 2a + 2bx - 2b^2
    3ax + 3bx^2 - 3b^2 x + 2a + 2bx - 2b^2 - x + b + 2x^2 - 2bx = 0
    x^2(3b + 2) + x(3a - 3b^2 + 2b - 1 - 2b) + (2a - 2b^2 + b) = 0

    Then 3b + 2 = 0 and 3a - 3b^2 - 1 = 0 and 2a - 2b^2 + b = 0

    3b + 2 = 0 => b = -2/3

    3a - 3(-2/3)^2 - 1 = 0 and 2a - 2(-2/3)^2 + (-2/3) = 0
    3a - 3(4/9) - 1 = 0 and 2a - 2(4/9) + -2/3 = 0
    3a - 12/9 - 9/9 = 0 and 2a - 8/9 - 6/9 = 0
    3a = 21/9 and 2a = 14/9
    a = 21/27 and a = 14/18
    a = 7/9

    Answer: a = 7/9 and b = -2/3
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Apr 2013
    From
    Australia
    Posts
    19

    Re: Basic algebra problems

    Thank you all for your assistance. You have been very helpful.

    mathguy25, you equate b = -a/b, would this not be b = -b/a?

    Which leads to -
    a2 = 1
    a = 1

    b = -b/a
    ba = -b
    if a = 1, b = -b, which makes no sense (unless I guess b = 0), therefore a = -1 (I don't know if I can just assume this).

    ax + b = (x - b)/a
    a2x + ba + b = x
    (-1)2x + b(-1) + b = x
    b = 1

    Solution f(x) = -x + 1
    Last edited by deSitter; April 11th 2013 at 04:36 AM.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Apr 2013
    From
    Green Bay
    Posts
    68
    Thanks
    16

    Re: Basic algebra problems

    Yes, I made a mistake there. It should be b = -b/a. Remember a = +/- 1. Thus,

    a = +/- 1 and b = -b/a
    a = +/- 1 and ab = -b
    a = +/- 1 and ab + b = 0
    a = +/- 1 and b(a + 1) = 0
    a = +/- 1 and b = 0 or a + 1 = 0

    If b = 0, then a = +/- 1 so f(x) = x and f(x) = -x would be solution.
    If b =/= 0, then a = -1. Then f(x) = -x + b is a solution for any number b =/= 0.

    Thus, either f(x) = x or f(x) = -x + b for any b (since f(x) = -x is the case where b = 0).

    Soln: {f(x) = x, f(x) = -x + b where b is any real number}
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Apr 2013
    From
    Australia
    Posts
    19

    Re: Basic algebra problems

    Great stuff, thanks again.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: March 6th 2013, 10:03 PM
  2. Replies: 2
    Last Post: March 6th 2013, 09:58 PM
  3. Replies: 3
    Last Post: September 28th 2009, 05:24 PM
  4. Replies: 3
    Last Post: October 2nd 2008, 07:09 PM
  5. Basic Algebra Word Problems (I got a few)
    Posted in the Algebra Forum
    Replies: 3
    Last Post: January 24th 2007, 12:25 AM

Search Tags


/mathhelpforum @mathhelpforum