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Homework Help: Basic algebra ex. prob., can't work out a step

  1. Apr 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Hi all,

    I'm not in school, but I am self-studying because I'm helping tutor relatives of mine. The book says at this point, "apply the diagonal rule". I just need a pointer, because I don't know where to go from here. The book contains the answer, and I understand the rest of the steps, but this one step, my solutions are no such thing, so I'm embarassed to try and put them down!

    \frac{34} {m} - \frac{1}{2} = \frac{34*4}{4m+1}

    68 + 271m - 4m^2 = 272m

    I try fiddling with the top eq., I can't make it work into the bottom.

    2. Relevant equations

    3. The attempt at a solution
    Last edited by a moderator: Apr 16, 2013
  2. jcsd
  3. Apr 16, 2013 #2


    Staff: Mentor

    I've never heard it called the diagonal rule, but this is what I think they're hinting at:

    The equation ##\frac{a}{b} = \frac{c}{d}## is equivalent to ad = bc.

    Combine 34/m and -1/2 into a single fraction, and then use what I wrote above. The least common denominator for the two fractions is 2m. Multiply the first fraction by 1 in the form of 2/2, and multiply the second fraction by 1 in the form of m/m. When you combine the two terms, you'll have a single fraction on the left side.
  4. Apr 16, 2013 #3


    User Avatar

    Staff: Mentor

    Welcome to the PF.

    The first thing I would do is put the lefthand side (LHS) over a common denominator. Do you see how to do that?

    Then use "cross multiplication" to get rid of the fractions on both sides of the = sign. If you multiply both sides of an equation by the same quantity, the two sides are still equal, right?

    So if I have this equation:

    [tex]\frac{A}{B} = \frac{C}{D}[/tex]

    Then I can multiply both sides of the equation by the same thing:

    [tex](B*D)\frac{A}{B} = (B*D)\frac{C}{D}[/tex]

    And then simplifying:

    A*D = B*C

    So the net effect is the same as if you took the denominator from the LHS up to the RHS numerator, and the denominator from the RHS and brought it up to the numerator of the LHS, and got rid of the denominators on both sides. That's where the term "cross-multiply" comes from.

    Make sesnse?

    EDIT -- Beat out by the fast-typing Mark44 again! :smile:
  5. Apr 16, 2013 #4
    OK, thanks for the hints, I'm going to work it out and come back and show my work.
  6. Apr 16, 2013 #5


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  7. Apr 21, 2013 #6
    ok, worked it out

    For completions sake, I'mi posting to show I got past my algebra brain-freeze ;)

    1st combine 34/m and -1/2 into a single fraction. This is where I was messing up, by not first doing this, but rather multiplying all sides by all the denominators save the one I was on. I was making it more complicated than it had to be. So, combining the lefthand side (LHS) yields:

    (68 - m)/2m

    then multiply both sides with ((4m+1)(2m))/1

    2m cancels on the lefthand side and 4m+1 cancels on the righthand side, yields:

    (4m + 1) (68 - m) = 68 - m + (-4m**2) + 272m = -4m^2 + 271m + 68

    (2m)(134) = 272m

    so we have:
    -4m^2 + 271m + 68 = 272m ## subtract 271 and 68 from both sides
    -4m^2 - m = -68 ## divide both sides by -4
    m^2 + m = 17
    m^2 + m -17 = 0
    ## now we can solve the 2nd degree polynomial w/ quadratic formula
    Last edited: Apr 21, 2013
  8. Apr 21, 2013 #7
    Thanks for the replies, btw, which got me back on the right track to understand the work to solve the problem.
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