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Help with provinga simple algebraic equation?

  1. Jun 21, 2009 #1
    1. The problem statement, all variables and given/known data
    [itex]ar_1 = b <=> r_1 = \frac{b}{a}[/itex] (equation1)
    [itex]cr_d = c <=> r_2 = \frac{d}{c}[/itex] (equation2)
    [itex]ar_1 + cr_2 = b + c <=> r_1 + r_2 = \frac{b}{a} + \frac{d}{c}[/itex] (add equation1 and equation2)

    3. The attempt at a solution
    I am trying to make the equation on the left equals to the equation on the right side. I have tried adding, subtracting and dividing values, also multiplying and dividing conjugates but nothing works.
     
  2. jcsd
  3. Jun 21, 2009 #2
    I do not understand something with the second equation.

    Should it be

    [itex]
    cr_2 = d <=> r_2 = \frac{d}{c}
    [/itex]
    ?

    What do you need to prove? What is the original text of the task?
     
  4. Jun 21, 2009 #3
    I have 2 equations.
    [itex]ar_1 = b[/itex] (equation1)
    and [itex]cr_d = c[/itex] (equation2)

    i just divided by 'a' in both sides in equation1
    and divided by 'c' in both sides in equation2
    so i get
    [itex]r_1 = \frac{b}{a}[/itex] (equation1)
    [itex]r_2 = \frac{d}{c}[/itex] (equation2)

    but they are both reversible so
    [itex]ar_1 = b <=> r_1 = \frac{b}{a}[/itex] (equation1)
    [itex]cr_d = c <=> r_2 = \frac{d}{c}[/itex] (equation2)

    if i add both equations i get:
    [itex]ar_1 + cr_2 = b + c <=> r_1 + r_2 = \frac{b}{a} + \frac{d}{c}[/itex]
    since both of the equations are the same, so should the sum be
    and that is what i am trying to prove.

    Any numerical examples i try, both expressions seem to be the same except for 'a' and 'c' equals to zero.
     
  5. Jun 21, 2009 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You can't prove it, it isn't true. Knowing that [itex]ar_1+ br_2= b+ c[/itex] does NOT tell you that [itex]ar_1= b[/itex] and [itex]ar_2= c[/itex] separately.

    Knowing the two things [itex]ar_1= b[/itex] and [itex]cr_d= c[/itex] tells you, by adding the equations, that [itex]ar_1+ cr_d= b+ c[/itex] but knowing that single fact, that [itex]ar_1+ cr_d= b+ c[/itex] does NOT tell you the two things.

    For example, If x= 3 and y= 2 then x+ y= 5. But knowing only that x+y= 5 does NOT tell you that x= 3 and y= 2.
     
  6. Jul 28, 2009 #5
    Then what about this case?

    The parametric equation of a ellipse is:
    [itex]x = acos \theta[/itex]
    [itex]y = bsin \theta[/itex]

    I first square both equations so I get:
    [itex]x^2 = a^2cos^2 \theta[/itex] (1)
    [itex]y^2 = b^2sin^2 \theta[/itex] (2)

    Then for equation (1) divide by a^2 and equation (2) divide by b^2:
    [itex]\frac{x^2}{a^2} = cos^2 \theta[/itex] (3)
    [itex]\frac{y^2}{b^2} = sin^2 \theta[/itex] (4)

    So if i were to add equations (1) and (2) i would get:
    [itex]x^2 + y^2 = a^2cos^2 \theta + b^2sin^2 \theta[/itex] (5)
    And adding equation (3) and (4)
    [itex]\frac{x^2}{a^2} + \frac{y^2}{b^2} = cos^2 \theta + sin^2 \theta[/itex] (6)

    Then putting them in my format i would get:
    [itex]x^2 + y^2 = a^2cos^2 \theta + b^2sin^2 \theta <=>[/itex]
    [itex]\frac{x^2}{a^2} + \frac{y^2}{b^2} = cos^2 \theta + sin^2 \theta[/itex]

    Which both are equations of the ellipse the only difference is that one I can remove
    the third variable of [itex]\theta[/itex] by using the trigonometry identity of [itex]sin^2\theta + cos^2\theta=1[/itex].
     
  7. Jul 28, 2009 #6

    Mark44

    Staff: Mentor

    It's fairly simple to go from the parametric equations for an ellipse
    [itex]x = acos \theta[/itex]
    [itex]y = bsin \theta[/itex]

    to the Cartesian form:
    [tex]x^2/a^2 + y^2/b^2 = 1[/tex]
    You almost got to this form using way too many steps.

    Your original equations were garbled and, I believe, confused several people responding. As far as I can tell, your equations should have been:
    [tex]ar_1 = b[/tex]
    [tex]cr_2 = d[/tex]
    [tex]ar_1 + cr_2 = b + d[/tex]
    (For your second equation you had [itex]cr_d = c[/itex] which I'm nearly certain is an error. Even though Дьявол asked you whether this was an oversight, you didn't respond to him and persisted with the erroneous equation.)
    Your 3rd equation essentially says that b + d = b + d, which is trivially true. There is no way to divide the first term in the third equation by a and the second term by c. IOW, you can't transform your third equation to [itex]r_1 + r_2 = <whatever>[/itex]
     
  8. Jul 28, 2009 #7
    Ya the second line was an error. I should have been [itex]cr_2 = d <=> r_2 = \frac{d}{c}[/itex].

    Sorry Дьявол. I did not notice that. And because i didn't notice, i was confused by his question of what needed to be proven; also copied the equations wrong for the second time.

    Ya, the ellipse it just an simple example. I tried to write it in general form at the beginning. So going back to the general examples, if you actually try to plug some numbers you see that in the third equation both sides are the same except for "a" and "c" = 0.
     
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