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Solve this equation

  1. Jun 2, 2010 #1
    I have a big favour to ask. I have posted this equation twice before on this site and the usually kindly, selfless people have replied and patiently explained to me how it is solved. However, I'm embarrassed to admit I still don't get it. I always learn things far more effectively if I can see a completed example and then work backwards from that. So I'd be really grateful if someone could not only explain how to solve this equation but actually solve it as well.
    Sorry to be so melodramatic but my exam is on Monday and I really need to learn this.

    Here's the equation:

    2/(n+3) + 7/(n+4) = 1

    Thank you
     
  2. jcsd
  3. Jun 2, 2010 #2

    Mark44

    Staff: Mentor

    Multiply the left and right sides of the equation by (n + 3)(n + 4), then solve the resulting quadratic equation.
     
  4. Jun 2, 2010 #3
    Hi Mark. Thanks for your reply.
    I'll be fine once I get to the quadratic stage but I'm still lost on the 1st part. The left side of the equation is:

    2/(n+3) + 7/(n+4)

    I don't know how to go about multiplying that by (n+3)(n+4). Could you please break it down a bit for me?
    Many thanks
     
  5. Jun 2, 2010 #4

    Mark44

    Staff: Mentor

    You really need to brush up on the arithmetic of fractions.
    [tex]\frac{(n+3)(n+4)}{1} * \frac{2}{n + 3}= 2(n + 4)[/tex]


    The n + 3 factor in the numerator cancels with the n + 3 factor in the denominator. You don't really need to have the 1 in the denominator that I showed. I added it to emphasize the fact that in the first rational expression, n + 3 is in the numerator.
     
  6. Jun 2, 2010 #5
    (i) Write out the question you are asking.

    [tex]\frac{2}{n+3}+\frac{7}{n+4}=1[/tex]

    (Note: [tex]n\neq-3[/tex] or [tex]n\neq-4[/tex].)

    (ii) Multiply both sides by (n+3)(n+4).

    (iii) Take away the bracket.

    (iv) Rearrange the expression on the left hand side.

    (v) Do the cancellation.

    (vi) Finish the multiplications.

    (vii) Get a quadratic equation.

    (viii) Solve the quadratic equation.

    (Note: You may use the formula [tex]x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/tex] for the quadratic equation [tex]ax^{2}+bx+c=0[/tex].)
     
    Last edited: Jun 3, 2010
  7. Jun 2, 2010 #6

    berkeman

    User Avatar

    Staff: Mentor

    Please take care not to do too much of the student's work for them. Per the PF Rules, they must do the bulk of the work. Thanks.
     
  8. Jun 2, 2010 #7

    Mentallic

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    Homework Helper

    Gringo, if you're given an equation [tex]\frac{a}{b}+\frac{c}{d}=1[/tex] and you're asked simplify this equation (this is the same as what you're meant to be doing), you multiply both sides of the equation by bd. It works like this:

    [tex]\frac{bd}{1}\left(\frac{a}{b}+\frac{c}{d}\right)=bd(1)[/tex]

    [tex]\frac{bd(a)}{b}+\frac{bd(c)}{d}\right)=bd[/tex]

    Notice how the fractions each have common factors which can cancel:

    [tex]\frac{da}{1}+\frac{bc}{1}=bd[/tex]

    [tex]da+bc=bd[/tex]

    But in your case, the only difference is that b=n+3 and such. Well all you have to do instead is multiply both sides by (n+3)(n+4).
     
  9. Jun 2, 2010 #8

    Mentallic

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    Ok I've been looking at your other posts and realize you need a better understanding of fractions. They are tripping you up badly.

    If you have [tex]\frac{a+1}{b+1}[/tex] you cannot cancel the ones to give [tex]\frac{a}{b}[/tex]. If this were true, then [tex]\frac{3+1}{4+1}=\frac{3}{4}[/tex] but it's not.

    What you can cancel are factors. Such as if you had [tex]\frac{a(x^2-1)}{a(x-1)}[/tex]. In this case you can cancel an 'a' from the top and bottom because they are both factors of the numerator and denominator.
    This is the same as saying [tex]\frac{2}{6}=\frac{1}{3}[/tex] because there is a common of 2 in both the numerator and denominator which you cancelled off (that is, you divided the numerator by 2 as well as the denominator by 2).

    Looking at that last example again, we now have [tex]\frac{x^2-1}{x-1}[/tex] and we can't do anything to this as it is... like... DO NOT cancel the 1's to give [tex]\frac{x^2}{x}=x[/tex] because they are not common factors (get this deep into your head, only cancel factors).

    What you can do however is factorize the top to give [tex]\frac{(x+1)(x-1)}{x-1}[/tex] because [itex]x^2-1[/itex] is a difference of two squares. Now you can cancel the common factor x-1 from the top and bottom to finally give the simplest answer of [itex]x+1[/itex].

    Note, if you had something like [tex]\frac{(x-1)(x+1)+1}{x-1}[/tex] you cannot cancel x-1 from the numerator and denominator because they are not complete factors.

    Now, say you were asked to simplify [tex]\frac{2}{3}+\frac{1}{4}[/tex]. What you would do is make the common denominator 12, because 3*4 gives 12. This is what we call a lowest common multiple.

    Using the same idea of factors being cancelled in the numerator and denominator, we can also multiply both numerator and denominator by the same factor (we cannot add the same number, but we can multiply the same number).

    So to solve it we would do this: [tex]=\frac{4}{4}\left(\frac{2}{3}\right)+\frac{3}{3}\left(\frac{1}{4}\right)[/tex].

    Notice how when you multiply something by 4/4=1 it doesn't change anything because multiplying by 1 keeps things the same.

    [tex]=\frac{8}{12}+\frac{3}{12}[/tex] and now you can add the numerators together to give [tex]\frac{8+3}{12}=\frac{11}{12}[/tex].

    The EXACT same process applies for your question [tex]\frac{2}{n+3}+\frac{7}{n+4}[/tex]

    The lowest common multiple of n+3 and n+4 is the two numbers multiplied together to give (n+3)(n+4). Try and do the same thing that I did in the example to this problem.
     
  10. Jun 3, 2010 #9
    Got it at last! x = -2.25
    Thanks everyone!
     
  11. Jun 3, 2010 #10

    Mentallic

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    For the equation [tex]\frac{2}{n+3}+\frac{7}{n+4}=1[/tex] ? no.

    First of all you should have two solutions, since turns into a quadratic equation, and what you gave isn't the correct answer.

    Try it yourself, plug in n=-2.25 and see if [tex]\frac{2}{n+3}+\frac{7}{n+4}=1[/tex]
     
  12. Jun 3, 2010 #11

    Mark44

    Staff: Mentor

    Show us the steps you took to get x = -2.25, and we can show you where you went astray.
     
  13. Jun 4, 2010 #12
    I 1st multiplied each side by (n+3)(n+4) giving:

    n(n+3)(n+4) / (n+3) + 7(n+3)(n+4) / (n+4) = (n+3)(n+4)

    Cancelling out common numerators and denominators that leaves us with:

    n(n+4) + 7(n+3) = (n+3)(n+4)
    by simplifing both terms and cancelling out like terms on each side I ended up with 4n = -9 hence n = -2.25
     
  14. Jun 4, 2010 #13

    Mentallic

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    Oh well you changed the question! The question you had in the first post was 2/(n+3) and you now have n/(n+3). What you did was right and seem to know what you're doing now :smile:
     
  15. Jun 4, 2010 #14

    Mark44

    Staff: Mentor

    It all depends on what's the problem to be worked. If it is 2/(n + 3) + 7/(n + 4) = 1 (as originally stated), n = -2.25 is incorrect and there are two answers. If the problem is n/(n + 3) + 7/(n + 4) = 1, then I agree with Mentallic that n = -2.25 is correct.

    Gringo, I'm glad to see that you are using parentheses appropriately!
     
  16. Jun 4, 2010 #15

    Mentallic

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    Isn't it heart-warming to know that some of the advice we give gets through to them :smile:
     
  17. Jun 4, 2010 #16

    Mark44

    Staff: Mentor

    Yep, it sure does!
     
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