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Math Help - Yet another fractions problem

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    Yet another fractions problem

    \frac{4}{y+2}=1-\frac{8}{y(y+2)}

    y(y+2) is what I should be multiplying both sides by?

    y(y+2)\frac{4}{y+2}=1-\frac{8}{y(y+2)}y(y+2)

    4y=-7

    y=-\frac{7}{4}

    Which I know is wrong, so then I try y+2 as the multiplier and that is also wrong.
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    Re: Yet another fractions problem

    Quote Originally Posted by uperkurk View Post
    \frac{4}{y+2}=1-\frac{8}{y(y+2)}

    y(y+2) is what I should be multiplying both sides by?

    y(y+2)\frac{4}{y+2}=1-\frac{8}{y(y+2)}y(y+2)

    4y=-7

    y=-\frac{7}{4}

    Which I know is wrong, so then I try y+2 as the multiplier and that is also wrong.
    y(y+2)\frac{4}{y+2}=\left[1-\frac{8}{y(y+2)}\right]y(y+2)=y(y+2)-8
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    Re: Yet another fractions problem

    Quote Originally Posted by Plato View Post
    \left[1-\frac{8}{y(y+2)}\right]y(y+2)=y(y+2)-8
    How so? By the looks of it \left[1-\frac{8}{y(y+2)}\right]y(y+2) looks like the y(y+2) cancel each other leaving \left[1-8\right] Explain why this is not the case please
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    Re: Yet another fractions problem

    Quote Originally Posted by uperkurk View Post
    How so? By the looks of it \left[1-\frac{8}{y(y+2)}\right]y(y+2) looks like the y(y+2) cancel each other leaving \left[1-8\right] Explain why this is not the case please
    Do you know the distribution rules?

    \left[(x+y)z\right]=xz+yz

    \left[1-\frac{8}{A}\right]A=A-8 no matter what the nonzero factor A happens to be.
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    Re: Yet another fractions problem

    With fractions, we have to have "the same denominators" before we can actually add them together. So let's get everything over a common denominator: y(y+2).

    One of our fractions already has this denominator, so let's focus on the other two:

    \frac{4}{y+2} = \frac{4y}{y(y+2)}

    1 = \frac{y(y+2)}{y(y+2)}

    So now our equation is:

    \frac{4y}{y(y+2)} = \frac{y(y+2)}{y(y+2)} - \frac{8}{y(y+2)}

    NOW, we can add the "tops" on the right-hand side:

    \frac{4y}{y(y+2)} = \frac{y^2 + 2y - 8}{y(y+2)}

    Now, if y = 0 or -2, that's bad news for everyone. So let's hope this doesn't turn out to be the case. If not, we can multiply both sides by y(y+2), to get:

    4y = y^2 + 2y - 8, or:

    y^2 - 2y - 8 = 0.

    Hopefully, you are able to factor this to get two possible choices for y. One makes sense, one doesn't.
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    Re: Yet another fractions problem

    Quote Originally Posted by uperkurk View Post
    \frac{4}{y+2}=1-\frac{8}{y(y+2)}

    y(y+2) is what I should be multiplying both sides by?

    y(y+2)\frac{4}{y+2}=1-\frac{8}{y(y+2)}y(y+2)
    In simpler words: you forgot to multiply the 1 with y(y+2).
    You can't leave any member outside for no reason.
    (the reason of course you have to multiply every member, is the properties of distributive law that Plato used)
    Last edited by ChessTal; August 3rd 2013 at 08:39 AM.
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    Re: Yet another fractions problem

    I think I'm just going to forget about learning fractions. I can't seem to get the hang of this, as soon as a question looks slightly different, I can't do it.

    \frac{a-1}{4}=\frac{8}{a+3}

    4(a+3)\left(\frac{a-1}{4}\right)=\left(\frac{8}{a+3}\right)4(a+3)

    So I can see the 4's on the left and the a+3's on the right cancel.

    (a-1)(a+3)=32


    I'm confused :/
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    Re: Yet another fractions problem

    Quote Originally Posted by uperkurk View Post
    I think I'm just going to forget about learning fractions. I can't seem to get the hang of this, as soon as a question looks slightly different, I can't do it.

    \frac{a-1}{4}=\frac{8}{a+3}

    4(a+3)\left(\frac{a-1}{4}\right)=\left(\frac{8}{a+3}\right)4(a+3)

    So I can see the 4's on the left and the a+3's on the right cancel.

    (a-1)(a+3)=32


    I'm confused :/
    What you did so far is fine, you just didn't finish. Now, you have to expand the left:

    (a-1)(a+3) = a(a+3) - 1(a+3) = a^2 + 3a - a - 3 = a^2 + 2a - 3.

    Since this equals 32, subtract 32 from both sides, to get:

    a^2 + 2a - 35 = 0.

    Factoring, this becomes:

    (a - 5)(a + 7) = 0, so either a = 5, or a = -7.
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    Re: Yet another fractions problem

    oooooo I see, I wasn't expecting this type of question because it was in a list of questions that just have 1 value for x, so I wasn't aware of it.
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    Re: Yet another fractions problem

    Quote Originally Posted by uperkurk View Post
    I think I'm just going to forget about learning fractions. I can't seem to get the hang of this, as soon as a question looks slightly different, I can't do it.

    \frac{a-1}{4}=\frac{8}{a+3}

    4(a+3)\left(\frac{a-1}{4}\right)=\left(\frac{8}{a+3}\right)4(a+3)

    So I can see the 4's on the left and the a+3's on the right cancel.

    (a-1)(a+3)=32


    I'm confused :/
    And what exactly was your problem anyway? You have transformed an equation with fractions into one with no fractions. That was the whole purpose of multiplying each side with LCM.
    So you should now continue solving the equation normally and get the appropriate result.... (as Deveno did)

    BUT BY ALL MEANS you should never forget that the original equation is not equivalent(in most times) with the values you get for the variable in the final step, because it may make one of the denominators zero.
    In other words the solution you get is a possible one but not a sure one. So you should verify that the final answer verifies the original equation.

    Example:
    How you would solve the following one?

    \frac{{{x^2} - 2x}}{{x - 3}} = \frac{1}{4} + \frac{{8x - 18}}{{2(x - 3)}}
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    Re: Yet another fractions problem

    Quote Originally Posted by ChessTal View Post
    And what exactly was your problem anyway? You have transformed an equation with fractions into one with no fractions. That was the whole purpose of multiplying each side with LCM.
    So you should now continue solving the equation normally and get the appropriate result.... (as Deveno did)

    BUT BY ALL MEANS you should never forget that the original equation is not equivalent(in most times) with the values you get for the variable in the final step, because it may make one of the denominators zero.
    In other words the solution you get is a possible one but not a sure one. So you should verify that the final answer verifies the original equation.

    Example:
    How you would solve the following one?

    \frac{{{x^2} - 2x}}{{x - 3}} = \frac{1}{4} + \frac{{8x - 18}}{{2(x - 3)}}
    I really don't know this is too difficult... all I know is to factor some stuff out.

    \frac{x(x-2)}{x-3}=\frac{1}{4}+\frac{8x-18}{2(x-3)}

    Then I guess would multiply all terms by 2(x-12) to be honest I really have no idea what I'm doing.

    Maybe I could move the terms around abit \frac{x(x-2)}{x-3}-\frac{8x-18}{2(x-3)}=\frac{1}{4}

    and now what? There are too many mixed things here... ok so I apply the \frac{a}{c}\cdot\frac{b}{d} = ad \cdot bc rule.

    So then we have some even more complicated looking expression. x(x-2)(2(x-3)) - (8x-18) forget it I'm lost I forgot everything. I can't do this question there are too many fraction, too many disimilarities and I dunno how to work it out. Sorry.
    Last edited by uperkurk; August 3rd 2013 at 03:35 PM.
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    Re: Yet another fractions problem

    After jotting down a bunch of random crap on my paper I found an expression that looks like this 2x^2-12x-18=\frac{x-3}{2} which is more than likely incorrect.
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    Re: Yet another fractions problem

    Quote Originally Posted by uperkurk View Post
    I really don't know this is too difficult... all I know is to factor some stuff out.

    \frac{x(x-2)}{x-3}=\frac{1}{4}+\frac{8x-18}{2(x-3)}
    Then I guess would multiply all terms by 2(x-12) to be honest I really have no idea what I'm doing.
    No, no 2(x-12). This is not the LCM.
    But first, to get an idea of what you should be doing follow every time these steps:

    a)Find the LCM of all the denominators.
    b)Multiply both sides of the equation with the LCM. (That means if you have a+7 = c+v+t-77^y and them LCM is Q then you should do: Q(a+7) = Q(c+v+t-77^y) <=> Q穉 + Q7 = Q穋 + Q穊 + Q穞 - Q77^y)
    c)After the simplifications all the denominators will be gone! There are no exceptions to this.
    d)Make as many simplifications in each side you can(e.g 3x-8x=-5x , or 3x^2+x^2 = 4x^2, or 4-65=-61, etc...) In this step you must make all the brackets to vanish!
    e)If the equation is of second degree(there is a variable in the form of e.g 9x^2 or 1.5a^2 or -y^2) then bring all the members of one side to the other and then solve according to the known rules for a 2nd degree equation. If the equation is of first degree then you isolate the numbers(or the known variables) to one side and in the other the unknown variable).
    f)Solve and find the unknown variable.
    g)Check which of the solution/solutions verify the original equation you have been asked to solve in the first place.

    Now about the LCM of the denominators of the problem, which are: x-3 , 4 , 2(x-3)
    To find it you must factorize the denominators.
    x-3 can't be factored anymore. Its factors is only one, the: (x-3) and it's has a degree of 1 since x-3=(x-3)^1
    4 is 2^2. Its factor is the 2 and its degree is 2 also.
    The 2(x-3) can't be factored more either(if it was in the 2x-6 form, then you should factor it to 2(x-3)). So its factors are 2 and (x-3) with degrees of 1 for both.

    The LCM consist of ALL common and non common factors of each denominator with the higher degree they appear.
    So since our factors are the x-3 and 2 only and the x-3 appears with a maximum degree of 1 and 2 with a maximum degree of 2, then the LCM is:
    (x-3)2^2 = 4(x-3)


    and now what? There are too many mixed things here... ok so I apply the \frac{a}{c}\cdot\frac{b}{d} = ad \cdot bc rule.
    Where did this rule come from?
    There is no such rule and the above equality is false(generally-for random variables).
    Last edited by ChessTal; August 4th 2013 at 02:12 AM.
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    Re: Yet another fractions problem

    OK, there are a couple of differing approaches being espoused here, and I want you to see that these both work out the same.

    Let's say we have an equation of the form:

    \frac{A}{B} = \frac{C}{D} + \frac{E}{F}.

    Here, any of the letters might be some complicated expression, for example we might have A = 2x + 3y3, or A = sin(x), or A = -8. What these expressions are, really doesn't matter, yet. Let's just deal with the "algebra of fractions".

    Approach number 1:

    step 1A: multiply each fraction by BDF (you CAN use the LCM(B,D,F) but you don't HAVE to. You can always cancel common factors later).

    This makes our new equation:

    ADF = BCF + BDE, which has no fractions in it.

    step 2A; solve without those pesky fractions in the way.

    Approach number 2 (this is really the same as approach #1, but shows "the steps in-between"):

    step 2A: get the right-hand side over a common denominator (in this case, DF):

    \frac{A}{B} = \frac{CF}{DF} + \frac{DE}{DF} = \frac{CF + DE}{DF}

    step 2B: cross-multiply (multiply the right side by B, and the left side by DF). THis gives:

    A(DF) = B(CF + DE) = BCF + BDE

    which is what we got with approach #1. Step 2C is the same as step 1B.

    Let's use one of these to solve the problem that ChessTal posed:

    \frac{x^2 - 2x}{x-3} = \frac{1}{4} + \frac{8x-18}{2(x -3)}.

    First, we look at the denominators!!! They are going to control our next move. We need a common denominator of some sort (the least common denominator is "optimal" but let's be inefficient).

    So let's just multiply ALL the denominators together. Our denominators are: x-3, 4 and 2(x-3), so we'll just multiply EVERYTHING by 8(x-3)2. Now we have:

    8(x-3)^2\cdot\frac{x^2 - 2x}{x-3} = 8(x-3)^2\cdot\frac{1}{4} + 8(x-3)^2\cdot\frac{8x-18}{2(x-3)}

    Now a lot of stuff is going to cancel. I'll show how each term cancels, before resuming:

    8(x-3)^2\cdot\frac{x^2 - 2x}{x - 3} = 8(x-3)(x^2 - 2x). This is our new left-hand side.

    8(x-3)^2\cdot\frac{1}{4} = 2(x-3)^2,

    8(x-3)^2\cdot\frac{8x-18}{2(x-3)} = 4(x-3)(8x-18).

    So our new equation is:

    8(x-3)(x^2 - 2x) = 2(x-3)^2 + 4(x-3)(8x-18) <---look! no fractions!!!!

    Let's get everything on one side, with 0 on the other:

    8(x-3)(x^2 - 2x) - 2(x-3)^2 - 4(x-3)(8x-18) = 0.

    Note that 2 divides every term, and so does x-3, so we will first divide everything by 2 (0/2 = 0, of course):

    4(x-3)(x^2 - 2x) - (x-3)^2 - 2(x-3)(8x-18) = 0.

    Now let's divide everything by x - 3 (this means we are ASSUMING x is not 3, or else we cannot do this):

    4(x^2 - 2x) - (x-3) - 2(8x-18) = 0. Now, we can solve "like normal":

    4x^2 - 8x - x + 3 - 16x + 36 = 0 = 0

    4x^2 - 25x + 39 = 0

    (4x - 13)(x - 3) = 0.

    We can obviously throw out the solution x = 3, so we have x= 13/4. Does this answer work? Check and see....
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    Re: Yet another fractions problem

    Quote Originally Posted by Deveno View Post
    \frac{x^2 - 2x}{x-3} = \frac{1}{4} + \frac{8x-18}{2(x -3)}.

    First, we look at the denominators!!! They are going to control our next move. We need a common denominator of some sort (the least common denominator is "optimal" but let's be inefficient).

    So let's just multiply ALL the denominators together. Our denominators are: x-3, 4 and 2(x-3), so we'll just multiply EVERYTHING by 8(x-3)2. Now we have:

    8(x-3)^2\cdot\frac{x^2 - 2x}{x-3} = 8(x-3)^2\cdot\frac{1}{4} + 8(x-3)^2\cdot\frac{8x-18}{2(x-3)}
    Everything in your post makes perfect sense to me and when I see it written like that I can see easily how each step leads to the next and I can't really believe how easy that question was. But my problem is knowing what to do with things depending on the look of the question.

    For example you multiplied the denominators to get 8(x-3)^2 are there certain rules you have to follow to multiply these like that? Because looking at it I would have done x-3\times4\times2(x-3) which is ofcourse wrong. How did you know in which order to multiply these denominators by?

    Great post btw you cleared up quite a lot of things for me. Thanks.

    Lets just assume that the denominators are now x-1 , 4 and x+2 can we simply just multiply all these together?
    Last edited by uperkurk; August 4th 2013 at 05:08 AM.
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