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Thread: Solving Rational Equations: 3y/3y+9 + 8y+16/2y+6 = y-4/y+3

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    Question Solving Rational Equations: 3y/3y+9 + 8y+16/2y+6 = y-4/y+3

    3y/3y+9 + 8y+16/2y+6 = y-4/y+3

    I understand that you have to get an LCD
    so when I factor the denominators I get

    3(y+3) 2(y+3) and (y+3) making the LCD: 6(y+3)

    at this point do you multiply 3y by 6(y+3)?

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    How would you do this?


    1/6 + 1/9 + 1/15

    ???
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

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    Quote Originally Posted by Rujaxso View Post
    3y/3y+9 + 8y+16/2y+6 = y-4/y+3

    I understand that you have to get an LCD
    so when I factor the denominators I get

    3(y+3) 2(y+3) and (y+3) making the LCD: 6(y+3)

    at this point do you multiply 3y by 6(y+3)?
    I suspect that you violated the order of operations in giving your problem.
    Last edited by JeffM; 11-08-2017 at 11:07 PM.

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    Quote Originally Posted by tkhunny View Post
    How would you do this?


    1/6 + 1/9 + 1/15

    ???
    Is this step one of a walk-through?

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    Quote Originally Posted by JeffM View Post
    I suspect that you violated the order of operations in giving your problem.
    Not sure that I follow. Do you mean I should reduce the coefficients after factoring what could be factored?

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    Quote Originally Posted by Rujaxso View Post
    3y/3y+9 + 8y+16/2y+6 = y-4/y+3
    Just so you know, you're missing some very very important grouping symbols. What you wrote literally means:

    [tex]\dfrac{3y}{3y}+9 + 8y+\dfrac{16}{2y}+6 = y-\dfrac{4}{y}+3[/tex]

    But what I think you meant was:

    [tex]\dfrac{3y}{3y+9} + \dfrac{8y+16}{2y+6} = \dfrac{y-4}{y+3}[/tex]

    Which would be type-set as: (3y)/(3y + 9) + (8y+16)/(2y + 6) = (y-4)/(y + 3)

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    Quote Originally Posted by ksdhart2 View Post
    Just so you know, you're missing some very very important grouping symbols. What you wrote literally means:

    [tex]\dfrac{3y}{3y}+9 + 8y+\dfrac{16}{2y}+6 = y-\dfrac{4}{y}+3[/tex]

    But what I think you meant was:

    [tex]\dfrac{3y}{3y+9} + \dfrac{8y+16}{2y+6} = \dfrac{y-4}{y+3}[/tex]

    Which would be type-set as: (3y)/(3y + 9) + (8y+16)/(2y + 6) = (y-4)/(y + 3)
    Your assumption is correct. I think Jeff wrote it out correctly too (before he edited his post that I was about to reply to) .

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    Quote Originally Posted by Rujaxso View Post
    3y/(3y+9) + (8y+16)/(2y+6) = (y-4)/(y+3)

    I understand that you have to get an LCD …
    Actually, in this exercise, you already have a common denominator. You just don't see it, yet, because you have factored only the denominators.

    Factor everything that you can (numerators and denominators), and then look for cancellations of common factors.

    For example:

    (17x - 34)/(68y + 17) reduces to (x - 2)/(4y + 1)

    because we can cancel a common factor. Factor both the numerator and denominator, to see it:

    [17(x - 2)]/[17(4y + 1)]

    The common factor 17 cancels.


    Now, in the same way, reduce the two algebraic ratios on the left-hand side of your given equation. You can then immediately add the reduced ratios, yes?

    Once you have a single, algebraic ratio on each side of the equation, ask yourself, "If the denominators are the same on each side, what does that say about the numerators?"
    Last edited by mmm4444bot; 11-09-2017 at 03:03 AM.
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

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    Quote Originally Posted by mmm4444bot View Post
    Actually, in this exercise, you already have a common denominator. You just don't see it, yet, because you have factored only the denominators.

    Factor everything that you can (numerators and denominators), and then look for cancellations of common factors.

    For example:

    (17x - 34)/(68y + 17) reduces to (x - 2)/(4y + 1)

    because we can cancel a common factor. Factor both the numerator and denominator, to see it:

    [17(x - 2)]/[17(4y + 1)]

    The common factor 17 cancels.


    Now, in the same way, reduce the two algebraic ratios on the left-hand side of your given equation. You can then immediately add the reduced ratios, yes?

    Once you have a single, algebraic ratio on each side of the equation, ask yourself, "If the denominators are the same on each side, what does that say about the numerators?"
    Okay, so I get (y+3) as the common denom.

    Whats the general rules for reducing after you factor? I think this is where I'm foggy on concepts.
    for instance why cant Y cancel out the Y in (Y+3) ---> (Y)/(Y+3) ?


    p.s thanks for the replies so far.

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    Quote Originally Posted by Rujaxso View Post
    Okay, so I get (y+3) as the common denom.

    Whats the general rules for reducing after you factor? I think this is where I'm foggy on concepts.
    for instance

    why cant Y cancel out the Y in (Y+3) ---> (Y)/(Y+3) ?


    p.s thanks for the replies so far.
    Do you think you could reduce [tex]\frac{2}{2+3}[/tex] and get [tex]\frac{1}{3}[/tex] by "cancelling" out the 2s in the numerator and denominator.

    The answer is "No" and I hope you "No" it!!

    The reason for not "cancelling" out Y in that expression is same.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

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