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Math Help - Solving Rational Equations...help!!

  1. #1
    Newbie mOoGaLy's Avatar
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    Solving Rational Equations...help!!

    Okay, I don't know how to do this what so ever and I'm getting frustrated because my book isn't showing in great detail what I'm suppose to be doing!

    I need help with 5 equations... this is one of them.

    8/x+2 + 8/2 = 5

    Please help me!!!
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  2. #2
    Eater of Worlds
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    Quote Originally Posted by mOoGaLy View Post
    Okay, I don't know how to do this what so ever and I'm getting frustrated because my book isn't showing in great detail what I'm suppose to be doing!

    I need help with 5 equations... this is one of them.

    8/x+2 + 8/2 = 5

    Please help me!!!
    You haven't used parentheses, so your post is rather ambiguous.

    Is this it:

    \frac{8}{x+2}+\frac{8}{2}=5

    If so, why not write 8/2 as 4?.

    Anyway, multiply by x+2:

    (x+2)\frac{8}{x+2}+4(x+2)=5(x+2)

    8+4(x+2)=5(x+2)

    Now, can you finish?.
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  3. #3
    Newbie mOoGaLy's Avatar
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    Quote Originally Posted by galactus View Post
    You haven't used parentheses, so your post is rather ambiguous.

    Is this it:

    \frac{8}{x+2}+\frac{8}{2}=5

    If so, why not write 8/2 as 4?.

    Anyway, multiply by x+2:

    (x+2)\frac{8}{x+2}+4(x+2)=5(x+2)

    8+4(x+2)=5(x+2)

    Now, can you finish?.
    Yeah, that's the equation. and what I did was...

    8+4(x+2)=5(x+2)

    12(x+2)=5(x+2)

    Now, do I multiply 12 to x & 2 to get
    12x + 24 = 5x + 10

    The answer I got is
    x=-2

    ?
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  4. #4
    Eater of Worlds
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    No, check yourself. If you put your answer back into the equation, do you get 5?. I don't think so.

    You got a little discombobulated on your distribution.

    Let's try your answer first:

    \underbrace{\frac{8}{\underbrace{-2+2}_{\text{oops, division by 0, a big no-no}}}+4=5}_{\text{bad, the universe will\\collapse in on itself}}

    Now, let's do it this way:

    8+4(x+2)=5(x+2)

    8+4x+8=5x+10

    16+4x=5x+10

    Subtract 4x from both sides:

    16=x+10

    Subtract 10 from both sides:

    x=6

    See?.

    Actually, you're not seeing the forest for the trees. Don't let the algebra and x's psyche you out.

    Look at it for a second. Follow:

    \frac{8}{x+2}+4=5

    What plus 4 equals 5. Well, you don't have to be a brilliant mathematician to see that. 4+1=5

    Therefore, \frac{8}{x+2} must be 1. What makes that 1?. When x=6. See?.

    Relax. It'll be OK.
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  5. #5
    Newbie mOoGaLy's Avatar
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    Ohh. Parenthasis first..
    That seems so easy! Thank you!!

    \frac{2}{3x}+\frac{2}{3}=\frac{8}{x+6}

    Do I multiply x+6 by all the equations?
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  6. #6
    Eater of Worlds
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    You're partially correct. You use the GCD, which is 3x(x+6)

    See?. You have more in the denominators than just x+6. There's a 3x there, also.

    (x+6)(3x)\frac{2}{3x}+(x+6)(3x)\frac{2}{3}=(x+6)(3  x)\frac{8}{x+6}


    Cancel what needs cancelled and you get:

    2(x+6)+2(x+6)(x)=8(3x)

    Distribute:

    2x+12+2x^{2}+12x=24x

    14x+12+2x^{2}=24x

    2x^{2}-10x+12=0

    Divide by 2:

    x^{2}-5x+6=0

    Now, can you solve the quadratic?.
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  7. #7
    Newbie mOoGaLy's Avatar
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    Woah woah, why do you divide by 2?

    Okay, well using the equation x^{2}-5x+6=0
    I used the quadratic formula and I am stuck at
    (doesn't know the codes to make this look correct)
    x = -5 plus/minus symbol 1 all over 2
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  8. #8
    Eater of Worlds
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    I divided by 2 because I could and wanted to. All the terms were divisible by 2


    The quadratic x^{2}-5x+6=0 can be factored.

    Can you factor?. You can use the quadratic formula, too, but this one is easily factorable.

    (x-3)(x-2)=0
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  9. #9
    Newbie mOoGaLy's Avatar
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    Quote Originally Posted by galactus View Post
    I divided by 2 because I could and wanted to. All the terms were divisible by 2


    The quadratic x^{2}-5x+6=0 can be factored.

    Can you factor?. You can use the quadratic formula, too, but this one is easily factorable.

    (x-3)(x-2)=0
    I forgot all about those.
    So x=3 or x=2 ?
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  10. #10
    Newbie mOoGaLy's Avatar
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    \frac{x-3}{x-4}+4=\frac{3x}{x}

    (x-4)(x)\frac{x-3}{x-4}+(x-4)(x)4=(x-4)(x)\frac{3x}{x}

    I multipied them

    x(x-3)+4(x-4)(x)=3x(x-4) but I had a feeling that I was suppose to cancel out the 4's but I chose not to..

    Next, x^{2}-3x+4x^{2}-16x=3x^{2}-12x

    And I moved the 3x^{2}-12x
    to the left side of the equation...

    x(x-3)+4(x-4)(x)=3x(x-4)=3x^{2}-12x

    5x^{2}-3x{2}-19x-12x

    I'm stuck at 2x^{2}-21x

    I did something wrong, didn't I? =( I am horrible at this!
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  11. #11
    Eater of Worlds
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    (x-4)(x)\frac{x-3}{x-4}+(x-4)(x)4=(x-4)(x)\frac{3x}{x}

    I multipied them

    x(x-3)+4(x-4)(x)=3x(x-4) but I had a feeling that I was suppose to cancel out the 4's but I chose not to..

    Next, x^{2}-3x+4x^{2}-16x=3x^{2}-12x

    And I moved the 3x^{2}-12x
    to the left side of the equation...

    x(x-3)+4(x-4)(x)=3x(x-4)=3x^{2}-12x \leftarrow\text{good to here}

    5x^{2}-19x=3x^{2}-12x

    2x^{2}-7x=0

    Factor:

    x(2x-7)=0

    See the solution now?.





    5x^{2}-3x{2}-19x12x

    I'm stuck at 2x^{2}-21x

    I did something wrong, didn't I? =( I am horrible at this!
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  12. #12
    Newbie mOoGaLy's Avatar
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    You lost me. How does x(2x-7)=0 end up into x = a number?
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  13. #13
    Newbie mOoGaLy's Avatar
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    2x^{2}=7 ??
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  14. #14
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mOoGaLy View Post
    You lost me. How does x(2x-7)=0 end up into x = a number?
    x(2x-7) = 0

    Since the RHS is 0, at least one of x or 2x - 7 must also be 0. Thus:
    x = 0 or 2x - 7 = 0

    Thus x = 0 or x = \frac{7}{2}

    Now be careful. Your original equation canNOT work with x = 0. (The RHS of your original equation has us dividing 0/0, which is not allowed.) So your only solution is x = \frac{7}{2}.

    -Dan
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  15. #15
    Newbie mOoGaLy's Avatar
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    Oh! Okay, thank you so much galactus & topsquark!! You've been loads of help.

    If I need anymore, I'll be back lol
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