Results 1 to 15 of 15

Thread: Solving Rational Equations...help!!

  1. #1
    Newbie mOoGaLy's Avatar
    Joined
    Jan 2007
    From
    At Home
    Posts
    9

    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!!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,002
    Thanks
    1
    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:

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

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

    Anyway, multiply by $\displaystyle x+2$:

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

    $\displaystyle 8+4(x+2)=5(x+2)$

    Now, can you finish?.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie mOoGaLy's Avatar
    Joined
    Jan 2007
    From
    At Home
    Posts
    9
    Quote Originally Posted by galactus View Post
    You haven't used parentheses, so your post is rather ambiguous.

    Is this it:

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

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

    Anyway, multiply by $\displaystyle x+2$:

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

    $\displaystyle 8+4(x+2)=5(x+2)$

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

    $\displaystyle 8+4(x+2)=5(x+2)$

    $\displaystyle 12(x+2)=5(x+2)$

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

    The answer I got is
    $\displaystyle x=-2$

    ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,002
    Thanks
    1
    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:

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

    Now, let's do it this way:

    $\displaystyle 8+4(x+2)=5(x+2)$

    $\displaystyle 8+4x+8=5x+10$

    $\displaystyle 16+4x=5x+10$

    Subtract 4x from both sides:

    $\displaystyle 16=x+10$

    Subtract 10 from both sides:

    $\displaystyle x=6$

    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:

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

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

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

    Relax. It'll be OK.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie mOoGaLy's Avatar
    Joined
    Jan 2007
    From
    At Home
    Posts
    9
    Ohh. Parenthasis first..
    That seems so easy! Thank you!!

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

    Do I multiply $\displaystyle x+6$ by all the equations?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,002
    Thanks
    1
    You're partially correct. You use the GCD, which is $\displaystyle 3x(x+6)$

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

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


    Cancel what needs cancelled and you get:

    $\displaystyle 2(x+6)+2(x+6)(x)=8(3x)$

    Distribute:

    $\displaystyle 2x+12+2x^{2}+12x=24x$

    $\displaystyle 14x+12+2x^{2}=24x$

    $\displaystyle 2x^{2}-10x+12=0$

    Divide by 2:

    $\displaystyle x^{2}-5x+6=0$

    Now, can you solve the quadratic?.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie mOoGaLy's Avatar
    Joined
    Jan 2007
    From
    At Home
    Posts
    9
    Woah woah, why do you divide by 2?

    Okay, well using the equation $\displaystyle x^{2}-5x+6=0$
    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
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,002
    Thanks
    1
    I divided by 2 because I could and wanted to. All the terms were divisible by 2


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

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

    $\displaystyle (x-3)(x-2)=0$
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie mOoGaLy's Avatar
    Joined
    Jan 2007
    From
    At Home
    Posts
    9
    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 $\displaystyle x^{2}-5x+6=0$ can be factored.

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

    $\displaystyle (x-3)(x-2)=0$
    I forgot all about those.
    So $\displaystyle x=3 or x=2$ ?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie mOoGaLy's Avatar
    Joined
    Jan 2007
    From
    At Home
    Posts
    9
    $\displaystyle \frac{x-3}{x-4}+4=\frac{3x}{x}$

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

    I multipied them

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

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

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

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

    $\displaystyle 5x^{2}-3x{2}-19x-12x$

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

    I did something wrong, didn't I? =( I am horrible at this!
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,002
    Thanks
    1
    $\displaystyle (x-4)(x)\frac{x-3}{x-4}+(x-4)(x)4=(x-4)(x)\frac{3x}{x}$

    I multipied them

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

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

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

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

    $\displaystyle 5x^{2}-19x=3x^{2}-12x$

    $\displaystyle 2x^{2}-7x=0$

    Factor:

    $\displaystyle x(2x-7)=0$

    See the solution now?.





    $\displaystyle 5x^{2}-3x{2}-19x12x$

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

    I did something wrong, didn't I? =( I am horrible at this!
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie mOoGaLy's Avatar
    Joined
    Jan 2007
    From
    At Home
    Posts
    9
    You lost me. How does $\displaystyle x(2x-7)=0$ end up into x = a number?
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Newbie mOoGaLy's Avatar
    Joined
    Jan 2007
    From
    At Home
    Posts
    9
    $\displaystyle 2x^{2}=7$ ??
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,163
    Thanks
    736
    Awards
    1
    Quote Originally Posted by mOoGaLy View Post
    You lost me. How does $\displaystyle x(2x-7)=0$ end up into x = a number?
    $\displaystyle x(2x-7) = 0$

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

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

    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 $\displaystyle x = \frac{7}{2}$.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Newbie mOoGaLy's Avatar
    Joined
    Jan 2007
    From
    At Home
    Posts
    9
    Oh! Okay, thank you so much galactus & topsquark!! You've been loads of help.

    If I need anymore, I'll be back lol
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Solving Rational equations
    Posted in the Algebra Forum
    Replies: 7
    Last Post: Sep 25th 2011, 06:32 AM
  2. Solving Rational Equations
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Sep 20th 2011, 03:19 AM
  3. Solving Rational equations.
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Nov 8th 2009, 04:28 PM
  4. I need help solving these rational equations.
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: Oct 27th 2009, 01:27 PM
  5. Solving rational equations
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Oct 10th 2008, 02:26 AM

Search Tags


/mathhelpforum @mathhelpforum