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Math Help - simple question but i need help

  1. #1
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    simple question but i need help

    solve the following linear equation

    x-1/x+4 = x+3/x-7



    thank you

    regards
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  2. #2
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    Quote Originally Posted by rpatel View Post
    solve the following linear equation

    x-1/x+4 = x+3/x-7



    thank you

    regards
    x = 4/11; solve for x, that is:

    Multiply both sides by x:

    x^2 + 4*x - 1 = x^2 - 7*x + 3

    Get x's to one side:

    Add +7*x to the left side and subtract the x^2's.

    11*x - 1 = 3

    11*x = 4

    x = 4/11.
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  3. #3
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    hello thank you for that

    but i still can't answer

    7x+2/5 = 4x-1/2

    i tried to use similar method to the answer above but it isn't working out.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rpatel View Post
    solve the following linear equation

    x-1/x+4 = x+3/x-7



    thank you

    regards
    Just in case:
    (x-1)/(x+4) = (x+3)/(x-7) (If it IS this use parathesis!!)

    Multiply both sides by (x+4)(x-7):
    The LHS is
    (x-1)(x-7) = x^2 - 8x + 7

    The RHS is
    (x+3)(x+4) = x^2 + 7x + 12

    So
    x^2 - 8x + 7 = x^2 + 7x + 12
    -8x + 7 = 7x + 12
    -15x = 5
    x = -1/3.

    -Dan
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rpatel View Post
    hello thank you for that

    but i still can't answer

    7x+2/5 = 4x-1/2

    i tried to use similar method to the answer above but it isn't working out.
    7x + 2/5 = 4x - 1/2
    3x = -2/5 - 1/2 = -4/10 - 5/10 = -9/10
    x = -3/10

    -Dan
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  6. #6
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    i think the answers above are incorrect.
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rpatel View Post
    i think the answers above are incorrect.
    Specifically, which posts? You can see numbers at the top right.

    -Dan
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  8. #8
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    Hello, rpate1!

    You really must learn to use parentheses ... (and some spaces wouldn't hurt).

    . . . . . . . . . . . . . 2x + 3
    You cannot write: --------- .as: .2x + 3/5
    . . . . . . . . . . . . . . . 5

    It should be obvious that it looks like "2x plus three-fifths".

    You must write: .(2x + 3)/5
    . . so everyone knows that it is "all over 5".


    7x + 2/5 .= .4x - 1/2

    Is that really "7x plus two-fifths equals 4x minus one-half" ??

    Or did you mean: .(7x + 2)/5 .= .(4x - 1)/2 ?

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  9. #9
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    (x-1)/(x+4) = (x+3)/(x-7)

    i still don't quite understand how you get the answer for this


    for the second one i mean

    (7x+2)/5 = (4x-1)/2
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  10. #10
    Forum Admin topsquark's Avatar
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    rpatel sent me a PM stating he thought my solution to (x-1)/(x+4) = (x+3)/(x-7) was wrong. This is my reply to him:

    Let's take the LHS of the equation first:
    (x - 1)/(x + 4) = ([-1/3] - 1)/([-1/3] + 4)

    = ([-1/3] - [3/3])/([-1/3] + [12/3])

    = ([-4/3])/([11/3]) = -4/11.

    The RHS:
    (x + 3)/(x - 7) = ([-1/3] + 3)/([-1/3] - 7)

    = ([-1/3] + [9/3])/([-1/3] - [21/3])

    = ([8/3])/([-22/3]) = -8/22 = -4/11.

    If any of this is unclear, please let me know.

    I'm going to post this response in your thread as well, in case anyone else is having the same question.

    -Dan
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  11. #11
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rpatel View Post
    (x-1)/(x+4) = (x+3)/(x-7)

    i still don't quite understand how you get the answer for this
    Let's take it step by step.

    The first thing I want to do is get rid of all the fractions. To do this on the LHS I have to multiply by (x + 4), to do it on the RHS I have to multiply by (x - 7). So I'm going to multiply both sides by (x + 4)(x - 7).

    LHS:
    [(x - 1)/(x + 4)]*(x + 4)(x - 7) Note that the two factors of (x + 4) will cancel, leaving:
    (x - 1)(x - 7) = x^2 - 8x + 7.

    RHS:
    [(x + 3)/(x - 7)]*(x + 4)(x - 7) Note that the two factors of (x - 7) will cancel, leaving:
    (x + 3)(x + 4) = x^2 + 7x + 12.

    The rest is equating the two quadratic expressions and solving for x.

    If you are having troubles with the expansion consider the general case:
    (x + a)(x + b). If we distribute the multiplication on the right we get
    x(x + b) + a(x + b) = x^2 + bx + ax + ab = x^2 + (a + b)x + ab.

    Please let me know if anything is still confusing, and what specifically is confusing you.

    -Dan
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  12. #12
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rpatel View Post
    (x-1)/(x+4) = (x+3)/(x-7)

    i still don't quite understand how you get the answer for this


    for the second one i mean

    (7x+2)/5 = (4x-1)/2
    Again, start by clearing the fractions by multiplying both sides by 10.
    [(7x + 2)/5]*10 = [(4x - 1)/2]*10

    (7x + 2)*2 = (4x - 1)*5

    14x + 4 = 20x - 5
    -6x = -9
    x = 3/2.

    -Dan
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  13. #13
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    thank you topsquark

    i have printed out your solution and i am studying it now.

    thank you.

    all help was appreichated, thanks again.

    i haven't had 3 months of maths now. i just had my first maths lesson today. i am going to have a test next week on algebra. thats why i am practicing now.

    regards

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  14. #14
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    Hello, rpatel


    . . . . . . . . x - 1 . . . x + 3
    We have: . ------ .= .------
    . . . . . . . . x + 4 . . .x - 7


    Multiply both sides by the LCD: (x + 4)(x - 7)

    . . . . . . . . . . . . x - 1 . . . . . . . . . . . . . . x + 3
    . . (x + 4)(x - 7) ------- . = . (x + 4)(x - 7) -------
    . . . . . . . . . . . . x + 4 . . . . . . . . . . . . . . x - 7


    Reduce: . (x - 7)(x - 1) .= .(x + 4)(x + 3)

    . . . . . . . . x - 8x + 7 .= .x + 7x + 12

    . . . . . . . . . . . . .-15x .= .5

    . . . . . . . . . . . . . . .x .= .-1/3

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    . . . . . . . . 7x + 2 . . . 4x - 1
    We have: . -------- .= .-------
    . . . . . . . . . . 5 . . . . . . 2


    Multiply both sides by the LCD: 10

    . . . . . 7x + 2 . . . . . 4x - 1
    . . . 10 -------- .= .10 -------
    . . . . . . . 5 . . . . . . . . .2


    Reduce: . 2(7x + 2) .= .5(4x - 1)

    . . . . . . . . 14x + 4 .= .20x - 5

    . . . . . . . . . . . -6x .= .-9

    . . . . . . . . . . . . .x .= .3/2

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  15. #15
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    Quote Originally Posted by AfterShock View Post
    x = 4/11; solve for x, that is:

    Multiply both sides by x:

    x^2 + 4*x - 1 = x^2 - 7*x + 3

    Get x's to one side:

    Add +7*x to the left side and subtract the x^2's.

    11*x - 1 = 3

    11*x = 4

    x = 4/11.
    You've done something wrong. Not sure what it is, but this is how I did it

    (x-1)/(x+4) = (x+3)/(x-7)

    (x-1)/(x+4) - (x+3)/(x-7) = 0

    ((x-1)(x-7) - (x+4)(x+3)) / ((x+4)(x-7)) = 0

    (x-1)/(x-7) - (x+4)/(x+3) = 0

    x^2 - 8x + 7 - (x^2 + 7x + 12) = 0

    -15x - 5 = 0

    -15x = 5

    x = -5/15 = -1/3
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