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Thread: linear functions and linear equations: 8x-3y=6-4x

  1. #11
    oh, okay! that makes a little bit more sense, so i gave it another try and heres what i got this time

    y= 1 1/3 x -2

    x= 1/4 y + 1/2

    and this time, im pretty sure of myself....though i probably did mess up somewhere a little....

  2. #12
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    Quote Originally Posted by c_ortiz92
    oh, okay! that makes a little bit more sense, so i gave it another try and heres what i got this time

    y= 1 1/3 x -2

    x= 1/4 y + 1/2

    and this time, im pretty sure of myself....though i probably did mess up somewhere a little....
    I don't see what problem you are solving - where did you get those.

    If you are solving your original problem - then those are well "totally misguided".
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  3. #13
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    Why are you solving for x and y? All you need is to get your original equation in standard form and plug x = 0 and y = 0 to find your y and x intercepts. Subhotosh Khan showed you how to put it in standard form for you. Now all you need to do is plug in x = 0 and y = 0.

  4. #14
    really? well can you thoroughly explain the original problem to me, which is

    8x-3y=6-4x

    supposed to find x and y intercept

  5. #15
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    Quote Originally Posted by c_ortiz92
    really? well can you thoroughly explain the original problem to me, which is

    8x-3y=6-4x

    supposed to find x and y intercept
    Again, I'll go back to my example (which was not your problem).


    13y - 9x = -3y - x - 12

    4y - 2x + 3 = 0 <--- standard form

    x-intercept(y=0)

    0 - 2x + 3 = 0

    x = 3/2

    So x-intercept is at (1.5,0)

    y-intercept (x =0)

    4y - 0 + 3 = 0

    y = -3/4

    So y-intercept is at (0,-0.75)
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  6. #16
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    Do you understand what I mean when I say "set x = 0" to find the y-intercept? Let's try an example question and see if you can apply it to yours.

    7x = 8y - 32

    For finding the y-intercept, we let x = 0:
    7(0) = 8y - 32
    0 = 8y - 32
    32 = 8y
    y = 4

    So for the equation 7x = 8y - 32, it crosses the y-intercept at y = 4. We do the same thing for the x-intercept except we set y = 0 instead.

    For putting it into standard form (Ax + Bx + C = 0), we move everything to the other side:
    7x - 8y + 32 = 8y - 32 - 8y + 32
    7x - 8y + 32 = 0

    And that's it.

  7. #17
    okay, so first of all, i think im getting this roght this time,....how many times have i said that? anyways.

    i have figured that the standard is

    4x-3y-6=0

    so

    x= 3/4 y + 6/4

  8. #18
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    Can you show us your steps? Something's wrong when you're moving your variables around. Also, why do you solve for x?

  9. #19
    shouldnt i be solving for x if im
    1)writing the equation in standadrd form
    2)determining the x intercept
    3)determining the y intercept
    ?
    well heres what ive done so far

    the original problem is

    8x-3y=6-4x

    next i get

    4x-3y-6=0

    i got this because if you have a negative 4x and you add it to a positive 8x, you are gonna get 4x. and then, because you are moving the 6 to the other side of the equal sign, you are going to change it to a negative...

    and i keep this as the standard form....

    4x-3y-6=0

    as for determining the x intercept

    you have the standard form, which is

    4x-3y-6=0

    so you are going to get rid of the 4 that is attatched to the x, which means that you are going to do this:
    4x=3y+6
    4........4

    and so as the answer, you get

    x=3/4 y + 6/4

    thats what i did,

  10. #20
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    8x - 3y = 6 - 4x

    To get rid of the -4x on the right side, you ADD 4x to both sides. Not subtract.

    8x + 4x - 3y = 6 - 4x + 4x
    12x - 3y = 6
    12x - 3y - 6 = 0

    Ok. When finding the x-intercept, you don't solve for x. You take your equation (in this case 12x - 3y - 6 = 0) and set y = 0:
    12x - 3(0) - 6 = 0
    12x - 6 = 0
    And you solve for x.

    Then, for finding the y-intercept, you don't solve for y. You take your equation again and set x = 0 and THEN solve for y. Don't waste your time doing unnecessary algebra.

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