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Thread: I cannot for the life of me solve this linear equation, help please!

  1. #1

    Unhappy I cannot for the life of me solve this linear equation, help please!

    I'm really stuck here. I tried to solve this but I can't get past step one. I'd really appreciate it if someone could solve this, show their work, and graph this. I'm pretty bad at math and I think this is a systems of equations problem or a linear equation. Thank you in advance.

    Problem: 1/2y= -1/3x

  2. #2
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    Quote Originally Posted by Greh223 View Post
    I'm really stuck here. I tried to solve this but I can't get past step one. I'd really appreciate it if someone could solve this, show their work, and graph this. I'm pretty bad at math and I think this is a systems of equations problem or a linear equation. Thank you in advance.

    Problem: 1/2y= -1/3x
    What exactly do you need to do? I suspect you need to graph this equation. The two denominators are 2 and 3. What is the firs common multiple of 2 and 3. The 1st few multiples of 2 are 2,4,6,8,10 and the 1st few multiples of 3 are 3,6,9,12. The first (that is least) common multiple on both lists is 6. So we multiply both sides my 6. Then we set this equation to 0. Try to do these two steps and show us your work.
    A mathematician is a blind man in a dark room looking for a black cat which isnít there. - Charles R. Darwin

  3. #3
    Quote Originally Posted by Jomo View Post
    What exactly do you need to do? I suspect you need to graph this equation. The two denominators are 2 and 3. What is the firs common multiple of 2 and 3. The 1st few multiples of 2 are 2,4,6,8,10 and the 1st few multiples of 3 are 3,6,9,12. The first (that is least) common multiple on both lists is 6. So we multiply both sides my 6. Then we set this equation to 0. Try to do these two steps and show us your work.
    Alright, so here's my work:

    1/2y = -1/3x

    3y=-2x (my thought process here is that one half of 6 is is 3 and -1/3 of 6 is -2. Please correct me if I'm wrong.)

    -2x - 3y = 0 (Here I subtracted 3 from the left side.)

    Is this okay?

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    Quote Originally Posted by Greh223 View Post
    Alright, so here's my work:

    1/2y = -1/3x

    3y=-2x (my thought process here is that one half of 6 is is 3 and -1/3 of 6 is -2. Please correct me if I'm wrong.)

    -2x - 3y = 0 (Here I subtracted 3 from the left side.)

    Is this okay?
    I would have brought everything to the left side just to have positive coefficients.

    Now let x = 0 and find y. Then let y=0 and find x. Please report back with your results.
    A mathematician is a blind man in a dark room looking for a black cat which isnít there. - Charles R. Darwin

  5. #5
    Quote Originally Posted by Jomo View Post
    I would have brought everything to the left side just to have positive coefficients.

    Now let x = 0 and find y. Then let y=0 and find x. Please report back with your results.

    Agh, yeah, that would've been better.

    So -2x - 3y = 0

    -2x - 3(0) = 0

    -2x/-2= 0 / -2

    x = 0 ; x-intercept is (0,0) (?)
    ~~~~~~~~~~~~~~~~~~~~~~~~~~

    -2(0) - 3y = 0
    0 - 3y = 0
    -3y/-3 = 0/-3

    y = 0 ; y-intercept is also (0,0) (?)

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    Quote Originally Posted by Greh223 View Post
    Agh, yeah, that would've been better.

    So -2x - 3y = 0

    -2x - 3(0) = 0

    -2x/-2= 0 / -2

    x = 0 ; x-intercept is (0,0) (?)
    ~~~~~~~~~~~~~~~~~~~~~~~~~~

    -2(0) - 3y = 0
    0 - 3y = 0
    -3y/-3 = 0/-3

    y = 0 ; y-intercept is also (0,0) (?)
    Yes, -2x- 3y= 0 or 2x+ 3y= 0 are the same as 3y= -2x, y= (-2/3)x.

    The graph of that passes through (0, 0), so both x and y intercepts are (0, 0), and has slope -2/3. So, for example, when x= 3, y= (-2/3)(3)= -2 and when x= -3, y= (-2/3)(-3)= -2. The graph is a straight line passing through the points (-2, 3), (0, 0), and (2, -3).

  7. #7
    Quote Originally Posted by HallsofIvy View Post
    Yes, -2x- 3y= 0 or 2x+ 3y= 0 are the same as 3y= -2x, y= (-2/3)x.

    The graph of that passes through (0, 0), so both x and y intercepts are (0, 0), and has slope -2/3. So, for example, when x= 3, y= (-2/3)(3)= -2 and when x= -3, y= (-2/3)(-3)= -2. The graph is a straight line passing through the points (-2, 3), (0, 0), and (2, -3).
    graph.png Oh, right! Thanks, so they graph would look like this, correct?

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    Elite Member stapel's Avatar
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    Cool

    Quote Originally Posted by Greh223 View Post
    I'm really stuck here. I tried to solve this but I can't get past step one. I'd really appreciate it if someone could solve this, show their work, and graph this. I'm pretty bad at math and I think this is a systems of equations problem or a linear equation. Thank you in advance.

    Problem: 1/2y= -1/3x
    Note: If the equation is actually as posted, it is not linear since, in its original form, technically you cannot plug in zero for either of x or y. So you'd have holes where the line crosses the axes.

    What did the instructions say? What has caused you to think that this rational equation is a linear equation, that this single equation is actually a system of equations, and that this "graphing" exercise (apparently?) is actually a "solving" exercise? What "work" have you done?

    Please be complete. Thank you!

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    Stapel's point is, I think, that what you originally wrote, "1/2x= -1/3y", would normally be interpreted as "1/(2x)= -1/(3y)", which is NOT linear, rather than "(1/2)x= (-1/3)y" which, from the work you showed, is what you intended.

    If the problem were actually "1/(2x)= -1/(3y)" then multiplying both sides by 2x and 3y, would give 3y= -2x as long as x and y are not 0. Its graph would be a straight line with a "hole" at (0, 0).

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