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Solve the following system.

  1. Apr 14, 2006 #1
    I was reading the news and I found question 11 to be confusing. I never learned how to solve systems.
    [​IMG]

    Question:
    Solve the following system for x and y
    1)[tex]\frac{x}{8}-\frac{y}{2}=1[/tex]-------------------------2)[tex]\frac{x}{3}=y+\frac{2}{3}[/tex]

    WORK:

    1)[tex]\frac{x}{8}-\frac{y}{2}=1[/tex]-------------------------2)[tex]\frac{x}{3}-\frac{2}{3}=y[/tex]

    1)[tex]=\frac{x}{4}-y=2[/tex]

    [tex]x-y=8[/tex]-------------------------[tex]x-2=3y[/tex]

    Edit: Changed the "news" link.
     
    Last edited: Apr 14, 2006
  2. jcsd
  3. Apr 14, 2006 #2

    arildno

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    Dearly Missed

    Not surprising, since what you've written is meaningless.

    You've got TWO numbers, called x and y, that simultaneously satisfies TWO equations.

    Please set up the ORIGINAL equations, your attempt of solving them is totally flawed.
     
  4. Apr 14, 2006 #3
    I am sorry that I had the wrong link. Now, it's corrected.
     
  5. Apr 14, 2006 #4

    berkeman

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    You've still got a lot of algebra errors in your GIF file and your post. It should look more like:

    -1-
    x/8 - y/2 = 1, which simplifies to x - 4y = 8

    -2-
    x/3 = y + 2/3, which simplifies to x - 3y = 2

    Then you combine the simplified equations 1 & 2 to eliminate one variable, so you can solve for the other one. Do you have an idea of how to combine the simplified 1 & 2 to solve for y? Hint -- think about subtracting equations....
     
  6. Apr 14, 2006 #5

    HallsofIvy

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    It looked like you were starting out correctly by solving the second equation for y: y= x/3- 2/3. But then your x/4- y= 2 is puzzling. I don't see where it comes from. Presumably the reason for solving for y was to substitute for y in the first equation: x/8- y/2= x/8 - (1/2)(x/3- 2/3)= x/8- x/6- 1/3= 1.
    Then 3x/24- 4x/24- 1/3= -x/24- 1/3= 1. Adding 1/3 to both sides,
    -x/24= 1+ 1/3= 4/3 so x= (4.24)/(3)= 8(4)= 32. Then y= 32/3- 2/3= 30/3= 10.
     
  7. Apr 14, 2006 #6
    I finally found how to do this.
    From equation 1, I found [tex]x=8+4y[/tex]
    I used that equation as a replacment of x in equation 2.
    I got x= -16 and y= -6.
     
  8. Apr 14, 2006 #7
    Yep that works perfectly.
     
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