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85295james

Solve the system of equations by graphing

  • one year ago
  • one year ago

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  1. 85295james
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    -1/3x + y = –1 y = 4+1/3 x

    • one year ago
  2. theEric
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    Do you know what it means to solve them by graphing? I'm unsure! I'm into calculus 3 and I can't think of what it means. Does this sound familiar? You're looking for a point (x,y) so that \[\frac{1}{3}x+y=-1y=4+\frac{1}{3}x\]

    • one year ago
  3. theEric
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    Because I would immediately think to use algebra. I'm looking online for a refresher. If you multiplied by (-1), though... \[(-1)\frac{1}{3}x+(-1)y=y=(-1)4+(-1)\frac{1}{3}x\]

    • one year ago
  4. theEric
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    But that would be changing values. I'm not sure right now, sorry, but I will research how to solve a syste of equations by graphing now.

    • one year ago
  5. theEric
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    Wait.... I should've asked - are those two separate equations? I thought they were, and then I decided maybe they weren't..

    • one year ago
  6. theEric
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    If you have two line equations and they share just one (x,y) point, then they intersect on a graph. You just have to be able to plot the lines on grid paper and look for the intersection. For really random lines you will need to calculate, but for math problems designed for learning, you should be fine with just graphing.

    • one year ago
  7. theEric
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    So \[-\frac{1}{3}x + y = -1\]and\[y=4+\frac{1}{3}x\] ?

    • one year ago
  8. theEric
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    THAT I can do for sure! :)

    • one year ago
  9. theEric
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    That would be a system of equations...

    • one year ago
  10. theEric
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    Sorry I delayed due to uncertainty. I'll guide you through it now. First, you should know how to plot a line, given it's equation. Second, you should rearange each equation so that you can easily plot it (I recommend making your equations look like y=mx+b) Third, you need to plot both lines. Fourth, you can the see where they intersect. So look at what the point is.

    • one year ago
  11. theEric
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    Of the four steps, where would you like to start? You need to do all four, but maybe you already know how to plot a line with it's equation. If you don't, let me know, please!

    • one year ago
  12. theEric
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    Actually there is a 5th step. You have to check to make sure that the (x,y) point you found to be the intersection actually IS on both lines buy using eliassaab's method for substituting.

    • one year ago
  13. 85295james
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    i got three pics to choos from and i dont know how to load them

    • one year ago
  14. 85295james
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    B. no solution C. no solution D. infinitely many solutions

    • one year ago
  15. theEric
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    Once I put the equations into y=mx+b form, which I will do in a moment, you will see that it must be "no solution". Once you can plot the two equations, you will see why!

    • one year ago
  16. theEric
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    For now, think about intersections. As simple as they are. They are the point or points where two things meet! If they never meet, then there are no intersections. If they are together at every point, there are infinately many points of intersection.

    • one year ago
  17. 85295james
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    |dw:1346094672501:dw|

    • one year ago
  18. 85295james
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    |dw:1346094737400:dw|

    • one year ago
  19. theEric
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    I'll trust that you can algebraically manipulate your equations to look like y=mx+b. \[-\frac{1}{3}x+y=-1\rightarrow y=\frac{1}{3}-1\]

    • one year ago
  20. 85295james
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    no

    • one year ago
  21. theEric
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    and \[y=\frac{1}{3}x + 4\]

    • one year ago
  22. theEric
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    No you can't do the algebra? Well lets look at the one line's equation: \[-\frac{1}{3}x+y=-1\]

    • one year ago
  23. theEric
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    Now, you have a goal. You want that "y" to be on its own side, all alone.

    • one year ago
  24. theEric
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    There's a rule of thumb, "if you do something to one side, do it to the other". This is so you do the same thing to all sides. By changing each side in the same way, each side will be different from what it was before. BUT each changed side will be equal to each other, so you know it's a legitamate equation! Here's an example, to help you understand what I mean. \[5=2+3\] add 10 to both sides \[5+(10) = 2+3+(10)\] and solve \[15=2+3+10\] \[15=15\]

    • one year ago
  25. theEric
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    Since both sides are equal, each side is okay to work with. And the variables are still the same too. \[x=1\]add 10 to both sides\[x+(10) = 1+(10)\]\[x+10=11\]x still equals 1!

    • one year ago
  26. theEric
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    So changing both sides is a great thing to do! I was adding 10 in those examples. To your equation, try adding \[\frac{1}{3}x\] to both sides. Here is the equation again: \[-\frac{1}{3}x+y=1\]

    • one year ago
  27. theEric
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    Tell me when your done, so I know you've seen the result, and we can talk about why I told you to add (1/3)x.

    • one year ago
  28. theEric
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    Sorry I've taken so long!

    • one year ago
  29. 85295james
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    it is cool

    • one year ago
  30. theEric
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    Thanks! So, have you done the addition? What did you get?

    • one year ago
  31. theEric
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    \[-\frac{1}{3}x+\frac{1}{3}x+y=-1+\frac{1}{3}x\]is what you want to simplify, here. I took the liberty of adding to both sides :P Just to be clear! :)

    • one year ago
  32. theEric
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    Okay, I'll go on without you... But I hope that you'll ask me about anything you don't understand or that you see might be wrong! I got this much, much more understandable equation,\[y=-1+\frac{1}{3}x=\frac{1}{3}x-1\] which looks like\[y=mx+b\] which is easier to plot by hand.

    • one year ago
  33. 85295james
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    so is it C

    • one year ago
  34. theEric
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    Nope! Just look at either equation, and the number multiplying x, known as the slope, is \[\frac{1}{3}\] The slope is, you go up one unit (along y) and you go over 3 units (along x). The slope goes up and to the right. So look for that.

    • one year ago
  35. theEric
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    Since they habe the same slope, they are parallel, and thus cannot intersect.

    • one year ago
  36. theEric
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    Thank you!

    • one year ago
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