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85295james

  • 2 years ago

Solve the system of equations by graphing

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  1. 85295james
    • 2 years ago
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    -1/3x + y = –1 y = 4+1/3 x

  2. theEric
    • 2 years ago
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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 \[\frac{1}{3}x+y=-1y=4+\frac{1}{3}x\]

  3. theEric
    • 2 years ago
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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... \[(-1)\frac{1}{3}x+(-1)y=y=(-1)4+(-1)\frac{1}{3}x\]

  4. theEric
    • 2 years ago
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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.

  5. theEric
    • 2 years ago
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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..

  6. theEric
    • 2 years ago
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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.

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

  8. theEric
    • 2 years ago
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    THAT I can do for sure! :)

  9. theEric
    • 2 years ago
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    That would be a system of equations...

  10. theEric
    • 2 years ago
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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.

  11. theEric
    • 2 years ago
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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!

  12. theEric
    • 2 years ago
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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.

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

  14. 85295james
    • 2 years ago
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    B. no solution C. no solution D. infinitely many solutions

  15. theEric
    • 2 years ago
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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!

  16. theEric
    • 2 years ago
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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.

  17. 85295james
    • 2 years ago
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    |dw:1346094672501:dw|

  18. 85295james
    • 2 years ago
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    |dw:1346094737400:dw|

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

  20. 85295james
    • 2 years ago
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    no

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

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

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

  24. theEric
    • 2 years ago
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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 \[15=2+3+10\] \[15=15\]

  25. theEric
    • 2 years ago
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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!

  26. theEric
    • 2 years ago
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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: \[-\frac{1}{3}x+y=1\]

  27. theEric
    • 2 years ago
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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.

  28. theEric
    • 2 years ago
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    Sorry I've taken so long!

  29. 85295james
    • 2 years ago
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    it is cool

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

  31. theEric
    • 2 years ago
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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! :)

  32. theEric
    • 2 years ago
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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.

  33. 85295james
    • 2 years ago
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    so is it C

  34. theEric
    • 2 years ago
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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.

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

  36. theEric
    • 2 years ago
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    Thank you!

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