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gjhfdfg

  • 3 years ago

Solving systems with the substitution method,

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  1. gjhfdfg
    • 3 years ago
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    ,|dw:1360809292870:dw|

  2. Zelda
    • 3 years ago
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    So do this: y = 12/x Now plug this in the second equation.

  3. gjhfdfg
    • 3 years ago
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    How do I plug this in? Like, x^2 + 12/x = 40?

  4. whpalmer4
    • 3 years ago
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    \[x^2 + y^2=40\] \[x^2+(12/x)^2 = 40\] \[x^2 + \frac{144}{x^2} = 40\]

  5. whpalmer4
    • 3 years ago
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    Then multiply everything by \(x^2\) to get rid of the fraction and solve.

  6. whpalmer4
    • 3 years ago
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    When you multiply through by \(x^2\) you get \[x^4 + 144 = 40x^2\]or\[x^4 -40x^2+144 =0\]How to solve that? Make a substitution \(u = x^2\) to turn it into a quadratic: \[u^2-40u+144 = 0\] Find the solutions for \(u\), then replace \(u\) with \(x^2\).

  7. whpalmer4
    • 3 years ago
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    Then solve those for \(x\). You'll end up with 4 solutions; use \(xy=12\) to get the accompanying values of \(y\).

  8. gjhfdfg
    • 3 years ago
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    I'm confused

  9. whpalmer4
    • 3 years ago
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    Why don't you tell me what part of that makes sense to you, and what part doesn't?

  10. whpalmer4
    • 3 years ago
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    You can also solve \[x^4-40x^2+144=0\]by factoring. You'll have \[(x^2-a)(x^2-b)=0\] where the values of a and b add to 40 and multiply to 144. 4 and 36 fit that description, so \[x^4-40x^2+144 = (x^2-36)(x^2-4) = 0\]Now find the values of \(x\) so that \[x^2=36\]\[x^2=4\](and don't forget about the negative values that solve those equations!)

  11. gjhfdfg
    • 3 years ago
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    First confusion is multiplying by x^2, you've shown the steps but I don't get how everything came out to be

  12. gjhfdfg
    • 3 years ago
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    Everything is a big ball of confusion, I don't get why anything of this not coming through to me

  13. whpalmer4
    • 3 years ago
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    Okay, I was just trying to get rid of the fraction. You can multiply both sides of the equation by x^2 without damaging the equality, and it eliminates the fraction.

  14. gjhfdfg
    • 3 years ago
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    ^ that doesnt make any sense. :/ I get (12/x)^2 = 144, except what happened to the x? Wouldnt it of made it 144x?

  15. whpalmer4
    • 3 years ago
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    Look at it a different way: \[x^2 + \frac{144}{x^2} = 40\]We want to make a common denominator of \(x^2\) so we can add the two terms on the left. So, we can multiply the first term by \(\frac{x^2}{x^2}\) to get \[\frac{x^2}{1}*\frac{x^2}{x^2} + \frac{144}{x^2} = 40\]or\[\frac{x^4+144}{x^2} = 40\]But to solve that, we'll end up multiplying both sides by \(x^2\) giving \[x^4+144 = 40x^2\]and so I just saved a step multiplying by \(x^2\) instead of multiplying the first term by 1 (remember, \(x^2/x^2 = 1\)) and then the entire equation by \(x^2\). \[(\frac{12}{x})^2 = \frac{12}{x}*\frac{12}{x} = \frac{12*12}{x*x} = \frac{144}{x^2}\]

  16. gjhfdfg
    • 3 years ago
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    Ok, I think I'm with you so far...

  17. whpalmer4
    • 3 years ago
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    Did you understand how I solved the equation by factoring?

  18. gjhfdfg
    • 3 years ago
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    Kinda, except how the 36 & 4 came in

  19. gjhfdfg
    • 3 years ago
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    Ah nevermind, 4 * 36 = 144, that was the factored out version of 144 right?

  20. whpalmer4
    • 3 years ago
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    Well, what are the factors of 144? 1*144 2*72 3*48 4*36 6*24 8*18 12*12

  21. gjhfdfg
    • 3 years ago
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    Got it but why 4 * 36?

  22. whpalmer4
    • 3 years ago
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    Because 4 and 36 add to 40 (the coefficient of the middle term) and multiply to 144 (the coefficient of the final term). \[(x^2-4)(x^2-36) = x^4 -36x^2 -4x^2+144 = x^4 - 40x^2 +144\]

  23. gjhfdfg
    • 3 years ago
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    Got that,

  24. whpalmer4
    • 3 years ago
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    Remember, when we multiply \[(x-a)(x-b) = x^2-ax-bx+ab = x^2-(a+b)x+ab\] Compare that with our equation (pretend it in x instead of x^2 for the moment) \[x^2-40x+144\] We can see that (a+b) = 40, and ab = 144. 36 and 4 are the two numbers that satisfy those equations.

  25. gjhfdfg
    • 3 years ago
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    Ok, so a=36, b= 4, a*b= 144, is what your saying right?

  26. whpalmer4
    • 3 years ago
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    Yes, that is correct. So that leaves us with solving \[(x^2-36)(x^2-4) = 0\]Clearly, for that product to be 0, either component equals 0, so \[x^2-36=0\]\[x^2-4=0\]will give us the answers. How would you solve them?

  27. gjhfdfg
    • 3 years ago
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    If its zero then wouldnt it make all the answers 0?

  28. whpalmer4
    • 3 years ago
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    No, because the value of x that makes x^2-36 = 0 isn't the same value as the value of x that makes x^2-4 = 0, is it?

  29. whpalmer4
    • 3 years ago
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    To find all the solutions, we have to find all the different values of x that can make that product = 0. There are 4 of them, because we have an x^4 in the equation.

  30. whpalmer4
    • 3 years ago
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    Looking at it a different way, 0*0 is not the only way to get 0 in a multiplication. 1*0 = 0, 2*0 = 0, 13423523493459823487234.6734234*0 = 0, etc.

  31. whpalmer4
    • 3 years ago
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    I'm not sure I've ever actually multiplied 13423523493459823487234.6734234*0 before now, but I'm still pretty sure it equals 0 :-)

  32. whpalmer4
    • 3 years ago
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    Please, no laughing, this is serious stuff ;-)

  33. gjhfdfg
    • 3 years ago
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    Oh no I'm not laughing lol. So I dont use the x^2 to find the answers?

  34. whpalmer4
    • 3 years ago
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    Yeah, you do use x^2 to find the answers. I'll do two, you can do two. \[x^2-36=0\]\[x^2=36\]Take square root of both sides \[x=\pm \sqrt{36} = \pm 6\] Remember that \(-6*-6 =36\) so we have to include it as one of the solutions. Now we'll find the values of \(y\) that go with those values of \(x\):\[xy=12\]\[y=\frac{12}x\]\[y=\frac{12}6 = 2\]\[y=\frac{12}{-6}=-2\]My two solutions are \[(6,2),(6,-2)\] What are your solutions for the other half? (\(x^2-4=0\))

  35. whpalmer4
    • 3 years ago
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    Ah, sorry, my solutions are (6,2) and (-6,-2)

  36. gjhfdfg
    • 3 years ago
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    Would it be (2,3) & (-2, -3) ?

  37. whpalmer4
    • 3 years ago
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    I agree with the x values, but I'm not so sure about the y values. How did you get them?

  38. gjhfdfg
    • 3 years ago
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    I divided 12 by 4,

  39. whpalmer4
    • 3 years ago
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    Why 4? 4 isn't the value you found for x...

  40. gjhfdfg
    • 3 years ago
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    Should I of done it by 2?

  41. whpalmer4
    • 3 years ago
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    Yes.

  42. gjhfdfg
    • 3 years ago
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    So then it should be (2,6) & (-2, -6)

  43. whpalmer4
    • 3 years ago
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    Right. Notice the symmetry with my solutions? There's a reason for that! Here's a picture:

  44. gjhfdfg
    • 3 years ago
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    Ah

  45. whpalmer4
    • 3 years ago
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    So are you comfortable with your understanding of what we did? Would you like another problem to try?

  46. gjhfdfg
    • 3 years ago
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    I think I'm still shaky with the substitution method, I could use another problem :P

  47. whpalmer4
    • 3 years ago
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    Just a minute, I want to make sure I don't give you one that is too awkward! Okay, how about this? You'll only have 2 solutions for this one: \[(x-5)^2+4y^2=100, x+y=10\]

  48. gjhfdfg
    • 3 years ago
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    Okay, I'll give this one a try,

  49. gjhfdfg
    • 3 years ago
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    Ok this looks more complicated. o_o so, (x-5) ^2 + 4y^2 = 100, x + y = 10, I take the square root of 5 which is 25, makes the problem (x-25) + 4y^2 = 100, Am I going in the right direction so far?

  50. whpalmer4
    • 3 years ago
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    Square root of 5 isn't 25, 5 is the square root of 25. I suggest that you first solve x+y=10 to give you x = 10-y. Now plug that in to the other formula in place of x...

  51. whpalmer4
    • 3 years ago
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    It might look more complicated, but I think it is a bit easier...

  52. whpalmer4
    • 3 years ago
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    "now plug that in ..." is the substitution process we're practicing...

  53. gjhfdfg
    • 3 years ago
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    How would I plug that in the x's place? (10-5) ^2 + 4y^2 = 100?

  54. whpalmer4
    • 3 years ago
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    \[(x-5)^2+4y^2=100\]\[x=10-y\]\[((10-y)-5)^2+4y^2=100\]\[(5-y)^2+4y^2=100\]Now expand \[(5-y)^2\]and continue on to the solution

  55. gjhfdfg
    • 3 years ago
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    You just lost me...

  56. whpalmer4
    • 3 years ago
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    \[x+y= 10 \implies x = 10-y\]Right?

  57. whpalmer4
    • 3 years ago
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    (the arrow means "implies")

  58. whpalmer4
    • 3 years ago
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    So we just take our equation and write (10-y) wherever we see x. \[(x-5)^2+4y^2=100\]\[((10-y)-5)^2 + 4y^2=100\]\[(10-y-5)^2 + 4y^2=100\]\[(5-y)^2+4y^2=100\]If we multiply out (expand) (5-y)^2 \[(5-y)^2=(5-y)(5-y) = 5*5-5y-5y+(-y)(-y) = 25-10y+y^2\]so our equation goes from \[(5-y)^2+4y^2=100\]to\[(25-10y+y^2)+4y^2=100\]and after rearranging\[5y^2-10y+25=100\]What would you do to solve that bad boy?

  59. gjhfdfg
    • 3 years ago
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    Yes,

  60. whpalmer4
    • 3 years ago
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    yes, you would solve it? I'm happy to hear that :-)

  61. whpalmer4
    • 3 years ago
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    I have to go out for a while, so I'll post the rest of the steps in my next post, don't look ahead until you've tried to solve it, okay?

  62. gjhfdfg
    • 3 years ago
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    Yes as in reply to, " x+y=10⟹x=10−y Right?" lol

  63. gjhfdfg
    • 3 years ago
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    Alrighty, take your time. It'll take a bit to get use to these problems

  64. gjhfdfg
    • 3 years ago
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    Everything makes sense up to step 4 where the 10 was taken out, How/why was it taken out?

  65. whpalmer4
    • 3 years ago
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    10-y-5 = 10-5-y = 5-y, no?

  66. gjhfdfg
    • 3 years ago
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    *face palm* Got it lol

  67. whpalmer4
    • 3 years ago
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    And probably a little extra care needs to be taken when you multiply things like (5-y)(5-y) since the order is different than you are used to...

  68. gjhfdfg
    • 3 years ago
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    25y^2?

  69. whpalmer4
    • 3 years ago
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    no...take another look.

  70. whpalmer4
    • 3 years ago
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    if you prefer, it's the same as \((-1)(y-5)(-1)(y-5)\) because \(5-y = -1(y-5)\) The \((-1)(-1) = 1\), and \((y-5)(y-5) = y^2-5y-5y+25 = y^2-10y+25\)

  71. gjhfdfg
    • 3 years ago
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    5*5= 25, 2 -'s = + 2 y's = y^2 I thought?

  72. whpalmer4
    • 3 years ago
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    \[(a-b)(a-b) = a(a-b) - b(a-b) = a^2 - ab - ba + b^2 = a^2 -2ab + b^2\]

  73. whpalmer4
    • 3 years ago
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    You are thinking of \[(a+b)(a-b) = a^2-ba + ba - b^2 = a^2-b^2\]I think...

  74. gjhfdfg
    • 3 years ago
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    Who knows what I was thinking...

  75. whpalmer4
    • 3 years ago
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    Today's unrelated, unsolicited life tip: when walking home in the dark, do not under any circumstances startle a skunk by nearly stepping on it. Guess what happened to me last night? :-)

  76. gjhfdfg
    • 3 years ago
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    Yikes & yuck aha. Hopefully you had some tomato juice on hand

  77. whpalmer4
    • 3 years ago
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    Hydrogen peroxide and baking soda is a better combo. Fortunately it was cold enough that I wasn't walking around in short sleeves, and just my clothing got zapped. Unfortunately, I seem to have brushed against many surfaces in the house on the way :-( If I don't respond, I'm probably busy cleaning something!

  78. whpalmer4
    • 3 years ago
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    Here are the final steps, don't read on until you're ready to check your work or completely stuck: How to solve \[5y^2-10y+25=100\] Let's move the 100 onto the same side as everything else: \[5y^2-10y+25-100=0\]\[5y^2-10y-75=0\]Now let's divide out the common factor of 5:\[y^2-2y-15=0\]Which two numbers multiply to -15 and add to -2? -5 and 3 do, so we can factor that as\[(y-5)(y+3)=0\]Now solve each of those for 0\[y-5=0, y=5\]\[y+3=0, y=-3\]And finally use our substitution equation to find the corresponding values of x:\[x=10-y\]\[x=10-5 = 5\]\[x=10-(-3)=13\]So our two solutions are (5,5) and (13,-3).

  79. gjhfdfg
    • 3 years ago
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    Hah no worries, its only a practice problem I'm in no rush. :-) I've had my share of ' skunk perfume' when I was younger. Had a dog who loved to catch them & bring them to me. Unfortunately(the odor), it was still alive & wanted to share its odor. I'm off to go grab a bite to eat, I'll annoy you with my questions later. :P

  80. gjhfdfg
    • 3 years ago
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    ^thats an easier way way to understand it, having it step by step by step

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