Here's the question you clicked on:

## gjhfdfg Solving systems with the substitution method, one year ago one year ago

• This Question is Closed
1. gjhfdfg

,|dw:1360809292870:dw|

2. Zelda

So do this: y = 12/x Now plug this in the second equation.

3. gjhfdfg

How do I plug this in? Like, x^2 + 12/x = 40?

4. whpalmer4

$x^2 + y^2=40$ $x^2+(12/x)^2 = 40$ $x^2 + \frac{144}{x^2} = 40$

5. whpalmer4

Then multiply everything by $$x^2$$ to get rid of the fraction and solve.

6. whpalmer4

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

Then solve those for $$x$$. You'll end up with 4 solutions; use $$xy=12$$ to get the accompanying values of $$y$$.

8. gjhfdfg

I'm confused

9. whpalmer4

Why don't you tell me what part of that makes sense to you, and what part doesn't?

10. whpalmer4

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

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

Everything is a big ball of confusion, I don't get why anything of this not coming through to me

13. whpalmer4

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

^ 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

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

Ok, I think I'm with you so far...

17. whpalmer4

Did you understand how I solved the equation by factoring?

18. gjhfdfg

Kinda, except how the 36 & 4 came in

19. gjhfdfg

Ah nevermind, 4 * 36 = 144, that was the factored out version of 144 right?

20. whpalmer4

Well, what are the factors of 144? 1*144 2*72 3*48 4*36 6*24 8*18 12*12

21. gjhfdfg

Got it but why 4 * 36?

22. whpalmer4

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

Got that,

24. whpalmer4

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

Ok, so a=36, b= 4, a*b= 144, is what your saying right?

26. whpalmer4

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

If its zero then wouldnt it make all the answers 0?

28. whpalmer4

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

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

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

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

Please, no laughing, this is serious stuff ;-)

33. gjhfdfg

Oh no I'm not laughing lol. So I dont use the x^2 to find the answers?

34. whpalmer4

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

Ah, sorry, my solutions are (6,2) and (-6,-2)

36. gjhfdfg

Would it be (2,3) & (-2, -3) ?

37. whpalmer4

I agree with the x values, but I'm not so sure about the y values. How did you get them?

38. gjhfdfg

I divided 12 by 4,

39. whpalmer4

Why 4? 4 isn't the value you found for x...

40. gjhfdfg

Should I of done it by 2?

41. whpalmer4

Yes.

42. gjhfdfg

So then it should be (2,6) & (-2, -6)

43. whpalmer4

Right. Notice the symmetry with my solutions? There's a reason for that! Here's a picture:

44. gjhfdfg

Ah

45. whpalmer4

So are you comfortable with your understanding of what we did? Would you like another problem to try?

46. gjhfdfg

I think I'm still shaky with the substitution method, I could use another problem :P

47. whpalmer4

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

Okay, I'll give this one a try,

49. gjhfdfg

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

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

It might look more complicated, but I think it is a bit easier...

52. whpalmer4

"now plug that in ..." is the substitution process we're practicing...

53. gjhfdfg

How would I plug that in the x's place? (10-5) ^2 + 4y^2 = 100?

54. whpalmer4

$(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

You just lost me...

56. whpalmer4

$x+y= 10 \implies x = 10-y$Right?

57. whpalmer4

(the arrow means "implies")

58. whpalmer4

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

Yes,

60. whpalmer4

yes, you would solve it? I'm happy to hear that :-)

61. whpalmer4

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

Yes as in reply to, " x+y=10⟹x=10−y Right?" lol

63. gjhfdfg

Alrighty, take your time. It'll take a bit to get use to these problems

64. gjhfdfg

Everything makes sense up to step 4 where the 10 was taken out, How/why was it taken out?

65. whpalmer4

10-y-5 = 10-5-y = 5-y, no?

66. gjhfdfg

*face palm* Got it lol

67. whpalmer4

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

25y^2?

69. whpalmer4

no...take another look.

70. whpalmer4

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

5*5= 25, 2 -'s = + 2 y's = y^2 I thought?

72. whpalmer4

$(a-b)(a-b) = a(a-b) - b(a-b) = a^2 - ab - ba + b^2 = a^2 -2ab + b^2$

73. whpalmer4

You are thinking of $(a+b)(a-b) = a^2-ba + ba - b^2 = a^2-b^2$I think...

74. gjhfdfg

Who knows what I was thinking...

75. whpalmer4

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

Yikes & yuck aha. Hopefully you had some tomato juice on hand

77. whpalmer4

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

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

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

^thats an easier way way to understand it, having it step by step by step