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gjhfdfg
 2 years ago
Solving systems with the substitution method,
gjhfdfg
 2 years ago
Solving systems with the substitution method,

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Zelda
 2 years ago
Best ResponseYou've already chosen the best response.0So do this: y = 12/x Now plug this in the second equation.

gjhfdfg
 2 years ago
Best ResponseYou've already chosen the best response.0How do I plug this in? Like, x^2 + 12/x = 40?

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1\[x^2 + y^2=40\] \[x^2+(12/x)^2 = 40\] \[x^2 + \frac{144}{x^2} = 40\]

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1Then multiply everything by \(x^2\) to get rid of the fraction and solve.

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1When 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: \[u^240u+144 = 0\] Find the solutions for \(u\), then replace \(u\) with \(x^2\).

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1Then solve those for \(x\). You'll end up with 4 solutions; use \(xy=12\) to get the accompanying values of \(y\).

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1Why don't you tell me what part of that makes sense to you, and what part doesn't?

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1You can also solve \[x^440x^2+144=0\]by factoring. You'll have \[(x^2a)(x^2b)=0\] where the values of a and b add to 40 and multiply to 144. 4 and 36 fit that description, so \[x^440x^2+144 = (x^236)(x^24) = 0\]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!)

gjhfdfg
 2 years ago
Best ResponseYou've already chosen the best response.0First confusion is multiplying by x^2, you've shown the steps but I don't get how everything came out to be

gjhfdfg
 2 years ago
Best ResponseYou've already chosen the best response.0Everything is a big ball of confusion, I don't get why anything of this not coming through to me

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1Okay, 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.

gjhfdfg
 2 years ago
Best ResponseYou've already chosen the best response.0^ that doesnt make any sense. :/ I get (12/x)^2 = 144, except what happened to the x? Wouldnt it of made it 144x?

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1Look 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 \(x^2\). \[(\frac{12}{x})^2 = \frac{12}{x}*\frac{12}{x} = \frac{12*12}{x*x} = \frac{144}{x^2}\]

gjhfdfg
 2 years ago
Best ResponseYou've already chosen the best response.0Ok, I think I'm with you so far...

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1Did you understand how I solved the equation by factoring?

gjhfdfg
 2 years ago
Best ResponseYou've already chosen the best response.0Kinda, except how the 36 & 4 came in

gjhfdfg
 2 years ago
Best ResponseYou've already chosen the best response.0Ah nevermind, 4 * 36 = 144, that was the factored out version of 144 right?

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1Well, what are the factors of 144? 1*144 2*72 3*48 4*36 6*24 8*18 12*12

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1Because 4 and 36 add to 40 (the coefficient of the middle term) and multiply to 144 (the coefficient of the final term). \[(x^24)(x^236) = x^4 36x^2 4x^2+144 = x^4  40x^2 +144\]

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1Remember, when we multiply \[(xa)(xb) = x^2axbx+ab = x^2(a+b)x+ab\] Compare that with our equation (pretend it in x instead of x^2 for the moment) \[x^240x+144\] We can see that (a+b) = 40, and ab = 144. 36 and 4 are the two numbers that satisfy those equations.

gjhfdfg
 2 years ago
Best ResponseYou've already chosen the best response.0Ok, so a=36, b= 4, a*b= 144, is what your saying right?

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1Yes, that is correct. So that leaves us with solving \[(x^236)(x^24) = 0\]Clearly, for that product to be 0, either component equals 0, so \[x^236=0\]\[x^24=0\]will give us the answers. How would you solve them?

gjhfdfg
 2 years ago
Best ResponseYou've already chosen the best response.0If its zero then wouldnt it make all the answers 0?

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1No, because the value of x that makes x^236 = 0 isn't the same value as the value of x that makes x^24 = 0, is it?

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1To 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.

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1Looking 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.

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1I'm not sure I've ever actually multiplied 13423523493459823487234.6734234*0 before now, but I'm still pretty sure it equals 0 :)

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1Please, no laughing, this is serious stuff ;)

gjhfdfg
 2 years ago
Best ResponseYou've already chosen the best response.0Oh no I'm not laughing lol. So I dont use the x^2 to find the answers?

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1Yeah, you do use x^2 to find the answers. I'll do two, you can do two. \[x^236=0\]\[x^2=36\]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? (\(x^24=0\))

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1Ah, sorry, my solutions are (6,2) and (6,2)

gjhfdfg
 2 years ago
Best ResponseYou've already chosen the best response.0Would it be (2,3) & (2, 3) ?

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1I agree with the x values, but I'm not so sure about the y values. How did you get them?

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1Why 4? 4 isn't the value you found for x...

gjhfdfg
 2 years ago
Best ResponseYou've already chosen the best response.0Should I of done it by 2?

gjhfdfg
 2 years ago
Best ResponseYou've already chosen the best response.0So then it should be (2,6) & (2, 6)

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1Right. Notice the symmetry with my solutions? There's a reason for that! Here's a picture:

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1So are you comfortable with your understanding of what we did? Would you like another problem to try?

gjhfdfg
 2 years ago
Best ResponseYou've already chosen the best response.0I think I'm still shaky with the substitution method, I could use another problem :P

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1Just 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: \[(x5)^2+4y^2=100, x+y=10\]

gjhfdfg
 2 years ago
Best ResponseYou've already chosen the best response.0Okay, I'll give this one a try,

gjhfdfg
 2 years ago
Best ResponseYou've already chosen the best response.0Ok this looks more complicated. o_o so, (x5) ^2 + 4y^2 = 100, x + y = 10, I take the square root of 5 which is 25, makes the problem (x25) + 4y^2 = 100, Am I going in the right direction so far?

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1Square 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 = 10y. Now plug that in to the other formula in place of x...

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1It might look more complicated, but I think it is a bit easier...

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1"now plug that in ..." is the substitution process we're practicing...

gjhfdfg
 2 years ago
Best ResponseYou've already chosen the best response.0How would I plug that in the x's place? (105) ^2 + 4y^2 = 100?

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1\[(x5)^2+4y^2=100\]\[x=10y\]\[((10y)5)^2+4y^2=100\]\[(5y)^2+4y^2=100\]Now expand \[(5y)^2\]and continue on to the solution

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1\[x+y= 10 \implies x = 10y\]Right?

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1(the arrow means "implies")

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1So we just take our equation and write (10y) wherever we see x. \[(x5)^2+4y^2=100\]\[((10y)5)^2 + 4y^2=100\]\[(10y5)^2 + 4y^2=100\]\[(5y)^2+4y^2=100\]If we multiply out (expand) (5y)^2 \[(5y)^2=(5y)(5y) = 5*55y5y+(y)(y) = 2510y+y^2\]so our equation goes from \[(5y)^2+4y^2=100\]to\[(2510y+y^2)+4y^2=100\]and after rearranging\[5y^210y+25=100\]What would you do to solve that bad boy?

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1yes, you would solve it? I'm happy to hear that :)

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1I 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?

gjhfdfg
 2 years ago
Best ResponseYou've already chosen the best response.0Yes as in reply to, " x+y=10⟹x=10−y Right?" lol

gjhfdfg
 2 years ago
Best ResponseYou've already chosen the best response.0Alrighty, take your time. It'll take a bit to get use to these problems

gjhfdfg
 2 years ago
Best ResponseYou've already chosen the best response.0Everything makes sense up to step 4 where the 10 was taken out, How/why was it taken out?

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.110y5 = 105y = 5y, no?

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1And probably a little extra care needs to be taken when you multiply things like (5y)(5y) since the order is different than you are used to...

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1no...take another look.

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1if you prefer, it's the same as \((1)(y5)(1)(y5)\) because \(5y = 1(y5)\) The \((1)(1) = 1\), and \((y5)(y5) = y^25y5y+25 = y^210y+25\)

gjhfdfg
 2 years ago
Best ResponseYou've already chosen the best response.05*5= 25, 2 's = + 2 y's = y^2 I thought?

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1\[(ab)(ab) = a(ab)  b(ab) = a^2  ab  ba + b^2 = a^2 2ab + b^2\]

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1You are thinking of \[(a+b)(ab) = a^2ba + ba  b^2 = a^2b^2\]I think...

gjhfdfg
 2 years ago
Best ResponseYou've already chosen the best response.0Who knows what I was thinking...

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1Today'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? :)

gjhfdfg
 2 years ago
Best ResponseYou've already chosen the best response.0Yikes & yuck aha. Hopefully you had some tomato juice on hand

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1Hydrogen 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!

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1Here are the final steps, don't read on until you're ready to check your work or completely stuck: How to solve \[5y^210y+25=100\] Let's move the 100 onto the same side as everything else: \[5y^210y+25100=0\]\[5y^210y75=0\]Now let's divide out the common factor of 5:\[y^22y15=0\]Which two numbers multiply to 15 and add to 2? 5 and 3 do, so we can factor that as\[(y5)(y+3)=0\]Now solve each of those for 0\[y5=0, y=5\]\[y+3=0, y=3\]And finally use our substitution equation to find the corresponding values of x:\[x=10y\]\[x=105 = 5\]\[x=10(3)=13\]So our two solutions are (5,5) and (13,3).

gjhfdfg
 2 years ago
Best ResponseYou've already chosen the best response.0Hah 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

gjhfdfg
 2 years ago
Best ResponseYou've already chosen the best response.0^thats an easier way way to understand it, having it step by step by step
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