A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
Solving systems with the substitution method,
anonymous
 3 years ago
Solving systems with the substitution method,

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0,dw:1360809292870:dw

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

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

whpalmer4
 3 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
 3 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
 3 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
 3 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
 3 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
 3 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!)

anonymous
 3 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

anonymous
 3 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
 3 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.

anonymous
 3 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
 3 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 \(x^2\). \[(\frac{12}{x})^2 = \frac{12}{x}*\frac{12}{x} = \frac{12*12}{x*x} = \frac{144}{x^2}\]

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

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

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

anonymous
 3 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
 3 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Got it but why 4 * 36?

whpalmer4
 3 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
 3 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.

anonymous
 3 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
 3 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?

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

whpalmer4
 3 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
 3 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
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1Please, no laughing, this is serious stuff ;)

anonymous
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1Ah, sorry, my solutions are (6,2) and (6,2)

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

whpalmer4
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1Why 4? 4 isn't the value you found for x...

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

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

whpalmer4
 3 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
 3 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?

anonymous
 3 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
 3 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: \[(x5)^2+4y^2=100, x+y=10\]

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

anonymous
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1It might look more complicated, but I think it is a bit easier...

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

anonymous
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1\[x+y= 10 \implies x = 10y\]Right?

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

whpalmer4
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1yes, you would solve it? I'm happy to hear that :)

whpalmer4
 3 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?

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

anonymous
 3 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

anonymous
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.110y5 = 105y = 5y, no?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0*face palm* Got it lol

whpalmer4
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1no...take another look.

whpalmer4
 3 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 \((y5)(y5) = y^25y5y+25 = y^210y+25\)

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

whpalmer4
 3 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
 3 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...

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

whpalmer4
 3 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? :)

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

whpalmer4
 3 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
 3 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).

anonymous
 3 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

anonymous
 3 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
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.