Solving systems with the substitution method,

- anonymous

Solving systems with the substitution method,

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- anonymous

,|dw:1360809292870:dw|

- anonymous

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

- anonymous

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

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## More answers

- whpalmer4

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

- whpalmer4

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

- 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 \(x^2\).

- whpalmer4

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

- anonymous

I'm confused

- whpalmer4

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

- 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!)

- anonymous

First confusion is multiplying by x^2, you've shown the steps but I don't get how everything came out to be

- anonymous

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

- 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.

- anonymous

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

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

- anonymous

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

- whpalmer4

Did you understand how I solved the equation by factoring?

- anonymous

Kinda, except how the 36 & 4 came in

- anonymous

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

- whpalmer4

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

- anonymous

Got it but why 4 * 36?

- whpalmer4

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

- anonymous

Got that,

- 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.

- anonymous

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

- 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?

- anonymous

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

- 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?

- 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.

- 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.

- whpalmer4

I'm not sure I've ever actually multiplied 13423523493459823487234.6734234*0 before now, but I'm still pretty sure it equals 0 :-)

- whpalmer4

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

- anonymous

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

- 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? (\(x^2-4=0\))

- whpalmer4

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

- anonymous

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

- whpalmer4

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

- anonymous

I divided 12 by 4,

- whpalmer4

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

- anonymous

Should I of done it by 2?

- whpalmer4

Yes.

- anonymous

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

- whpalmer4

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

##### 1 Attachment

- anonymous

Ah

- whpalmer4

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

- anonymous

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

- 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:
\[(x-5)^2+4y^2=100, x+y=10\]

- anonymous

Okay, I'll give this one a try,

- anonymous

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?

- 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...

- whpalmer4

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

- whpalmer4

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

- anonymous

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

- 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

- anonymous

You just lost me...

- whpalmer4

\[x+y= 10 \implies x = 10-y\]Right?

- whpalmer4

(the arrow means "implies")

- 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?

- anonymous

Yes,

- whpalmer4

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

- 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?

- anonymous

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

- anonymous

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

- anonymous

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

- whpalmer4

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

- anonymous

*face palm* Got it lol

- 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...

- anonymous

25y^2?

- whpalmer4

no...take another look.

- 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 \((y-5)(y-5) = y^2-5y-5y+25 = y^2-10y+25\)

- anonymous

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

- whpalmer4

\[(a-b)(a-b) = a(a-b) - b(a-b) = a^2 - ab - ba + b^2 = a^2 -2ab + b^2\]

- whpalmer4

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

- anonymous

Who knows what I was thinking...

- 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? :-)

- anonymous

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

- 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!

- 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).

- anonymous

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

- anonymous

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

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