Question

Solve this system.

Answer 1

x^2 + y^2 + 2x + 2y = 0
x^2 + y^2 + 4x + 6y + 12 = 0

Answer 2

I know they are both circles. Usually I use substitution or elimination for problems like these, but I don't see how I could use either of those in this case...

Answer 3

let's label the two equations A and B
A is the first one on top
B is the one below

Answer 4

subtract the equations A - B

notice how we have these like terms match up
x^2 - x^2 = 0x^2 = 0
y^2 - y^2 = 0y^2 = 0
so the x^2 and y^2 terms cancel out

then we have these other like terms
2x - 4x = -2x
2y - 6y = -4y

there is no like term that pairs up with the 12 in equation B, so we just have -12 when we subtract

Answer 5

so after subtracting the two equations A-B we end up with
-2x-4y-12 = 0

Answer 6

Let's call this equation C: -2x-4y-12 = 0

Answer 7

Now go back to equation A
x^2 + y^2 + 2x + 2y = 0

isolate the x^2+y^2 portion to get
x^2 + y^2 = -2x - 2y

then move to equation B and replace the "x^2+y^2" with "-2x - 2y" to go from
x^2 + y^2 + 4x + 6y + 12 = 0
to
-2x - 2y + 4x + 6y + 12 = 0
which simplifies to
2x+4y+12 = 0

Answer 8

let's call this equation D: 2x+4y+12 = 0

Answer 9

Whew. Lots of work, but I'm following. :) What next?

Answer 10

hmm one moment

Answer 11

I"m running into a roadblock

Answer 12

haha okay

Answer 13

C and D are basically the same thing, just opposite signs

Answer 14

yeah I wasn't thinking

Answer 15

Equation A: x^2 + y^2 + 2x + 2y = 0
Equation B: x^2 + y^2 + 4x + 6y + 12 = 0


Let's complete the square for the x terms in equation A

x^2 + y^2 + 2x + 2y = 0
x^2 + 2x+0 + y^2 + 2y = 0
x^2 + 2x+1-1 + y^2 + 2y = 0
(x^2 + 2x+1)-1 + y^2 + 2y = 0
(x+1)^2-1 + y^2 + 2y = 0

Now let's isolate x

(x+1)^2-1 + y^2 + 2y = 0
(x+1)^2 = -y^2 - 2y + 1
x+1 = sqrt(-y^2 - 2y + 1) ... or ... x+1 = -sqrt(-y^2 - 2y + 1)
x = -1+sqrt(-y^2 - 2y + 1) ... or ... x = -1-sqrt(-y^2 - 2y + 1)

I have a feeling I'm making it more complicated than it has to be. I'm not sure. Are you allowed to use graphing techonology? That would make things much easier.

Answer 16

Hmmm yeah I don't think it would be that complicated... I'm supposed to do them by hand and show work. I don't have too much space either. hmm

Answer 17

Maybe we could do something with that equation C we found earlier?
-2x-4y-12 = 0
Maybe isolate one of the variables and plug it into one of the original equations perhaps?

Answer 18

oh right, wasn't thinking

So let's solve for y in equation C to get

-2x-4y-12 = 0
-2x-12 = 4y
4y = -2x-12
y = (-2x-12)/4
y = -0.5x - 3

then plug this into equation A to get

x^2 + y^2 + 2x + 2y = 0
x^2 + (-0.5x - 3)^2 + 2x + 2(-0.5x - 3) = 0

at this point, you have one variable to solve for. Expand and simplify everything out. Then use the quadratic formula

Answer 19

I'll let you know what I get! Thanks so much

Answer 20

I simplified it to 5/4x^2 + 4x + 3
Then used the quadratic equation to get -6/5 and -2
Does this look right so far? It still seems a lot more complicated that it should... I haven't done problems like this the whole lesson.
I'll plug those x values in and see what I get for y!

Answer 21

yep you got both x values correct

x = -6/5 = -1.2
or
x = -2

Answer 22

I'm having a bit of trouble simplifying the quadratic equation when I plugged -6/5 in for x in the first equation.
Right now I have it down to:
\[-1 \pm \sqrt{116/5}\] ^ the sqrt 116/5 is over 2

Answer 23

Do you know how to simplify the stuff in the square root? The y-coordinate should be a fraction.

Answer 24

x^2 + y^2 + 2x + 2y = 0
(-1.2)^2 + y^2 + 2(-1.2) + 2y = 0 ... plug in x = -1.2
1.44 + y^2 - 2.4 + 2y = 0
y^2 + 2y - 0.96 = 0

Do you have this so far?

Answer 25

I've been keeping everything in fractions...
(-6/5)^2 + 2(-6/5) + y^2 + 2y = 0
36/25 -12/5 + y^2 + 2y = 0
-24/5 + y^2 + 2y= 0
Then I plugged into the quadratic formula.

Answer 26

Although it looks like I did something wrong, because -24/5 is -4.8, not -0.96....

Answer 27

36/25 -12/5 = 36/25 -12/5*5/5
36/25 -12/5 = 36/25 -60/25
36/25 -12/5 = (36 -60)/25
36/25 -12/5 = -24/25 ... you wrote 5 in the denominator when it should be 25
36/25 -12/5 = -0.96

Answer 28

Good catch!

From there, I plugged into the quadratic equation and got 23/5 and -33/5.

-2 +/- sqrt(4 - 4*(-24/25) everything over 2
-1 +/- sqrt (196/25) < over 2
-1 +/- (14/5)/2
-1 +/- 28/5

How does it look?

Answer 29

the (14/5)/2 portion should turn into 14/10 = 7/5

Answer 30

wait a minute, why not just say this?

y = -0.5x - 3
y = -0.5*(-1.2) - 3 ... plug in x = -1.2
y = 0.6 - 3
y = -2.4

Answer 31

so when x = -6/5 = -1.2, the value of y is y = -12/5 = -2.4

Answer 32

Hahahahah. Oh my gosh. That is indeed much simpler...

Perfect! Thank you * a million for helping me figure this out.

Answer 33

you're welcome

Answer 34

Wait a minute... the way I was doing it, if (14/5)/2 becomes 7/5, I would have:
-1 +/- 7/5
Which is different from -12/5.....