Question

okay..still don't understand these...solve the nonlinear system of equations
{x^2+y^2=16
{x+y=4

Answer 1

ok u take the 2nd eqyation that is x+y=4 take the value of x and plug it in the first one

Answer 2

That's one way to do it

Answer 3

is there another way?...one that's really easy perhaps? lol

Answer 4

There's another way, but it isn't necessarily "easier"

Answer 5

There are several ways: gaussian elimination, substitution, matrices....but I think substitution (as shown above) is pretty easy with simple ones like this.

Answer 6

would anyone mind writing out some of the work?....not the answer but the beginning of the work showing the substitution

Answer 7

Look at the second equation and solve for say, y:
y=4-x

Plug that into the first equation and solve for x. Once you know the value of x, you can use one of the original equations to solve for y,\.

Answer 8

Plugging it in as I mentioned would look like this:\[x^2+(4-x)^2=16\]

Answer 9

{x^2+y^2=16
{x+y=4

1. Square both sides on the second equation:
\((x+y)^2 = 4^2 \)

2. Expand the left and right sides:
\(x^2 + 2xy + y^2 = 16\)

3. Set 16 = 16
\(x^2 + y^2 = x^2 + 2xy + y^2\)

subtract \(x^2\) and \(y^2\) from both sides:
\(0 = 2xy\)

5. Divide both sides by 2:
\(0 = xy\)

6. Therefore,
(x,y) = (4,0)
(x,y) = (0,4)

Answer 10

oooo okay i get it now....thanks so much for all of your help!!!

Answer 11

That's just an approach. Not saying it's the most logical one.

Answer 12

well it makes sense...it was what my teacher was trying (and failing) to teach us hahaha

Answer 13

I don't think your teacher taught you this way. Probably some other method.

Answer 14

Do you have any more of these?

Answer 15

ya i have 2 more

Answer 16

Let me see them

Answer 17

well this one is slightly different...
1. solve the system to determine the number of real solutions
{x^2+y^2=16
{4x^2-9y^2=36

2. solve....
{x^2+y^2=63
{x^2-3y^2=27

Answer 18

okay...i think i got it, thanks again for your help Hero!

Answer 19

Actually, it would probably be better if you solved for y^2 and set y^2 = y^2

Answer 20

For example for the first one, you should get

\(y^2 = y^2\)
\(16-x^2 = \frac{4x^2 - 36}{9}\)

Answer 21

Then you can just cross multiply to get

\(9(16 - x^2) = 4x^2 - 36\)

Answer 22

\(144 - 9x^2 = 4x^2 - 36\)
\(144 +36 = 4x^2 + 9x^2\)
\(180 = 13x^2\)
\(\large\frac{180}{13} = x^2\)

Answer 23

You'd have to square root both sides to have x

Answer 24

then substitute x back in to find y

Answer 25

That's by far the simplest way to do it

Answer 26

I hope that makes sense

Answer 27

lol took me a minute to figure out what your were doing but ya now i see..it makes sense...a lot more sense than my teacher was making haha

Answer 28

Well, at least you were able to understand it

Answer 29

Do you want me to show you my exact steps in vyew?