Question

give all the solutions of the nonlinear system of euqations, including those with nonreal complex components:

Answer 1

2x+|y|=4
x^2 + y^2 = 5

Answer 2

Can I use substitution??

Answer 3

@jim_thompson

Answer 4

@jim_thompson5910

Answer 5

@zepdrix

Answer 6

@Luigi0210

Answer 7

@Jhannybean

Answer 8

I don't see why not.

Answer 9

The absolute is confusing me

Answer 10

2x+|y|=4 implies |y|=4-2x so y=4-2x and y=2x-4

Answer 11

So now I sub which one? Both??

Answer 12

x^2 + (2x-4)^2 = 5
and
x^2 + (4-2x)^2 = 5

Answer 13

both

Answer 14

got it?

Answer 15

Yes! So I get two solutions for x?

Answer 16

prob more, two in each equation

Answer 17

I got 5x^2-16x+11 = 0 for both

Answer 18

Correct??

Answer 19

seems so to me

Answer 20

hmm sort of strange because this gives no complex solutions

Answer 21

you may also try to represent both graphically

Answer 22

hmm just looking at this I see that 2, -2, and 1 are solutions. I am confused...

Answer 23

What??

Answer 24

lets try
x^2 + y^2 = 5
so
y=+-sqrt(5-x^2)
so
2x+sqrt(5-x^2)=4
sqrt(5-x^2)=4-2x
5-x^2=(4-2x)^2 = 16-16x+4x^2
5x^2-16x-11=0, again the same solutions
but I know 2 is a solution....

Answer 25

Ok wait so...What? Lol

Answer 26

I think you lost me, so is solution x= 2 , -2, 1??

Answer 27

what are you doing in class? what class is this for?

Answer 28

well I know that those are 3 solutions, and I don't know how to algebraically get 2 or -2

Answer 29

Isn't it 5x2-16x+11??

Answer 30

that equation has two solutions, namely 1 and 11/5 and I know 2 is a solution from looking at it. So we are missing something. I hope @shubhamsrg has an idea.

Answer 31

x=2 is not a solution

Answer 32

that equation(5x2-16x+11) has two solutions, namely 1 and 11/5 and I know 2 is a solution to the original system from looking at it. So we are missing something. I hope @shubhamsrg has an idea.

Answer 33

x=1 is the only solution. which will give 2 values of y

Answer 34

How @husbandry ??

Answer 35

sorry its late....

Answer 36

@shubhamsrg

Answer 37

ok so why is 11/5 not a solution.?

Answer 38

graphically it'll show you it should have only 2 real solution
(1,2) and (1,-2) are the ones
also, you may check 11/5 does not satisfy both

Answer 39

and late ?

Answer 40

I know, but using the substitution method, we get x=11/5 when we solve the quadratic. So im just curious.

Answer 41

can you graph what you are speaking of?

Answer 42

x=11/5 does not even fall on 2x + |y| = 4

Answer 43

I'll wolfram this..
and give you its graph..

Answer 44

http://prntscr.com/1d3j3l

Answer 45

2x+|y|=4 implies |y|=4-2x so y=4-2x and y=2x-4
plugging this in we get
5x^2-16x+11 = 0 this has two solutions 1 and 11/5

Answer 46

I understand It does not satisfy the first, but im confused then why are we getting it as a solution to the second equation depending on the first equation.

Answer 47

I have not done much work with systems of non linear equations.

Answer 48

y^2 = |y|^2
and |y| = 4-2x

since (4-2x)^2 = (2x-4)^2 , we get the result we are right now getting, i.e 11/5 as solution
had the linear eqn been simply 2x+ y = 4 or 2x-y = 4, 11/5 would have been another correct answer .

Answer 49

Me too, zz I am unfamiliar

Answer 50

im just trying to figure out a way to see that without plugging in the solution to see if it works, or graphing it.

Answer 51

was I helpful in clearing the doubt ?

Answer 52

there is no doupt, I still would like an analytic way to know that a solution is a solution without having to plug it in.

Answer 53

Yes! I just can't see how to get the solutions, did you use sub or graph???

Answer 54

well we get 2 solutions from 5x^2-16x+11 = 0
1,11/5
plug in 11/5 for x
2*11/5+|y|=4
so |y| = 4-2*11/5 <0 has no solutions
but
let x=1
and
we see y=+-2

Answer 55

because 2*1+|y|=4 implies |y| = 2 implies y=+-2

Answer 56

do you understand how we can get the solutions?

Answer 57

I would just like to know a way that skips the plug and check at the end.

Answer 58

or logic....

Answer 59

I got it. Can you help with just ONE more. I want to finish hw before class

Answer 60

I can try:)

Answer 61

Ok here or new box??

Answer 62

maybe new one just in case im confused again:)