Question about Intersecting Equations

[The Student]

The question wants me to find the intersection/s for y = 2x and x^2 + y^2 = 1.

I know that this is just a line going through a circle, but a problem arises when I actually work out the solution set.

So I start by substituting 2x with the y in x^2 + y^2 = 1. That gives x^2 + 4x^2 = 1. Then isolating x I get x = +/-(1/5)^(1/2).

Now I will plug in the negative value for x in x^2 + y^2 = 1. (-(1/5)^(1/2))^2 + y^2 = 1. Then y^2 = 4/5. Finally y = +/-(4/5)^(1/2)

But on the graph, we know that one of the solution set cannot be ( -(1/5)^(1/2), (4/5)^(1/2) ) .

I get the correct solution set when I plug the values that I found for x into y = 2x.

Why do the solutions that I found for x work in y = 2x but not in x^2 + y^2 = 1?

 

[HallsofIvy]

Yes, \(\displaystyle x= \pm \sqrt{\frac{1}{5}}= \pm\frac{1}{\sqrt{5}}= \pm\frac{\sqrt{5}}{5}\).

And, you can show that \(\displaystyle y= \pm\sqrt{4/5}\).
But both values of y do NOT "go with" both values of x. You are given that y= 2x so if \(\displaystyle x= \frac{\sqrt{5}}{5}\) then \(\displaystyle y= \frac{2\sqrt{5}}{5}\) and if \(\displaystyle x= -\frac{\sqrt{5}}{5}\) then \(\displaystyle y= -\frac{2\sqrt{5}}{5}\). That is, the two points of intersection are \(\displaystyle \left(\frac{\sqrt{5}}{5}, \frac{2\sqrt{5}}{5}\right)\) and \(\displaystyle \left(-\frac{5}{5}, -\frac{2\sqrt{5}}{5}\right)\).

 

[The Student]

Yes, \(\displaystyle x= \pm \sqrt{1}{5}= \pm\frac{1}{\sqrt{5}}= \pm\frac{\sqrt{5}}{5}\).

And, you can show that \(\displaystyle y= \pm\sqrt{4/5}\).
But both values of y do NOT "go with" both values of x. You are given that y= 2x so if \(\displaystyle x= \frac{\sqrt{5}}{5}\) then \(\displaystyle y= \frac{2\sqrt{5}}{5}\) and if \(\displaystyle x= -\frac{\sqrt{5}}{5}\) then \(\displaystyle y= -\frac{2\sqrt{5}}{5}\). That is, the two points of intersection are \(\displaystyle \left(\frac{\sqrt{5}}{5}, \frac{2\sqrt{5}}{5}\right)\) and \(\displaystyle \left(-\frac{5}{5}, -\frac{2\sqrt{5}}{5}\right)\).

Is there a reason that the equation for the solution set gives wrong answers? Did I make an error somewhere?

 

[HallsofIvy]

What "equation for the solution set" are you talking about?

Your question before was "Why do the solutions that I found for x work in y = 2x but not in x^2 + y^2 = 1?" which I couldn't answer because they obviously do "work in x^2+ y^2= 1". If you meant it the other way around, "Why do the solutions that I found for x work in x^2 + y^2 = 1 but not in y = 2x?" it is because "x^2+ y^2= 1" does not distinguish between "x" and "-x" nor between "y" and "-y". Replacing y with 2x in x^2+ y^2= 1 gives two values for x and two values for y which can be combined to make four points. Two combinations are where y= 2x crosses the circle, the other two are where y= -2x crosses the circle.

 

[The Student]

What "equation for the solution set" are you talking about?

Your question before was "Why do the solutions that I found for x work in y = 2x but not in x^2 + y^2 = 1?" which I couldn't answer because they obviously do "work in x^2+ y^2= 1". If you meant it the other way around, "Why do the solutions that I found for x work in x^2 + y^2 = 1 but not in y = 2x?" it is because "x^2+ y^2= 1" does not distinguish between "x" and "-x" nor between "y" and "-y". Replacing y with 2x in x^2+ y^2= 1 gives two values for x and two values for y which can be combined to make four points. Two combinations are where y= 2x crosses the circle, the other two are where y= -2x crosses the circle.

I guess my main problem is that I don't know the method for doing this question even though the method that I was using for other questions like these was working.

I am lost. I don't know how I would have known the right answer if it weren't for the graph being obvious.

I guess I am just not looking at math problems properly. This is why I am always on here trying to get past the fundamentals.

 

[HallsofIvy]

What you did originally was perfectly good- use one of the equations (y= 2x) to replace one of the unknown variables in the other equation getting one equation in a single variable. Solve that equation for the variable, then use the other equation to solve for the corresponding other variable. The difficulty seems to be that you are treating x and y as completely separate. As I said before, replacing y with 2x gives a quadratic equation for x that you can solve to get \(\displaystyle x= \pm\frac{\sqrt{5}}{5}\). Then \(\displaystyle y= 2x= \pm\frac{2\sqrt{5}}{5}\).

I'm having a little difficult understanding exactly what your question is! I think you are trying to put both y values together with both x values getting four points altogether. You can't do that. The two x values are \(\displaystyle \frac{\sqrt{5}}{5}\) and \(\displaystyle -\frac{\sqrt{5}}{5}\). We know that y= 2x. If \(\displaystyle x= \frac{\sqrt{5}}{5}\) then \(\displaystyle y= \frac{2\sqrt{5}}{5}\) so one point of intersection is \(\displaystyle \left(\frac{\sqrt{5}}{5}, \frac{2\sqrt{5}}{5}\right)\). If \(\displaystyle x= -\frac{\sqrt{5}}{5}\) then \(\displaystyle y= -\frac{2\sqrt{5}}{5}\) so the other point of intersection is \(\displaystyle \left(-\frac{\sqrt{5}}{5}, -\frac{2\sqrt{5}}{5}\right)\).

But we cannot put If \(\displaystyle x= \frac{\sqrt{5}}{5}\) together with \(\displaystyle y= -\frac{2\sqrt{5}}{5}\) because they do not satisfy y= 2x.

 

[The Student]

What you did originally was perfectly good- use one of the equations (y= 2x) to replace one of the unknown variables in the other equation getting one equation in a single variable. Solve that equation for the variable, then use the other equation to solve for the corresponding other variable. The difficulty seems to be that you are treating x and y as completely separate.

What do you mean when you say that I am treating x and y separately?

As I said before, replacing y with 2x gives a quadratic equation for x that you can solve to get \(\displaystyle x= \pm\frac{\sqrt{5}}{5}\). Then \(\displaystyle y= 2x= \pm\frac{2\sqrt{5}}{5}\).

It's like there is something "unsymmetrical" about the logic of finding these values. We plug the one y equation into the other equation to get the x values. Then we plug in x values into the y equation. I just feel like this is a random procedure that just works somehow; in my mind, it isn't obvious, expected or consistent. I see the big picture, but I feel like I am missing a deeper understanding.

I'm having a little difficult understanding exactly what your question is!

I guess I am trying to figure out why I am so uncomfortable with this question. My main problem is that there isn't really anything that I don't understand about all of this, but for some reason I feel really uncomfortable with it.

I think you are trying to put both y values together with both x values getting four points altogether. You can't do that.

I know why I can't in terms of this specific question, but I feel like I am missing a deeper reason.

Or do I just have accept that there certain exceptions that emerge, and that I just hope that I can adjust and find them?

 

[Ishuda]

...I know why I can't in terms of this specific question, but I feel like I am missing a deeper reason. ...

It is the same reasoning for all questions of this type. You started with y = 2x, so for any value you get for x, you must have y=2x to satisfy the complete solution. That is obtain an x and then y must be twice that (for this example). Suppose the equations had been y = x and x² + 2xy + y² = 1. Then y = x and x² + 2x² + x² = 4 x² = 1 or x=\(\displaystyle \pm \frac{1}{2}\). Suppose x = \(\displaystyle +\frac{1}{2}\), then since we had y = x to get that, we must also have y = \(\displaystyle +\frac{1}{2}\). If x = \(\displaystyle -\frac{1}{2}\), then since to have the complete solution y must also equal x, y =\(\displaystyle -\frac{1}{2}\)

It is the same sort of thing if the solutions are linear. Suppose
x + y = 2
2x - y = 1

We can use the second equation to get
y = 2x - 1
put that in the first equation to get
x + 2x - 1 = 2 ==> 3 x = 3
or
x = 1.
Now we go back to the y = 2x - 1 to get
y = 1

 

[The Student]

Thanks HallsofIvy and Ishuda for trying. I am not clear about this yet, but at least I have these posts that I can refer to in the future.

 

[daon2]

Thanks HallsofIvy and Ishuda for trying. I am not clear about this yet, but at least I have these posts that I can refer to in the future.

What they explained might be clearer in a picture

The black dots are the solutions to \(\displaystyle x^2+y^2=1\) where \(\displaystyle y=2x\)

 

[The Student]

What they explained might be clearer in a picture

The black dots are the solutions to \(\displaystyle x^2+y^2=1\) where \(\displaystyle y=2x\)

Beautiful illustration, but I did draw this graph already, and that was the only way that I new the answer for sure.

My issue is that I don't feel like I know formally why I can't plug the x values into the circle equation. I understand that it leaves y = +/- for either sign of x which makes all 4 points true, but I don't understand why using y = 2x should work over y^2 + x^2 = 1. I feel like I am missing something that would lead a another math student to see this naturally.

 

[daon2]

Beautiful illustration, but I did draw this graph already, and that was the only way that I new the answer for sure.

My issue is that I don't feel like I know formally why I can't plug the x values into the circle equation. I understand that it leaves y = +/- for either sign of x which makes all 4 points true, but I don't understand why using y = 2x should work over y^2 + x^2 = 1. I feel like I am missing something that would lead a another math student to see this naturally.

When you plug, for instance, x=+1/sqrt(5) into the circle equation, you can see there are two points on the circle that have this x-coordinate, one black and one orange. The black one represents a solution to the equation. The orange one does not satisfy y=2x.

 

[Ishuda]

Beautiful illustration, but I did draw this graph already, and that was the only way that I new the answer for sure.

My issue is that I don't feel like I know formally why I can't plug the x values into the circle equation. I understand that it leaves y = +/- for either sign of x which makes all 4 points true, but I don't understand why using y = 2x should work over y^2 + x^2 = 1. I feel like I am missing something that would lead a another math student to see this naturally.

It isn't a matter of one equation works over the other, it is a matter of both equations must be satisfied.

 

[The Student]

It isn't a matter of one equation works over the other, it is a matter of both equations must be satisfied.

Ohhhh, I see. Thank-you so much!!!!