Why Do Some Solutions Satisfy One Equation But Not the System?

[Niles]

Homework Statement

I wish to solve the following two equations, which are related:

$$h(x,y) = y \quad \text{and} \quad g(x,y) = y+x^2-bx.$$

First I set h=0, which gives me y=0. Then I insert y=0 in g, where I find x=b and x=0. All is good here.

*********

Now I wish to solve the same system, but I write it as follows:

$$h(x,y) = y =0\cdot x + y \quad \text{and} \quad g(x,y) = y+x^2-bx.$$

Here I see that all x and y=0 satisfy h(x,y)=0, why I insert these values in g. But all x and y=0 do not satisfy g(x,y)=0 now. I know my reasoning is wrong somewhere, but I cannot see where. Can you shed some light on this simple problem?

Thanks in advance.

 

[HallsofIvy]

Homework Statement

I wish to solve the following two equations, which are related:

$$h(x,y) = y \quad \text{and} \quad g(x,y) = y+x^2-bx.$$

First I set h=0, which gives me y=0. Then I insert y=0 in g, where I find x=b and x=0. All is good here.

*********

Now I wish to solve the same system, but I write it as follows:

$$h(x,y) = y =0\cdot x + y \quad \text{and} \quad g(x,y) = y+x^2-bx.$$

Here I see that all x and y=0 satisfy h(x,y)=0, why I insert these values in g. But all x and y=0 do not satisfy g(x,y)=0 now. I know my reasoning is wrong somewhere, but I cannot see where. Can you shed some light on this simple problem?

Thanks in advance.

I don't understand what you asking or what "problem" you have but a few comments. First, you titled this "Systems of linear equations" but the second is not a linear equation. Second, you set "h= 0", why? Are you saying that you actually are solving "y= 0, y+ x²- bx= 0"? If so why define "h" and "g" at all?

Finally, I don't understand why you think that "all x and y= 0" should satisfy both equations. Adding a second equation restricts the possible solutions. Any solution to the both equations must satisfy the first equation by not the other way around: some solutions to one equation will satisfy both. Yes, "all x and y=0" satisfy y= 0. That's true whether you write "+ 0*x" or not. Geometrically, that is the vertical, y, axis. y+ x²- bx= 0 is the same as y= bx- x² and is satisfied by all points on the parabola. Points that satisffy both are the points where they intersect: x= b, y= 0 and x= 0, y= 0.

 

[Niles]

My question is that when I have to solve a non-linear and/or linear system of equations, then what is the most efficient way of doing it, and when I am I allowed to insert the solution that satisfies 1 of the equations into the others.

1) Yeah, the topic should have been non-linear. My mistake.

2) Yes, I guess it is overkill defining h(x,y) and g(x,y) in the first place.

3) If adding a second equation restricts the possible solutions, then how is it that in the first example in post #1 I can just insert y=0 in the second equation (i.e. the equation for g(x,y))?

Thanks for helping.