College Algebra: 2X2 Non-Linear Systems.
I need to solve this system algebraically and then graph both of the equations on the same coordinate system to support my solution. I'll write the problem in comments.

Question

College Algebra: 2X2 Non-Linear Systems.
I need to solve this system algebraically and then graph both of the equations on the same coordinate system to support my solution. I'll write the problem in comments.

anonymous · Answer

$$y=\sqrt{x}$$
$$y=2x$$

johnweldon1993 · Answer

Okay, so lets look at what we have

$$\large y = \sqrt{x}$$
and
$$\large y = 2x$$

Just thinking about it: If y equals the square root of x

AND

y equal 2x

it would only make sense that the square root of x is equal to 2x right?

Since y equals both things...then those "things" must be equal

so to solve your system we can set

$$\large y = y$$
or (replacing what they equal
$$\large \sqrt{x} = 2x$$

making sense?

anonymous · Answer

You haven't lost me yet.

johnweldon1993 · Answer

Lol good :)

so how would we solve that?

$$\large \sqrt{x} = 2x$$

we need to solve for 'x'

lets try squaring both sides

$$\large x = 4x^2$$

make it any easier to see what values for 'x' would make those equal?

johnweldon1993 · Answer

(if this is hard to grasp, there's another way to go about it too, so no worries if you cant see it yet :)

anonymous · Answer

Hmm, I see why x would equal 4x^2. 
For values, wouldn't 0 and 1 be good?

johnweldon1993 · Answer

Well, 0 yes!

but what happens when we plug in x = 1?

$$\large x = 4x^2$$
$$\large 1 = 4(1)^2$$
$$\large 1 = 4$$

not quite

johnweldon1993 · Answer

But, what if I simplified this a little bit more?

$$\large x = 4x^2$$

What if I divided both sides by 'x'?

$$\large \frac{x}{x} = \frac{4x^2}{x}$$
$$\large 1 = \frac{4x\cancel{^2}}{\cancel{x}}$$
$$\large 1 = 4x$$

now what would you say could be another value for 'x'

anonymous · Answer

So, 1 is a bad value to use; however, 0 seems good like you said.
$$x=4x^{2}$$
$$x=4(0)^{2}$$
$$x=4(0)$$
$$x=0$$

johnweldon1993 · Answer

Right, 0 was perfect, but that is only 1 solution to our system here, there is another, check out that above post^ :)

anonymous · Answer

Wow, got me there. Wouldn't you divide 4 on both sides?

johnweldon1993 · Answer

Correct!

which would make x = ?

anonymous · Answer

$$\frac{ 1 }{ 4 }$$

johnweldon1993 · Answer

Lol latex and everything :P

yes perfect!

So now

x = 0  x = 1/4 are solutions to this system!

What does that mean in terms of a graph? at x = 0 and x = 1/4, WHAT will happen in the graph?

anonymous · Answer

Sorry for being late, I have siblings lolx 
Hmm, well since there is no y yet. I'm going to assume it's (0,0) and (1/4, 0) on the graph.

johnweldon1993 · Answer

Lol no problem, I'm studying for finals inbetween responses so its fine lol

Okay, we'll see what happens on the graph

You'll see that the solutions we just found, are where the 2 lines will intersect :)

So, what does the graph of y = 2x looks like? If you could draw it in the draw bar below, it would work best :)

anonymous · Answer

|dw:1418085955944:dw|
Like this?

anonymous · Answer

And yeah I'm studying for my Chapter 5 Test tomorrow and my College Algebra Final next Tuesday.

johnweldon1993 · Answer

Hmm not quite, at least not from first sight (and yeah, calc 4 final on monday :P)

If I told you 

$$\large y = 2x$$When x = 1, what does y equal?

anonymous · Answer

2

johnweldon1993 · Answer

Sorry just realized we're gonna have to spread out the 'x' axis a bit to make this look better

|dw:1418086270753:dw|

look good? when x = 1, its up on y = 2