Acclerated Integrated Geometry

Johnson220 said:
How would I solve each system: X^2+Y^2=4 and X+Y=2?
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What geometric shapes you will get out of those equations?

Where would those intersect?

What are your thoughts/work?

Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.
 
Johnson220 said:
The two systems are for a circle and a line. I don't know how to solve systems.
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Did you plot those functions (on same screen)?

How many solutions would you have (expect)?

x + y = 2

x = 2 - y............................(1)

next given equation:

x2 + y2 = 4

Using (1) above,

(2 - y)2 + y2 = 4

Now continue.....
 
ok, if i solve the problem, I come to 2Y(Y-2)=0... What do I do to solve for y? set 2y=0 and y-2=0 so y=0 and y=2, then add it back to the original formula and then solve for x?
 
I come to 2Y(Y-2)=0... What do I do to solve for y? set 2y=0 and y-2=0 so y=0 and y=2, then add it back to the original formula and then solve for x?
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Yes (except please say "substitute" back into the original equations rather than "add"). Also, please graph this, and the meaning of your result will become obvious.
 
Help please...

are the solutions y=0 and y=-2? Do I now substitute the values into the original equation to solve for x? How do I check my work?
 
Johnson220 said:
are the solutions y=0 and y=-2? Do I now substitute the values into the original equation to solve for x? How do I check my work?
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No

y - 2 = 0 → y = 2

So you would have

y = 0

x = 2 - y → x = 2

and for y = 2

x = 2 - y → x = 0

You have co-ordinates of two points. To check your answer you need to prove that both of these points lie on both the graphs (of the 2 given functions).
 
Accelerated Intergrated Geometry

How do I "prove" without graphing. I must show my work.
 
You started out with two equations.

Solving the system of equations gave you two points.

You "prove" or check your solution the way you always learned (I HOPE) in algebra.

Substitute the coordinates of the two points into each of the equations. Do the coordinate make both equations true? If so, then the points with those coordinates lie on the graphs represented by each of the equations.

You started with the equation of a circle and the equation of a line. If you came up with two points represented by two pairs of coordinates, and each point satisfies each of the equations, then said points must lie on the circle, and also on the line.

Ok?
 
Johnson220 said:
How would I solve each system: X^2+Y^2=4 and X+Y=2?
Click to expand...


So we need to solve this system of equations:

\(\displaystyle x^{2} + y^{2} = 4\)
\(\displaystyle x + y = 2\)


There are a ton of ways to do this, but let's start by rewriting the second equation:

\(\displaystyle y = 2 - x\)

And then substitute y for 2-x in the first equation:

\(\displaystyle x^{2} + (2 - x)^{2} = 4\)

Then solving for x:

\(\displaystyle x^{2} + x^{2} - 4x + 4 = 4 \to 2x^{2} - 4x = 0 \to 2x(x-2) = 0\)

If \(\displaystyle ab = 0\), then \(\displaystyle a\) and \(\displaystyle b = 0\) means:

\(\displaystyle x = 0\), and \(\displaystyle x = 2\)


We know \(\displaystyle y = 2-x\), so \(\displaystyle y = 2 - 0 = 2\) and \(\displaystyle y = 2 - 2 = 0\)

Solutions to this system are then \(\displaystyle (x,y) = (0,2)\) and \(\displaystyle (2,0)\).



Check:

\(\displaystyle 2^{2} + 0^{2} = 4\)

\(\displaystyle 2 + 0 = 2\)



\(\displaystyle 0^{2} + 2^{2} = 4\)

\(\displaystyle 0 + 2 = 2\)


All of which are true.
 
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