Answer/Discussion to 1a
A(3, 1), B(-2, -1/2), C(2, -2), and D(0,1)

A(3, 1)  lies in quadrant I.

B(-2, -1/2)  lies in quadrant III.

C(2, -2)  lies in quadrant IV.

D(0, 1)  lies on the y axis.

(return to problem 1a)

Which number is the x value and which one is the y value?  If you said x = 0 and y = -10, you are correct! 

Let's plug (0, -10) into the equation and see what we get:

Now, let's take a look at (1, -14). 

Which number is the x value and which one is the y value?    If you said x = 1 and y = -14, you are right! 

Let's plug (1, -14) into the equation and see what we get:

Now, let's take a look at (-1, -14). 

Which number is the x value and which one is the y value?    If you said x = -1 and y = -14, you are right! 

Let's plug (-1, -14) into the equation and see what we get:

(return to problem 2a)

Since we can write it in the standard form, Ax + By = C, then we have a linear equation. 

This means that we will have a line when we go to graph this. 

The three x values I'm going to use are -1, 0, and 1.  (Note that you can pick ANY three x values that you want.  You do not have to use the values that I picked.) You want to keep it as simple as possible.  The following is the chart I ended up with after plugging in the values I mentioned for x.
 
 

(return to problem 3a)

If we add  x squared to both sides we would end up with .   Is this a linear equation?  Note how we have an x squared as opposed to x to the one power. 

It looks like we cannot write it in the form Ax + By = C, because the x has to be to the one power, not squared.  So this is not a linear equation. 

However, we can still graph it.

The seven x values that I'm going to use are -3, -2, -1, 0, 1, 2, and 3.  (Note that you can pick ANY x values that you want.  You do not have to use the values that I picked.) You want to keep it as simple as possible.  The following is the chart I ended up with after plugging in the values I mentioned for x.
 
 

x		(x, y)
-3	y = -(-3)^2 = -9	(-3, -9)
-2	y = -(-2)^2  = -4	(-2, -4)
-1	y = -(-1)^2 = -1	(-1, -1)
0	y = -(0)^2 = 0	(0, 0)
1	y = -(1)^2 = -1	(1, -1)
2	y = -(2)^2 = -4	(2, -4)
3	y = -(3)^2 = -9	(3, -9)

(return to problem 3b)

In other words, we can't write it in the form Ax + By = C.  This means that this equation is not a linear equation.

The seven x values that I'm going to use are -3, -2, -1, 0, 1, 2, and 3.  (Note that you can pick ANY x values that you want.  You do not have to use the values that I picked.) You want to keep it as simple as possible.  The following is the chart I ended up with after plugging in the values I mentioned for x.
 
 

x		(x, y)
-3	y = |-3| - 1 = 2	(-3, 2)
-2	y = |-2| - 1 = 1	(-2, 1)
-1	y = |-1| - 1 = 0	(-1, 0)
0	y = |0| - 1 = -1	(0, -1)
1	y = |1| - 1 = 0	(1, 0)
2	y = |2| - 1 = 1	(2, 1)
3	y = |3| - 1 = 2	(3, 2)

(return to problem 3c)