Beginning Algebra
Answer/Discussion to Practice
Problems
on Graphing Linear Equations
 Answer/Discussion
to 1a
y = 2x - 1 |
| If we subtract 2x from both sides, then
we can write the given equation as -2x + y
= -1.
Since we can write it in the standard form, Ax
+ By = C, then we have a linear equation.
(return to problem
1a) |
| Answer/Discussion
to 1b
|
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.
(return to problem
1b) |
| Answer/Discussion
to 2a
y = 2x
- 1 |
| Step 1: Find
three ordered pair solutions.
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.
|
x
|
y = 2x -
1
|
(x, y)
|
|
-1
|
y = 2(-1) - 1 = -3
|
(-1, -3)
|
|
0
|
y = 2(0) - 1 = -1
|
(0, -1)
|
|
1
|
y = 2(1) - 1 = 1
|
(1, 1)
|
|
| Answer/Discussion
to 2b
|
| Step 1: Find
three ordered pair solutions.
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.
|
x
|
y = -1/2x
|
(x, y)
|
|
-1
|
y = -1/2(-1) = 1/2
|
(-1, 1/2)
|
|
0
|
y = -1/2(0) = 0
|
(0, 0)
|
|
1
|
y = -1/2(1) = -1/2
|
(1, -1/2)
|
|
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Last revised on Jan. 9, 2002 by Kim Seward. |
