Beginning Algebra
Answer/Discussion to Practice
Problems
on Solving Linear Inequalities
 Answer/Discussion
to 1a |
Graph:
 |
*Inv. of add 3 is sub. 3
*Inv. of mult. by -2 is div. both sides by
-2, so reverse inequality sign
*Visual showing all numbers less than -2 on
the number line
|
| Notice how our variable was on the right side of the inequality.
It doesn't matter what side you have the variable on, as long as it by
itself on one side and everything else is on the other side. What
you do have to be careful about is graphing it properly. It is almost
like reading it backwards this way. So, if you feel more comfortable
writing it with your variable on the left side, by all means, go ahead
and do that.
Graph:
Since we needed to indicate all values less than -2, the part of the
number line that was to the left of -2 was darkened.
Since we are not including where it is equal to, an open hole was used.
(return to problem
1a) |
| Answer/Discussion
to 1b |
Graph:
 |
*Distributive property
*Get x terms on one side, constants
on the other side
*Inv. of mult. by 2 is div. both sides by 2
*Visual showing all numbers less than or equal
to 4 on the number line
|
Graph:
Since we needed to indicate all values less than or equal to 4, the
part of the number line that was to the left of 4 was darkened.
Since we are including where it is equal to, a closed hole was used.
(return to problem
1b) |
| Answer/Discussion
to 1c |
Graph:
 |
*Mult. both sides by the LCD
*Get x terms on
one side, constants on the other
*Visual showing all numbers greater than or
equal to 3 on the number line
|
Graph:
Since we needed to indicate all values greater than or equal to 3,
the part of the number line that was to the right of 3 was darkened.
Since we are including where it is equal to, a closed hole was used.
(return to problem
1c) |
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Last revised on Jan. 9, 2002 by Kim Seward. |
