Re: inequalities
marshall1432 said:
1. Express the inequality x > 0 using interval notation. Use INF for infinity.
(INF,1)
x > 0 is true when x has any value that is greater than 0. In interval notation, the set of all numbers greater than 0 is
(0, inf)
2. Solve the inequality 3(x - 2) < 4x + 7. Give your solution in interval notation, using INF for infinity.
3x-6<4x+7
3x<4x+13
-x<13
?
What you have done so far is ok. Now, you want x, not -x, so you'll need to multiply both sides of the inequality by -1. Remember that when you multiply or divide both sides of an inequality by a negative number, you must also
change the direction of the inequality symbol:
(-1)(-x) > (-1)(13)
x > -13
This is true for any value of x which is greater than -13; in interval notation,
(-13, inf)
3. Solve the inequality -5 < 2(3 - x) < 1. Give your solution in interval notation.
-5<6-2x<1
11-2x<1
-2x=-10
x=5,0
BIG problems on this one!! Your original problem had TWO inequality symbols in it, and you managed to "lose" both of them.
Here's how I would do this:
-5 < 6 - 2x < 1
Add -6 to each part of the inequality:
-5 + (-6) < 6 - 2x + (-6) < 1 + (-6)
-11 < -2x < -5
Now, divide each part by -2; remember to change the direction of each inequality symbol because you're dividing by a negative:
(-11)/(-2) > (-2x)/(-2) > (-5)/(-2)
(11/2) > x > (5/2)
This is true for any value of x which is greater than 5/2 but less than 11/2. In interval notation,
(5/2, 11/2)
4. Solve the inequality | 4 - 2x | > 3. Give your solution in interval notation, using INF for infinity.
|-2x|>-1
x=-2,-1
Huh? Again, you changed an inequality to an equation. You can't do that!
This inequality states that the absolute value of "something" is greater than -1. You are expected to realize that the absolute value of any number must be non-negative (greater or equal to 0). So, no matter what x is, |-2x | will be greater than or equal to 0, and thus greater than -1. This is true for ANY value of x:
(-inf, inf)
5. Solve the inequality | 4 - 3x | < 10. Give your solution in interval notation, using INF for infinity.
|-3x|<6
x=2,0
You can't subtract 4 because it is inside the absolute value symbol.
If the absolute value of "something" is less than 10, that "something" must be less than 10 units from 0 on the number line. That "something" must be greater than -10, AND that something must be less than 10. So,
| 4 - 3x | < 10
means
4 - 3x > -10 AND 4 - 3x < 10
Now, solve each of those inequalities for x:
-3x > -14 AND -3x < 6
You're going to divide both sides of each inequality by -3; remember to switch the direction of each inequality symbol:
x < 14/3 AND x > -2
So, x must be greater than -2 AND it must be less than 14/3. The solution set, in interval notation, is
(-2, 14/3)
Am I doing any of this right? thanks!
I'm afraid not. You need to have a serious talk with your teacher.
