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The absolute value of a number measures its distance to the origin on the real number line.
Since 5 is at 5 units distance from the origin 0, the absolute value of 5 is 5, |5|=5. Since -5 is also at a distance of 5 units from the origin, the absolute value of -5 is 5, |-5|=5
Inequality We are ready for our first inequality. Find the set of solutions for |x|<5. Translate into English: we are looking for those real numbers x whose distance from the origin is less than 5 units. Obviously we are talking about the interval (-5,5): so all the values of X are -4 , -3 , -2 , -1 , 0 , 1, 2 , 3 , 4 . For the equation |x| â‰¤ 5, the solution will include 5 & -5 with the set of other numbers.
Solving linear inequalities is very similar to solving linear equations, except for one detail: you flip the inequality sign whenever you multiply or divide the inequality by a negative. The easiest way to show this is with some examples
X + 3 < 2 add -3 to both the site
- 3 -3
X < -1
The only difference here is that you have a "less than" sign, instead of an "equals" sign
a < b implies that b > a implies that b-a > 0
a â‰¤ b implies that b â‰¥ a implies that b-a > 0 or b = a.
Here are some rules for ordering real numbers. Try proving these yourself
Let a, b, c and d be real numbers:
1)a < 0 and b < 0 implies that ab > 0.
2)a < b and b < c implies that a < c.
3)a < b implies that a + c < b + c.
4)a < b and c > 0 implies that ac < bc
5)a < b and c < 0 implies that ac > bc. Multiplying by a negative number reverses the inequality.
6)a > 1 implies that a2 > a.
7)0 < a < 1 implies that a2 < a
8)0 a < b implies that a2 < b2 9)
0 a, 0 b, and a2 < b2 implies that a < b.
Now that we have all of these rules, we can start solving inequalities. This is done by manipulating the inequality into a form that has the variable on one side and has a real expression on the other side of the inequality. For instance, if the variable is x and the real expression is represented by a then the final form of the inequality is one of the following:
x > a
x < a
In this section, you will learn how so solve inequalities. "Solving'' an inequality means finding all of its solutions. A "solution'' of an inequality is a number which when substituted for the variable makes the inequality a true statement.
Here is an example: Consider the inequality X â€“ 2 > 5.
When we substitute 8 for x, the inequality becomes 8-2 > 5. Thus, x=8 is a solution of the inequality. On the other hand, substituting -2 for x yields the false statement (-2)-2 > 5. Thus x = -2 is NOT a solution of the inequality. Inequalities usually have many solutions.
As in the case of solving equations, there are certain manipulations of the inequality which do not change the solutions. Here is a list of "permissible'' manipulations:
Rule 1. Adding/subtracting the same number on both sides.
Example: The inequality x-2>5 has the same solutions as the inequality x > 7. (The second inequality was obtained from the first one by adding 2 on both sides.)
Rule 2. Switching sides and changing the orientation of the inequality sign .
Example: The inequality 5-x> 4 has the same solutions as the inequality 4 < 5 - x. (We have switched sides and turned the ``>'' into a ``<'').
Last, but not least, the operation which is at the source of all the trouble with inequalities:
Rule 3a. Multiplying/dividing by the same POSITIVE number on both sides.
Rule 3b. Multiplying/dividing by the same NEGATIVE number on both sides AND changing the orientation of the inequality sign.
Examples: This sounds harmless enough. The inequality 2x<6 has the same solutions as the inequality x<3. (We divided by +2 on both sides).
The inequality -2x > 4 has the same solutions as the inequality x< -2. (We divided by (-2) on both sides and switched ">'' to "<''.)
But Rule 3 prohibits fancier moves : The inequality xÂ² > x DOES NOT have the same solutions as the inequality x > 1. (We were planning on dividing both sides by x, but we can't, because we do not know at this point whether x will be positive or negative!) In fact, it is easy to check that x = -2 solves the first inequality, but does not solve the second inequality.
Only ``easy'' inequalities are solved using these three rules; most inequalities are solved by using different techniques.
Let's solve some inequalities:
Example :Consider the inequality 2x + 5 < 7
The basic strategy for inequalities and equations is the same: isolate x on one side, and put the "other stuff" on the other side. Following this strategy, let's move +5 to the right side. We accomplish this by subtracting 5 on both sides (Rule 1) to obtain
(2x + 5 ) â€“ 5 < 7 â€“ 5,
after simplification we obtain 2x < 2
Once we divide by +2 on both sides (Rule 3a), we have succeeded in isolating x on the left: 2x/2 < 2/2 or simplified, x < 1
All real numbers less than 1 solve the inequality. We say that the "set of solutions'' of the inequality consists of all real numbers less than 1. In interval notation , the set of solutions is the interval (-âˆž , 1).
When the inequalities involve absolte values, you must be very careful with the use of the words and . and or . They end up giving quite different results. Take a look here to see what we mean. When we have the absolute values, the and . condition applies when we have < or signs and the or . condition applies when we have > or signs. So we have:
|x| < a
means -a < x < a (which is the same as -a < x and x < a).
means â€“a â‰¤ x â‰¥ a (which is the same as -a x and x a)
|x| > a
means x < -a or x > a
means x â‰¤ -a or x â‰¥ a
Our method fails for more contrived examples.
Let us consider the inequality
It's back to basic algebra with a twist. The standard definition for the absolute value function is given by: |x| = x , if x â‰¥ 0
= -x , if xâ‰¤0
Thus we can get rid of the | | sign in our inequality if we know whether the expression inside, x-3, is positive or negative. We will do exactly that!
Let's first consider only those values of x for which x â€“ 3 â‰¥ 0.
Case 1 : xâ‰¥3 In this case we know that |x-3|=x-3, so our inequality becomes
X-3 < 2x-4
Solving the inequality, we obtain
We have found some solutions to our inequality: x is a solution if x â‰¥ 3and x>1 at the same time! We're talking about numbers X â‰¥3. What if x-3<0?
Case 2 : x<3 This time x-3<0, so |x-3|=-(x-3)=3-x, so our inequality reads as
3-x < 2x-4.
Applying the standard techniques, this can be simplified to x > 7/3
Our inequality has some more solutions: Under our case assumption x<3, solutions are those real numbers which satisfy x > 7/3 .
We're talking about numbers in the interval (7/3, 3) .Combining the solutions we found for both cases, we conclude that the set of solutions for the inequality
|x-3|<2x-4 are the numbers in the interval (7/3 , +âˆž).
Solve x 2 â€“ 3x + 2 > 0
Factoring, we get y = x2 â€“ 3x + 2 = (x â€“ 2)(x â€“ 1). Look at each of these factors separately. The factor x â€“ 1 is positive for x > 1 ; similarly, x â€“ 2 is positive for x > 2 . Thinking back to when you first learned about negatives, you know that (plus)Ã—(plus) = (plus), (minus)Ã—(minus) = (plus), and (minus)Ã—(plus) = (minus). So, to compute the sign on y = x2 â€“ 3x + 2 , we only really need to know the signs on the factors. So the solution of x 2 â€“ 3x + 2 > 0 are the two intervals
(Negative infinity, 1) and (2, positive infinity).
Solve x /(x â€“ 3) < 2 .
First off, you have to remember that you can't begin solving until you have the inequality in " = 0 " format. Here's how the problem works: x/(x-3) â‰¤ 2
â‰ˆ x / (x-3) - 2 â‰¤ 0
Now convert to a common denominator: x / (x â€“ 3) â€“ 2(x-3)/(x-3) â‰¤ 0
...and simplify: [ x- 2 ( x â€“ 3 ) ] / (x â€“ 3 ) â‰¤ 0 same as
(x- 2x + 6 )/ (x â€“ 3) â‰¤ 0 same as
(-x + 6 )/ (x â€“ 3) â‰¤ 0
The two factors are â€“x + 6 and x â€“ 3 . Note that x cannot equal 3 , or else you would be dividing by zero, which is not allowed. The first factor, â€“x + 6 , equals zero when x = 6 . The other factor, x â€“ 3 , equals zero when x = 3 . Now, x cannot actually equal 3 , so this endpoint will not be included in any solution interval (even though this is an "or equal to" inequality), but we need the value in order to figure out what our intervals are. In this case, our intervals are (negative infinity, 3 ), (3, 6] , and [6 , positive infinity). Note the use of brackets to indicate that 6 can be included in the solution, but that 3 cannot.