The best way to think of inequalities is as equations with slightly less specific information. Instead of providing an exact location, they provide a range of locations, all of which are satisfactory. With “=” instead of“<”, we can solve for the exact number. In an inequality, that number will be used as an endpoint.
Find the Range
To find the range, you can solve for each inequality as if it were an equation, and then look at the direction of the sign. For example:
1/2 < 2x + 3 < 8
To find the range of values for x, we can subtract 3 from all three expressions, and then divide by 2.
(1/2 – 3) < 2x < (8 – 3)
-5/2 < 2x < 5
-5/4 < x < 5/2
If the question had asked if x³ < 0 (if x is negative), then we would not have definitive proof. Inequalities literally provide a range of options, so be sure what critical points (0, +1, -1 etc) these ranges include.
How many Ranges? It Depends.
When you have one greater/less than sign in an inequality with a linear variable, there is only one range created. Here are your options:
x>2 OR
x<4
These both provide one range stretching to infinity or to negative infinity. But what if you had x² < 9?
x² < 9 actually provides a finite range with two endpoints, even though there is only one < sign. The endpoints in this case are +/-3. Any x value between these two points will satisfy the equation. Also note that the expression x² < 9 can also be written as |x| < 3.
Be careful how many endpoints are created, to recognize whether the range is finite or infinite.
Multiplication/Division of a Negative Number
Make sure you are not multiplying or dividing by a negative number. In these situations you must FLIP the signs. If you aren’t sure whether the number is positive or negative (as in a variable) then you cannot perform the action.
Inequalities and Absolute Values
Since an absolute value simply expresses a distance from a certain point, if we used an inequality instead of an equal sign, we know if a distance is more or less than the specified amount. Take a look at the following example:
Which inequality below most accurately represents the range of possible values for x?
A. The absolute value of x is less than or equal to 4.
B. The absolute value of x is less than or equal to 5.
C. The absolute value of (x + 2) is less than or equal to 2.
D. The absolute value of (x – 1) is less than or equal to 3.
E. The absolute value of (x + 1) is less than or equal to 3.
The first thing we want to do is determine the midpoint between the endpoints. What this does is create an equal distance in the negative and positive directions from that center location. For this question, the center point between -2 and +4 is +1.
From +1, our range of possible values for x extends at most 3 in either direction, but can also be 1, 1.5, 2.8 etc in either direction. This means that the distance from +1 is LESS THAN 3.
Looking at our answer choices, we now have to decide between D and E. With absolute values, as with some advanced pre-calculus equations, the shift within the equation is opposite of the shift on the graph. A good way to quickly test this is to plug in numbers toward the extremes and see which fit.
For (E), if we plug in x = 3.5 (which we know is within the given range) into | x + 1 | ≤ 3, we get 4.5 ≤ 3, which is false.
Any x within the given range meets the inequality | x – 1 | ≤ 3 in Choice D. Be sure to think of ranges in terms of both inequalities and absolute values, as these come up in about 1-2 questions per test.
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