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为了帮助大家高效备考SAT,新东方在线SAT频道为大家带来SAT2数学考试题目:Inequalities,希望对大家SAT备考有所帮助。更多精彩尽请关注新东方在线SAT频道!
Before you get too comfortable with expressions and equations, we should introduce inequalities. An inequality is like an equation, but instead of relating equal quantities, it specifies exactly how two quantities are not equal. There are four types of inequalities:
1. x > y: “x is greater than y.”
2. x < y: “x is less than y.”
3. x ≥ y: “x is greater than or equal to y.”
4. x ≤ y: “x is less than or equal to y.”
Solving inequalities is exactly like solving equations except for one very important difference: when both sides of an inequality are multiplied or divided by a negative number, the direction of the inequality switches.
Here are a few examples:
Solve for x in the inequality – 3 < 2y.
Solve for x in the inequality ≥ –2.
Notice that in the last example, the inequality had to be reversed. Another way to express the solution is x ≥ –2. To help remember that multiplication or division by a negative number reverses the direction of the inequality, remember that if x > y, then –x < –y, just as 5 > 4 and –5 < –4. Intuitively, this idea makes sense, and it might help you remember this special rule of inequalities.
Absolute Value and Inequalities
When absolute values are included in inequalities, the solutions come in two varieties.
1. If the absolute value is less than a given quantity, then the solution is a single range, with a lower and an upper bound. For example,
Solve for x in the inequality |2x – 4| ≤ 6.
? First, solve for the upper bound:
? Second, solve for the lower bound:
? Now, combine the two bounds into a range of values for x. –1 ≤ x ≤ 5 is the solution.
1. The other solution for an absolute value inequality involves two disjoint ranges: one whose lower bound is negative infinity and whose upper bound is a real number, and one whose lower bound is a real number and whose upper bound is infinity. This occurs when the absolute value is greater than a given quantity. For example,
Solve for x in the inequality |3x + 4| > 16.
? First, solve for the upper range:
? Then, solve for the lower range:
? Now combine the two ranges to form the solution, which is two disjoint ranges: –∞ < x < –20?3 or 4 < x < ∞.
When working with absolute values, it is important to first isolate the expression within absolute value brackets. Then, and only then, should you solve separately for the cases in which the quantity is positive and negative.
Ranges
Inequalities are also used to express the range of values that a variable can take. a < x < b means that the value of x is greater than a and less than b. Consider the following word-problem example:
A very complicated board game has the following recommendation on the box: “This game is only appropriate for people older than 40 but no older than 65.” What is the range of the age of people for which the board game is appropriate?
Let a be the age of people for which the board game is appropriate. The lower bound of a is 40, and the upper bound is 65. The range of a does not include its lower bound (it is appropriate for people “older than 40”), but it does include its upper bound (“no older than 65”, i.e., 65 is appropriate, but 66 is not). Therefore, the range of the age of people for which the board game is appropriate can be expressed by the inequality:
Here is another example:
A company manufactures car parts. As is the case with any system of mass production, small errors occur on virtually every part. The key for this company to succeed in making viable car parts is to keep the errors within a specific range. The company knows that a particular piece they manufacture will not work if it weighs less than 98% of its target weight or more than 102% of its target weight. If the target weight of this piece is 21.5 grams, in what range of weights must the piece measure for it to function?
The boundary weights of this car part are .98 21.5 = 21.07 and 1.02 21.5 = 21.93 grams. The problem states that the piece cannot weigh less than the minimum weight or more than the maximum weight in order for it to work. This means that the part will function at boundary weights themselves, and the lower and upper bounds are included. The answer to the problem is 21.07 ≤ x ≤ 21.93, where x is the weight of the part in grams.
Finding the range of a particular variable is essentially an exercise in close reading. Every time you come across a question involving ranges, you should carefully peruse the problem to pick out whether a particular variable’s range includes its bounds or not. This inclusion is the difference between “less than or equal to” and simply “less than.”
Operations on Ranges
Operations like addition, subtraction, and multiplication can be performed on ranges just like they can be performed on variables. For example:
If 4 < x < 7, what is the range of 2x + 3?
To solve this problem, simply manipulate the range like an inequality until you have a solution. Begin with the original range:
Then multiply the inequality by 2:
Add 3 to the inequality, and you have the answer:
There is one crucial rule you need to know about multiplying ranges: if you multiply a range by a negative number, you must flip the greater-than or less-than signs. For instance, if you multiply the range 2 < x < 8 by –1, the new range will be –2 > x > –8. Math IC questions that ask you to perform operations on ranges of one variable will often test your alertness by making you multiply the range by a negative number.
Some range problems on the Math IC will be made slightly more difficult by the inclusion of more than one variable. In general, the same basic procedures for dealing with one-variable ranges applies to adding, subtracting, and multiplying two-variable ranges.
Addition with Ranges of Two or More Variables
If –2 < x < 8 and 0 < y < 5, what is the range of x + y?
Simply add the ranges. The lower bound is –2 + 0 = –2. The upper bound is 8 + 5 = 13. Therefore, –2 < x + y < 13.
Subtraction with Ranges of Two or More Variables
Suppose 4 < s < 7 and –3 < t < –1. What is the range of s – t?
In this case, you have to find the range of –t. By multiplying the range of t by –1 and reversing the direction of the inequalities, we find that 1 < –t < 3. Now we can simply add the ranges again to find the range of s – t. 4 + 1 = 5, and 7 + 3 = 10. Therefore, 5 < s – t < 10.
In general, to subtract ranges, find the range of the opposite of the variable being subtracted, and then add the ranges as usual.
Multiplication with Ranges of Two or More Variables
If –1 < j < 4 and 6 < k < 12, what is the range of jk?
First, multiply the lower bound of one variable by the lower and upper bounds of the other variable:
Then, multiply the upper bound of one variable with both bounds of the other variable:
The least of these four products becomes the lower bound, and the greatest is the upper bound. Therefore, –12 < jk < 48.
Let’s try one more example of performing operations on ranges:
If 3 ≤ x < 7 and , what is the range of 2(x + y)?
The first step is to find the range of x + y. Notice that the range of y is written backward, with the upper bound to the left of the variable. Rewrite it first:
Next add the ranges to find the range of x + y:
We have our bounds for the range of x + y, but are they included in the range? In other words, is the range 0 < x + y < 11, 0 ≤ x + y ≤ 11, or some combination of these two?
The rule to answer this question is the following: if either of the bounds that are being added, subtracted, or multiplied is non-inclusive (< or >), then the resulting bound is non-inclusive. Only when both bounds being added, subtracted, or multiplied are inclusive (≤ or ≥) is the resulting bound also inclusive.
The range of x includes its lower bound, 3, but not its upper bound, 7. The range of y includes both its bounds. Therefore, the range of x + y is 0 ≤ x + y < 11, and the range of 2(x + y) is 0 ≤ 2(x + y) < 22.
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