章节大纲

  • 常规

    全部折叠 展开全部

  • 解决复合不平等问题的办法

    What if you had a compound inequality like 0 ≤ 2 x + 6 ≤ 6 ? How could you solve it and graph its solution set ?
    ::如果你有像 02x+66这样的复合不平等呢? 你怎么能解决它并绘制解决方案集的图表呢?

    Solutions to Compound Inequalities
    ::解决复合不平等问题的办法

    When we solve , we separate the inequalities and solve each of them separately. Then, we combine the solutions at the end.
    ::当我们解决的时候,我们把不平等分开,分别解决每一个不平等。然后,我们把解决方案放在最后。

    Solve the following compound inequalities and graph the solution set.
    ::解决以下复杂的不平等问题,并绘制解决方案的图表。

    a)  − 2 < 4 x − 5 ≤ 11
    ::a)-2<2<4x-511

    First we re-write the compound inequality as two separate inequalities with and . Then solve each inequality separately.
    ::首先,我们将复合不平等重新写成两种不同的不平等。然后分别解决每一种不平等。

    − 2 < 4 x − 5 4 x − 5 ≤ 11 3 < 4 x a n d 4 x ≤ 16 3 4 < x   x ≤ 4

    ::-2 < 2 < 4x- 54x-5113 < 4xand4x_1634 < x×4

      3 4 < x and x ≤ 4 . This can be written as 3 4 < x ≤ 4 .
    ::34 <x和 x4。 这可以写为 34 < x4 。

    b)  3 x − 5 < x + 9 ≤ 5 x + 13
    ::b) 3x-5 < x+95x+13

     Re-write the compound inequality as two separate inequalities with and . Then solve each inequality separately.
    ::将复合不平等重写为两种不同的不平等,然后分别解决每一种不平等。

    3 x − 5 < x + 9     x + 9 ≤ 5 x + 13     2 x < 14 a n d − 4 ≤ 4 x x < 7   − 1 ≤ x

    ::3x-5 < X+9 x+9+9=5x+13 2x < 14和-44xx < 7 - 1xx

      x < 7 and x ≥ − 1 . This can be written as: − 1 ≤ x < 7 .
    ::x7 和 x*\\\\\\\\\\\\\ x7\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\可以可以可以可以写写写写写写写写写写写写写写写写写写写写写写写写后写写写写写后写后写后写后写后写后来来来为-\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\xxxxxxx\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

    Solve the following compound inequalities and graph the solution set.
    ::解决以下复杂的不平等问题,并绘制解决方案的图表。

    a) 9 − 2 x ≤ 3 or 3 x + 10 ≤ 6 − x
    :伤心a) 9-2x3或3x+106-x

    Solve each inequality separately:
    ::分别解决每一种不平等:

    9 − 2 x ≤ 3 3 x + 10 ≤ 6 − x   − 2 x ≤ − 6 o r     4 x ≤ − 4 x ≥ 3       x ≤ − 1

    ::9 - 2x33x+10_6-x - 2x_%6or 4x4x3x1

    x ≥ 3 or x ≤ − 1
    ::x% 3 或 x% 1

    b) x − 2 6 ≤ 2 x − 4 or x − 2 6 > x + 5
    :伤心b) x- 262x-4 或 x- 26> x+5

    Solve each inequality separately:
    ::分别解决每一种不平等:

    x − 2 6 ≤ 2 x − 4 x − 2 6 > x + 5 x − 2 ≤ 6 ( 2 x − 4 ) x − 2 > 6 ( x + 5 ) x − 2 ≤ 12 x − 24 o r   x − 2 > 6 x + 30   22 ≤ 11 x   − 32 > 5 x       2 ≤ x − 6.4 > x

    ::x-26(x+5)x-2=6(x+5)x-2(x+5)x-2(x+5)x-212x-24or x-2(x+5)x-2(12)x-24or x-2(6)x+30 22-11x-32>5x-2__x-6.4>x

    x ≥ 2 or x < − 6.4
    ::x% 2 或 x% 6. 4

    One thing you may notice in the video for this Concept is that in the second problem, the two solutions joined with “or” overlap, and so the solution ends up being the set of all real numbers, or ( − ∞ , ∞ ) . This happens sometimes with compound inequalities that involve “or”; for example, if the solution to an inequality ended up being “ x < 5 or x > 1 ,” the solution set would be all real numbers. This makes sense if you think about it: all real numbers are either a) less than 5, or b) greater than or equal to 5, and the ones that are greater than or equal to 5 are also greater than 1—so all real numbers are either a) less than 5 or b) greater than 1.
    ::这个概念的视频中可以注意到的一件事,就是在第二个问题中,两个解决方案与“或”重叠相联,因此,解决方案最终是所有实际数字或(,,)的组合。这有时发生在涉及“或”的复合不平等方面;例如,如果不平等的解决方案最终是“x<5或 x>1”的,那么设定的解决方案将是所有真实数字。如果你想起来,这很有意义:所有真实数字都(a)小于5或(b)大于或等于5,而大于或等于5的,而高于或等于5的,也大于1——因此所有实际数字都大于1——因此,所有实际数字都小于(a)小于5或(b)大于1。

    Compound inequalities with “and,” meanwhile, can turn out to have no solutions. For example, the inequality “ x < 3 and x > 4 ” has no solutions: no number is both greater than 4 and less than 3. If we write it as 4 < x < 3 it’s even more obvious that it has no solutions; 4 < x < 3 implies that 4 < 3 , which is false.
    ::与此同时,“和”的复合不平等可能会变成没有解决方案。 比如,不平等“ x < 3 和 x>4 ” 就没有解决方案:任何数字都大于4, 小于3。 如果我们把它写成 4 < x < 3 , 更明显的是它没有解决方案; 4 < x < 3 意味着4 < 3 是虚假的。

    Solve Real-World Problems Using Compound Inequalities
    ::利用多重不平等解决现实世界问题

    Many application problems require the use of compound inequalities to find the solution.
    ::许多应用问题需要利用复合不平等来找到解决办法。

    Solve the Real-World Problem
    ::解决现实世界问题

    The speed of a golf ball in the air is given by the formula v = − 32 t + 80 . When is the ball traveling between 20 ft/sec and 30 ft/sec?
    ::公式 v32t+80 给出了高尔夫球在空中的速度。 球何时在 20 英尺/ sec 和 30 英尺/ sec 之间?

    First we set up the inequality 20 ≤ v ≤ 30 , and then replace v with the formula v = − 32 t + 80 to get 20 ≤ − 32 t + 80 ≤ 30 .
    ::首先我们设置了不平等 20v30, 然后用公式 v32t+80 取代 vvv32t+80, 以获得 2032t+80}30。

    Then we separate the compound inequality and solve each separate inequality:
    ::然后我们把复合不平等分开 解决每一个不同的不平等

      20 ≤ − 32 t + 80       − 32 t + 80 ≤ 30 32 t ≤ 60 and   50 ≤ 32 t t ≤ 1.875 1.56 ≤ t

    ::2032t+80 - 32t+803032t60 和 5032t1.8751.56_t

      1.56 ≤ t ≤ 1.875
    ::1.561.875

    To check the answer, we plug in the minimum and maximum values of t into the formula for the speed.
    ::为了检查答案,我们将 t 的最小值和最大值插入速度公式中。

    For t = 1.56 ,   v = − 32 t + 80 = − 32 ( 1.56 ) + 80 = 30   f t / s e c
    ::t=1.56, v32t+8032(1.56)+80=30英尺/秒

    For t = 1.875 ,   v = − 32 t + 80 = − 32 ( 1.875 ) + 80 = 20   f t / s e c
    ::t=1.875, v32t+8032(1.875)+80=20英尺/秒

    So the speed is between 20 and 30 ft/sec. The answer checks out.
    ::速度在20到30英尺/秒之间

    Example
    ::示例示例示例示例

    William’s pick-up truck gets between 18 to 22 miles per gallon of gasoline. His gas tank can hold 15 gallons of gasoline. If he drives at an average speed of 40 miles per hour, how much driving time does he get on a full tank of gas?
    ::威廉的皮卡车每加仑汽油在18到22英里之间。 他的油箱可以持有15加仑汽油。 如果他每小时平均开车40英里,那么他有多少驾驶时间坐满油箱?

    Let t = driving time. We can use to get from time per tank to miles per gallon:
    ::驾驶时间到每罐时间到每加仑英里:

    t   h o u r s 1   t a n k × 1   t a n k 15   g a l l o n s × 40   m i l e s 1   h o u r × 40 t 15 m i l e s g a l l o n

    ::1坦克×1坦克15加仑×40英里1小时×40吨15英里

    Since the truck gets between 18 and 22 miles/gallon, we set up the compound inequality 18 ≤ 40 t 15 ≤ 22 . Then we separate the compound inequality and solve each inequality separately:
    ::由于卡车每加仑在18到22英里之间,我们建立了复合不平等1840t1522。然后我们将复合不平等分开,分别解决每一种不平等:

        18 ≤ 40 t 15   40 t 15 ≤ 22   270 ≤ 40 t and 40 t ≤ 330 6.75 ≤ t     t ≤ 8.25

    ::18040t15 40t1522 27040t和40t3306.75t t8.25

      6.75 ≤ t ≤ 8.25 .
    ::6.758.25。

    Andrew can drive between 6.75 and 8.25 hours on a full tank of gas.
    ::安德鲁可以在6.75小时到8.25小时之间 开满油箱的汽油

    If we plug in t = 6.75 we get 40 t 15 = 40 ( 6.75 ) 15 = 18   m i l e s   p e r   g a l l o n .
    ::如果我们插入t=6.75,我们得到40t15=40(6.7515)=18英里每加仑。

    If we plug in t = 8.25 we get 40 t 15 = 40 ( 8.25 ) 15 = 22   m i l e s   p e r   g a l l o n .
    ::如果我们插入t=8.25,我们得到40t15=40(8.25)15=22英里每加仑。

    The answer checks out.
    ::答案检查出来。

    Review
    ::回顾

    Solve the following compound inequalities and graph the solution on a number line.
    ::解决以下的复合不平等,用数字线绘制解决方案图。

    1. − 5 ≤ x − 4 ≤ 13
      ::- 5x-413
    2. 1 ≤ 3 x + 5 ≤ 4
      ::13x+54
    3. − 12 ≤ 2 − 5 x ≤ 7
      ::- 122-5x=7
    4. 3 4 ≤ 2 x + 9 ≤ 3 2
      ::342x+932
    5. − 2 ≤ 2 x − 1 3 < − 1
      ::-22x-131
    6. 4 x − 1 ≥ 7 or 9 x 2 < 3
      ::4-1-17 或 9x2 < 3
    7. 3 − x < − 4 or 3 − x > 10
      ::3 - x% 4 或 3 - x > 10
    8. 2 x + 3 4 < 2 or − x 5 + 3 < 2 5
      ::2x+34 <2或-x5+3 < 25
    9. 2 x − 7 ≤ − 3 or 2 x − 3 > 11
      ::2 - 73 或 2 - 3 > 11
    10. 4 x + 3 < 9 or − 5 x + 4 ≤ − 12
      ::4x+3 < 9 或- 5x+412
    11. How would you express the answer to problem 1 as a set?
      ::你将如何表达对问题1的回答?
    12. How would you express the answer to problem 1 as an interval?
      ::您如何以间隔时间表达问题1的答案?
    13. How would you express the answer to problem 6 as a set?
      ::你将如何表达对问题6的回答?
    14. Could you express the answer to problem 6 as a single interval? Why or why not?
      1. How would you express the first part of the solution in interval form?
        ::您如何以间隔形式表达解决方案的第一部分 ?
      2. How would you express the second part of the solution in interval form?
        ::您如何以间隔形式表达解决方案的第二部分?

      ::您能否用一个间隔来表达对问题6的答案? 为什么不行? 您如何以间隔形式表达解决方案的第一部分? 您如何以间隔形式表达解决方案的第二部分?
    15. Express the answers to problems 2 through 5 in interval notation.
      ::以中间编号表示对问题2至问题5的答复。
    16. Solve the inequality “ x ≥ − 3 or x < 1 ” and express the answer in interval notation.
      ::解决“x3 或 x < 1” 不平等,以间距符号表示答案。
    17. How many solutions does the inequality “ x ≥ 2 and x ≤ 2 ” have?
      ::“x%2”和“x%2”的不平等有多少解决办法?
    18. To get a grade of B in her Algebra class, Stacey must have an average grade greater than or equal to 80 and less than 90. She received the grades of 92, 78, 85 on her first three tests.
      1. Between which scores must her grade on the final test fall if she is to receive a grade of B for the class? (Assume all four tests are weighted the same.)
        ::如果她要获得班级B级的分数,在最后测试的分数之间必须分多少分? (假定所有四次测试均加权相同。 )
      2. What range of scores on the final test would give her an overall grade of C, if a C grade requires an average score greater than or equal to 70 and less than 80?
        ::如果C级平均得分大于或等于70分或小于80分,最后考试的得分范围如何?
      3. If an A grade requires a score of at least 90, and the maximum score on a single test is 100, is it possible for her to get an A in this class? (Hint: look again at your answer to part a)
        ::如果一个年级至少需要90分,而单次考试的最大分数是100分,她能否在这个年级获得A分? (提示:再看看你对A部分的答复)

      ::为了在代数班中获得B级,Stacey必须有一个平均等级大于或等于80或小于90的B级。如果她要获得班级B级的分数,那么她头三次考试的分数必须是92、78、85;如果她要获得班级B级的分数必须是最后考试的分数中的哪一分数?(假定所有四个测验都是加权的。 )如果C级需要平均分大于或等于70或小于80分,那么最后考试的分数范围是多少?如果A级需要至少90分,而单次考试的最高分数是100分,她能否在这个班里拿到A级的分数? (hint:再次看一下你对A部分的回答)

    Review (Answers)
    ::回顾(答复)

    Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
    ::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键 。