Section: 绝对绝对值 | 6.8 绝对值

Section outline

Absolute Value Equations
::绝对绝对值

In the previous concept, we saw how to solve simple absolute value equations. In this concept, you will see how to solve more complicated absolute value equations.
::在前一个概念中,我们看到了如何解决简单的绝对值方程式。在这个概念中,你会看到如何解决更复杂的绝对值方程式。

Solving Absolute Value Equations
::解决绝对值

1. Solve the equation $\begin{aligned} | x - 4 | = 5 \end{aligned}$ and interpret the answers.
::1. 解答方程式 X - 4 - 5 并解释答案。

We consider two possibilities: the expression inside the absolute value sign is non-negative or is negative. Then we solve each equation separately.
::我们考虑两种可能性:绝对值符号内的表达是非负值或负值。然后我们分别解决每个方程式。

$\begin{aligned} x - 4 = 5 and x - 4 = - 5 \\ x = 9 x = - 1 \end{aligned}$

::x4=5和x-45 x=9 x%1

$\begin{aligned} x = 9 \end{aligned}$ and $\begin{aligned} x = - 1 \end{aligned}$ are the solutions.
::x=9 和 x1 是解决方案。

The equation $\begin{aligned} | x - 4 | = 5 \end{aligned}$ can be interpreted as “what numbers on the number line are 5 units away from the number 4?” If we draw the number line we see that there are two possibilities: 9 and -1.
::等式 *ux-45可被解释为“数字线上的5个单位离数字4是多少数字?” 如果我们绘制数字线,我们看到有两种可能性:9和-1。

2. Solve the equation $\begin{aligned} | x + 3 | = 2 \end{aligned}$ and interpret the answers.
::2. 解决方程式 'x+3 ' 2 ' 并解释答案。

Solve the two equations:
::解决两个方程式:

$\begin{aligned} x + 3 = 2 and x + 3 = - 2 \\ x = - 1 x = - 5 \end{aligned}$

::x+3=2和 x+3 @%2 x%1 x%5

$\begin{aligned} x = - 5 \end{aligned}$ and $\begin{aligned} x = - 1 \end{aligned}$ are the answers.
::答案是 x5 和 x% 1 。

The equation $\begin{aligned} | x + 3 | = 2 \end{aligned}$ can be re-written as: $\begin{aligned} | x - (- 3) | = 2 \end{aligned}$ . We can interpret this as “what numbers on the number line are 2 units away from -3?” There are two possibilities: -5 and -1.
::等式 {x+3}%2 可以重写为 {x_(-3)}}\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Solve Real-World Problems Using Absolute Value Equations
::使用绝对价值等量解决现实世界问题

Real-World Application: Packing Coffee
::真实世界应用程序:包装咖啡

A company packs coffee beans in airtight bags. Each bag should weigh 16 ounces, but it is hard to fill each bag to the exact weight. After being filled, each bag is weighed; if it is more than 0.25 ounces overweight or underweight, it is emptied and repacked. What are the lightest and heaviest acceptable bags?
::公司用密封袋包装咖啡豆,每袋重16盎司,但很难填满每袋的准确重量。装满后,每袋都要称重;如果超过0.25盎司超重或体重不足,则清空并重新包装。最轻、最重的袋是什么?

The weight of each bag is allowed to be 0.25 ounces away from 16 ounces; in other words, the difference between the bag’s weight and 16 ounces is allowed to be 0.25 ounces. So if $\begin{aligned} x \end{aligned}$ is the weight of a bag in ounces, then the equation that describes this problem is $\begin{aligned} | x - 16 | = 0.25 \end{aligned}$ .
::每个包的重量允许在16盎司之外为0.25盎司;换句话说,包的重量与16盎司之间的差值允许为0.25盎司。因此,如果x是盎司袋的重量,那么说明这一问题的方程式是x-160.25。

Now we must consider the positive and negative options and solve each equation separately:
::现在我们必须考虑积极的和消极的选择,并分别解决每个方程式:

$\begin{aligned} x - 16 = 0.25 and x - 16 = - 0.25 \\ x = 16.25 x = 15.75 \end{aligned}$

::x- 16=0. 25 和x- 16= 16= 0. 25x = 16. 25 x = 15. 75

The lightest acceptable bag weighs 15.75 ounces and the heaviest weighs 16.25 ounces.
::最轻可接受的袋子重15.75盎司,最重的袋重16.25盎司。

We see that $\begin{aligned} 16.25 - 16 = 0.25 o u n c e s \end{aligned}$ and $\begin{aligned} 16 - 15.75 = 0.25 o u n c e s \end{aligned}$ . The answers are 0.25 ounces bigger and smaller than 16 ounces respectively.
::我们看到16.25-16=0.25盎司和16-15.75=0.25盎司,答案分别为0.25盎司和小于16盎司。

The answer checks out.
::答案检查出来。

The answer you just found describes the lightest and heaviest acceptable bags of coffee beans. But how do we describe the total possible range of acceptable weights? That’s where inequalities become useful once again.
::你刚刚找到的答案描述了最轻和最重可接受的咖啡豆袋。但我们如何描述可能接受的重量的总范围? 这就是不平等再次变得有用的地方。

Example
::示例示例示例示例

Example 1
::例1

Solve the equation $\begin{aligned} | 2 x - 7 | = 6 \end{aligned}$ and interpret the answers.
::解析方程式 #% 2x- 7# 6, 并解释答案。

Solve the two equations:
::解决两个方程式:

$\begin{aligned} 2 x - 7 = 6 2 x - 7 = - 6 \\ 2 x = 13 and 2 x = 1 \\ x = \frac{13}{2} x = \frac{1}{2} \end{aligned}$

::2x-7=62x-76 2x=13和 2x=1 x1 x=132 x=12

Answer: $\begin{aligned} x = \frac{13}{2} \end{aligned}$ and $\begin{aligned} x = \frac{1}{2} \end{aligned}$ .
::答复:x=132和x=12。

The interpretation of this problem is clearer if the equation $\begin{aligned} | 2 x - 7 | = 6 \end{aligned}$ is divided by 2 on both sides to get $\begin{aligned} \frac{1}{2} | 2 x - 7 | = 3 \end{aligned}$ . Because $\begin{aligned} \frac{1}{2} \end{aligned}$ is nonnegative, we can distribute it over the absolute value sign to get $\begin{aligned} | x - \frac{7}{2} | = 3 \end{aligned}$ . The question then becomes “What numbers on the number line are 3 units away from $\begin{aligned} \frac{7}{2} \end{aligned}$ ?” There are two answers: $\begin{aligned} \frac{13}{2} \end{aligned}$ and $\begin{aligned} \frac{1}{2} \end{aligned}$ .
::这个问题的解释更加清楚,如果方程式+2x-76除以2, 双方获得 122x-73。由于12是非负数的,我们可以在绝对值符号上分配它,以获得x-723。问题就变成“数字线上的数字是多少,3个单位离72是多少?” 有两个答案:132个和12个。

Review
::回顾

Solve the absolute value equations and interpret the results by graphing the solutions on the number line.
::解决绝对值方程式,并通过在数字线上绘制解决方案的图形来解释结果。

$\begin{aligned} | x - 5 | = 10 \end{aligned}$
::~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$\begin{aligned} | x + 2 | = 6 \end{aligned}$
::X+26

$\begin{aligned} | 5 x - 2 | = 3 \end{aligned}$
::5x-23

$\begin{aligned} | x - 4 | = - 3 \end{aligned}$
::X - 43

$\begin{aligned} | 2 x - \frac{1}{2} | = 10 \end{aligned}$
::2x-1210

$\begin{aligned} | - x + 5 | = \frac{1}{5} \end{aligned}$
::X+5+15

$\begin{aligned} | \frac{1}{2} x - 5 | = 100 \end{aligned}$
::12x-5100

$\begin{aligned} | 10 x - 5 | = 15 \end{aligned}$
::10x -515

$\begin{aligned} | 0.1 x + 3 | = 0.015 \end{aligned}$
::0.1x+30.015

$\begin{aligned} | 27 - 2 x | = 3 x + 2 \end{aligned}$
::27 - 2x3x+2

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。

Section outline

Absolute Value Equations ::绝对绝对值

Solving Absolute Value Equations ::解决绝对值

Solve Real-World Problems Using Absolute Value Equations ::使用绝对价值等量解决现实世界问题

Real-World Application: Packing Coffee ::真实世界应用程序:包装咖啡

Example ::示例示例示例示例

Example 1 ::例1

Review ::回顾

Review (Answers) ::回顾(答复)

Absolute Value Equations
::绝对绝对值

Solving Absolute Value Equations
::解决绝对值

Solve Real-World Problems Using Absolute Value Equations
::使用绝对价值等量解决现实世界问题

Real-World Application: Packing Coffee
::真实世界应用程序:包装咖啡

Example
::示例示例示例示例

Example 1
::例1

Review
::回顾

Review (Answers)
::回顾(答复)