节: 增加和减减逻辑表达式的应用 | 12.11 增加和减减逻辑表达式的应用

章节大纲

Applications of Adding and Subtracting Rational Expressions
::增加和减减逻辑表达式的应用

In the previous section, you learned how to add and subtract rational expressions. In this section, you will use those tools to solve real-world problems.
::在上一节中,您学会了如何增减理性表达方式。在本节中,您将使用这些工具来解决现实世界的问题。

Real-World Application: Total Resistance
::现实世界应用:完全抗药性

In an electrical circuit with two resistors placed in parallel, the reciprocal of the total resistance is equal to the sum of the reciprocals of each resistance: $\begin{aligned} \frac{1}{R_{t o t}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} \end{aligned}$ . Find an expression for the total resistance, $\begin{aligned} R_{t o t} \end{aligned}$ .
::在一条电路中,两台阻力器平行安装,完全抗力的对等等于每次阻力:1Rt=1R1+1R2的对等相和,找到完全抗力的表达方式,Rtot。

Let's simplify the expression $\begin{aligned} \frac{1}{R_{1}} + \frac{1}{R_{2}} \end{aligned}$ .
::让我们简化表达式 1R1+1R2 。

The lowest common denominator is $\begin{aligned} R_{1} R_{2} \end{aligned}$ , so we multiply the first fraction by $\begin{aligned} \frac{R_{2}}{R_{2}} \end{aligned}$ and the second fraction by $\begin{aligned} \frac{R_{1}}{R_{1}} \end{aligned}$ :
::最低公分母是R1R2, 所以我们将第一个分位乘以R2R2, 第二个分位乘以R1R1:

\begin{aligned} \frac{R_{2}}{R_{2}} \cdot \frac{1}{R_{1}} + \frac{R_{1}}{R_{1}} \cdot \frac{1}{R_{2}} \\ Simplify: & \frac{R_{2} + R_{1}}{R_{1} R_{2}} \\ The total resistance is the reciprocal of this expression: & R_{t o t} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} A n s w e r \end{aligned}

::R2R2R11R1+R1R1R11R2 简化: R2+R1R1R1R2 总阻力是这个表达式的对等性: Rtot=R1R2R1+R2Aswer

Finding Unknown Numbers
::查找未知数字

The sum of a number and its reciprocal is $\begin{aligned} \frac{53}{14} \end{aligned}$ . Find the numbers.
::数字之和和和对数是5314 找出数字

Define variables:
::定义变量 :

Let $\begin{aligned} x \end{aligned}$ be the number; then its reciprocal is $\begin{aligned} \frac{1}{x} \end{aligned}$ .
::数字为 x; 对应为 1x 。

Set up an equation :
::设置方程式 :

The equation that describes the relationship between the numbers is $\begin{aligned} x + \frac{1}{x} = \frac{53}{14} \end{aligned}$
::描述数字之间的关系的方程式是 x+1x=5314

Solve the equation:
::解决方程式:

\begin{aligned} Find the lowest common denominator: & LCM = 14 x \\ Multiply all terms by 14 x : & 14 x \cdot x + 14 x \cdot \frac{1}{x} = 14 x \cdot \frac{53}{14} \end{aligned}

::查找最小公分母 : LCM=14xUltiply 全部条件 14x: 14xxxx+14xx1x=14x5314

(Notice that we’re multiplying the terms by $\begin{aligned} 14 x \end{aligned}$ instead of by $\begin{aligned} \frac{14 x}{14 x} \end{aligned}$ . We can do this because we’re multiplying both sides of the equation by the same thing, so we don’t have to keep the actual values of the terms the same. We could also multiply by $\begin{aligned} \frac{14 x}{14 x} \end{aligned}$ , but then the denominators would just cancel out a couple of steps later.)
:我们注意到我们把条件乘以14xx而不是14x14x。我们可以这样做是因为我们把等式的两边乘以同样的事情,所以我们不必将条件的实际值维持不变。我们还可以乘以14x14x,但是分母会只是后来取消几个步骤。 )

\begin{aligned} Cancel common factors in each term: & 14 x \cdot x + 14 x \cdot \frac{1}{x} = 14 x \cdot \frac{53}{14} \\ Simplify: & 14 x^{2} + 14 = 53 x \\ Write all terms on one side of the equation: & 14 x^{2} - 53 x + 14 = 0 \\ Factor: & (7 x - 2) (2 x - 7) = 0 \\ x = \frac{2}{7} and x = \frac{7}{2} \end{aligned}

::每个术语 : 14xxxxx+14xxxxx%1x=14x=14x=5314 简化 : 14x2+14=14x14=53x write 方程式的一边所有条件 : 14x2- 53xx+14=0 Factory: (7x-2)(2x-7)=0x=27和x=72

Notice there are two answers for $\begin{aligned} x \end{aligned}$ , but they are really parts of the same solution. One answer represents the number and the other answer represents its reciprocal.
::请注意 x 有两个答案, 但其实是同一解决方案的一部分。一个答案代表数字,另一个答案代表对等答案。

Check:
::检查 :

$\begin{aligned} \frac{2}{7} + \frac{7}{2} = \frac{4 + 49}{14} = \frac{53}{14} \end{aligned}$ . The answer checks out.
::27+72=4+4914=5314 答案验证。

Work problems are problems where two people or two machines work together to complete a job. Work problems often contain rational expressions. Typically we set up such problems by looking at the part of the task completed by each person or machine. The completed task is the sum of the parts of the tasks completed by each individual or each machine.
::工作问题是指两个人或两台机器一起工作以完成工作的问题,工作的问题往往含有合理的表达方式,我们通常通过查看每个人或机器完成的任务部分来设置这类问题,完成的任务是每个人或每台机器完成的任务部分的总和。

To determine the part of the task completed by each person or machine we use the following fact:
::为确定每个人或每个机器完成的任务部分,我们使用以下事实:

\begin{aligned} Part of the task completed = rate of work \times time spent on the task \end{aligned}

::完成的部分任务=任务所花费的工作时间

It’s usually useful to set up a table where we can list all the known and unknown variables for each person or machine and then combine the parts of the task completed by each person or machine at the end.
::通常应该设置一张表格, 列出每个人或机器的所有已知和未知变量, 然后将每个人或机器完成的任务部分合并到最后。

Real-World Application: Work Problems
::现实世界应用:工作问题

Mary can paint a house by herself in 12 hours. John can paint a house by himself in 16 hours. How long would it take them to paint the house if they worked together?
::玛丽在12小时内可以自己画房子,约翰在16小时内可以自己画房子,如果他们一起工作,他们需要多久才能画房子?

Define variables:
::定义变量 :

Let $\begin{aligned} t = \end{aligned}$ the time it takes Mary and John to paint the house together.
::让我们花时间玛丽和约翰一起画房子

Construct a table:
::构造表格 :

Since Mary takes 12 hours to paint the house by herself, in one hour she paints $\begin{aligned} \frac{1}{12} \end{aligned}$ of the house.
::因为Mary用12小时的时间自己画房子, 一小时后她画112个房子。

Since John takes 16 hours to pain the house by himself, in one hour he paints $\begin{aligned} \frac{1}{16} \end{aligned}$ of the house.
::John用16小时的时间一个人来折磨房子, 一小时后他画了116个房子。

Mary and John work together for $\begin{aligned} t \end{aligned}$ hours to paint the house together. Using
::玛丽和约翰一起工作了几个小时一起画房子

\begin{aligned} P a r t o f t h e t a s k c o m p l e t e d = r a t e o f w o r k \cdot t i m e s p e n t o n t h e t a s k \end{aligned}

::完成的部分任务完成的部分任务完成部分任务完成部分任务完成部分任务完成部分任务完成部分任务完成部分任务完成部分任务完成部分任务完成部分任务完成部分任务

we can write that Mary completed $\begin{aligned} \frac{t}{12} \end{aligned}$ of the house and John completed $\begin{aligned} \frac{t}{16} \end{aligned}$ of the house in this time.
::我们可以写到玛丽完成了12号房 John完成了16号房

This information is nicely summarized in the table below:
::下表对这方面的信息作了简洁的概述:

Painter	Rate of work (per hour)	Time worked	Part of task
Mary	$\begin{aligned} \frac{1}{12} \end{aligned}$	$\begin{aligned} t \end{aligned}$	$\begin{aligned} \frac{t}{12} \end{aligned}$
John	$\begin{aligned} \frac{1}{16} \end{aligned}$	$\begin{aligned} t \end{aligned}$	$\begin{aligned} \frac{t}{16} \end{aligned}$

Set up an equation:
::设置方程式 :

In $\begin{aligned} t \end{aligned}$ hours, Mary painted $\begin{aligned} \frac{t}{12} \end{aligned}$ of the house and John painted $\begin{aligned} \frac{t}{16} \end{aligned}$ of the house, and together they painted 1 whole house. So our equation is $\begin{aligned} \frac{t}{12} + \frac{t}{16} = 1 \end{aligned}$ .
::几小时后,玛丽画了房子的图12,约翰画了房子的图16,他们一起画了整个房子。所以我们的方程式是t12+T16=1。

Solve the equation:
::解决方程式:

\begin{aligned} Find the lowest common denominator: & LCM = 48 \\ Multiply all terms in the equation by the LCM: & 48 \cdot \frac{t}{12} + 48 \cdot \frac{t}{16} = 48 \cdot 1 \\ Cancel common factors in each term: & 4 \cdot \frac{t}{1} + 3 \cdot \frac{t}{1} = 48 \cdot 1 \\ Simplify: & 4 t + 3 t = 48 \\ 7 t = 48 \Rightarrow t = \frac{48}{7} = 6.86 h o u r s \end{aligned}

::查找最小公分母 : LCM=48Ultiply 方程式中的所有条件, 以 LCM : 48t12+48t16=481Cencel 常见系数 : 4t1+3t1=481=481 简化: 4t+3t=487t=48t=487=487=6.86小时

Check: The answer is reasonable. We’d expect the job to take more than half the time Mary would take by herself but less than half the time John would take, since Mary works faster than John.
::选中 : 答案是合理的。我们期望这项工作花费玛丽本人承担一半以上的时间,但约翰承担的时间不到一半,因为玛丽的工作速度比约翰要快。

Example
::示例示例示例示例

Example 1
::例1

Suzie and Mike take two hours to mow a lawn when they work together. It takes Suzie 3.5 hours to mow the same lawn if she works by herself. How long would it take Mike to mow the same lawn if he worked alone?
::Suzie和Mike一起工作需要两个小时才能修剪草坪。如果Suzie自己单独工作,那么Suzie需要3.5小时才能修剪同样的草坪。 Mike单独工作需要多长时间才能修剪同样的草坪?

Define variables:
::定义变量 :

Let $\begin{aligned} t = \end{aligned}$ the time it takes Mike to mow the lawn by himself.
::让我们花时间让迈克自己修草坪

Construct a table:
::构造表格 :

Painter	Rate of work (per hour)	Time worked	Part of Task
Suzie	$\begin{aligned} \frac{1}{3.5} = \frac{2}{7} \end{aligned}$	2	$\begin{aligned} \frac{4}{7} \end{aligned}$
Mike	$\begin{aligned} \frac{1}{t} \end{aligned}$	2	$\begin{aligned} \frac{2}{t} \end{aligned}$

Set up an equation:
::设置方程式 :

Since Suzie completed $\begin{aligned} \frac{4}{7} \end{aligned}$ of the lawn and Mike completed $\begin{aligned} \frac{2}{t} \end{aligned}$ of the lawn and together they mowed the lawn in 2 hours, we can write the equation: $\begin{aligned} \frac{4}{7} + \frac{2}{t} = 1 \end{aligned}$
::既然苏西完成了47个草坪,迈克完成了2个草坪, 并在2小时内一起修剪草坪, 我们可以写一个方程式: 47+2t=1

Solve the equation:
::解决方程式:

\begin{aligned} Find the lowest common denominator: & LCM = 7 t \\ Multiply all terms in the equation by the LCM: & 7 t \cdot \frac{4}{7} + 7 t \cdot \frac{2}{t} = 7 t \cdot 1 \\ Cancel common factors in each term: & t \cdot \frac{4}{1} + 7 \cdot \frac{2}{1} = 7 t \cdot 1 \\ Simplify: & 4 t + 14 = 7 t \\ 3 t = 14 \Rightarrow t = \frac{14}{3} = 4 \frac{2}{3} h o u r s \end{aligned}

::查找最小公分母 : LCM= 7tBlut 方程式中的所有条件, 以 LCM : 7t47+7t2t= 7t1Cancel 共同系数 : t41+721= 7t1 简化: 4t+14= 7t3t=14} t=143=423 小时

Check: The answer is reasonable. We’d expect Mike to work slower.
::检查:答案是合理的。我们期望迈克的工作速度会放慢。

Review
::回顾

For 1-5, perform the indicated operation. Leave the denominator in factored form.
::对于1-5, 执行指定的操作, 将分母保留为系数形式。

$\begin{aligned} \frac{4 x}{x + 1} - \frac{2}{2 (x + 1)} \end{aligned}$
::4xx+1-222(x+1)
$\begin{aligned} \frac{10}{21} + \frac{9}{35} \end{aligned}$

$\begin{aligned} \frac{2 x}{x - 4} + \frac{x}{4 - x} \end{aligned}$
::2xx-4+x4-xxx

$\begin{aligned} \frac{5}{2 x + 3} - 3 \end{aligned}$
::52x+3-3-3

$\begin{aligned} \frac{5 x + 1}{x + 4} + 2 \end{aligned}$
::5x+1x+4+2

For 6-8, find the missing resistance.
::6 -8,找到缺失的抵抗力量

$\begin{aligned} R_{1} = 4, R_{2} = 6, R_{t o t} = ? \end{aligned}$
::R1=4,R2=6,Rtot=?

$\begin{aligned} R_{1} = 1, R_{2} = ?, R_{t o t} = \frac{2}{3} \end{aligned}$
::R1=1,R2=?

$\begin{aligned} R_{1} = ?, R_{2} = 12, R_{t o t} = \frac{36}{15} \end{aligned}$
::R1=? R2=12,Rtot=3615

Solve the following work problems.
::解决以下的工作问题。

Andrea can wash the windows on their house in 30 minutes and Jorge can wash the windows on their house in 40 minutes. How long will it take for them to wash the windows together?
::Andrea可以在30分钟内洗他们家的窗户,而Jorge可以在40分钟内洗他们家的窗户。

A pool can be filled by one pipe in 5 hours and by a different pipe in 7 hours. How long will it take the pool to fill using both pipes?
::水池可以在5小时内用一个管子填充,在7小时内用另一个管子填充。用两个管子填充水池需要多长时间?

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。

章节大纲

Applications of Adding and Subtracting Rational Expressions ::增加和减减逻辑表达式的应用

Real-World Application: Total Resistance ::现实世界应用:完全抗药性

Finding Unknown Numbers ::查找未知数字

Real-World Application: Work Problems ::现实世界应用:工作问题

Example ::示例示例示例示例

Example 1 ::例1

Review ::回顾

Review (Answers) ::回顾(答复)

Applications of Adding and Subtracting Rational Expressions
::增加和减减逻辑表达式的应用

Real-World Application: Total Resistance
::现实世界应用:完全抗药性

Finding Unknown Numbers
::查找未知数字

Real-World Application: Work Problems
::现实世界应用:工作问题

Example
::示例示例示例示例

Example 1
::例1

Review
::回顾

Review (Answers)
::回顾(答复)