Section: 使用合理等量的应用 | 12.13 使用合理等等数的应用

Section outline

Applications Using Rational Equations
::使用合理等量的应用

A motion problem with no acceleration is described by the formula $\begin{aligned} d i s t a n c e = s p e e d \times t i m e \end{aligned}$ . Since speed is a measure of rate , the formula may also appear as $\begin{aligned} d i s t a n c e = r a t e \times t i m e \end{aligned}$ . These problems can involve the addition and subtraction of rational expressions.
::没有加速的动作问题由公式距离=速度x时间描述。由于速度是速度的尺度,公式也可以作为距离=ratex时间出现。这些问题可能涉及增加和减去合理表达方式。

Real-World Application: Canoeing
::Real- World 应用程序: 独木舟

Last weekend Nadia went canoeing on the Snake River. The current of the river is three miles per hour. It took Nadia the same amount of time to travel 12 miles downstream as it did to travel 3 miles upstream. Determine how fast Nadia’s canoe would travel in still water.
::上周末,Nadia在蛇河上划了独木舟。河水每小时3英里。 Nadia在下游行12英里的时间与在上游行驶3英里的时间相同。确定Nadia的独木舟在静水中行驶的速度。

Define variables:
::定义变量 :

Let $\begin{aligned} s = \end{aligned}$ speed (rate of travel) of the canoe in still water
::Lets=静水独木舟的速度(旅行速度)

Then, $\begin{aligned} s + 3 = \end{aligned}$ the speed of the canoe traveling downstream
::然后,S+3=下游独木舟的速度

$\begin{aligned} s - 3 = \end{aligned}$ the speed of the canoe traveling upstream
::s-3 = 向上游行驶的独木舟的速度

Construct a table:
::构造表格 :

Direction	Distance (miles)	Rate	Time
Downstream	12	$\begin{aligned} s + 3 \end{aligned}$	$\begin{aligned} t \end{aligned}$
Upstream	3	$\begin{aligned} s - 3 \end{aligned}$	$\begin{aligned} t \end{aligned}$

Write an equation :
::写一个方程式 :

Since $\begin{aligned} d i s t a n c e = r a t e \times t i m e \end{aligned}$ , we can say that $\begin{aligned} t i m e = \frac{d i s t a n c e}{r a t e} \end{aligned}$ .
::从距离=时间,我们可以说时间=距离。

\begin{aligned} The time to go downstream is: t = \frac{12}{s + 3} \\ The time to go upstream is: t = \frac{3}{s - 3} \\ Since the time it takes to go upstream and downstream are the same, we have: \frac{3}{s - 3} = \frac{12}{s + 3} \end{aligned}

::下游的时间是:t=12s+3 上上游的时间是:t=3s-3S-3 由于上上下游的时间相同,我们有:3s-3=12s+3

Solve the equation:
::解决方程式:

\begin{aligned} Cross-multiply: 3 (s + 3) = 12 (s - 3) \\ Simplify: 3 s + 9 = 12 s - 36 \\ Solve: s = 5 m i / h \end{aligned}

::跨倍数: 3(s+3)=12(s-3) 简化: 3s+9=12s- 36Solve:s=5 mi/h

Check: Upstream: $\begin{aligned} t = \frac{12}{8} = 1 \frac{1}{2} h o u r \end{aligned}$ ; downstream: $\begin{aligned} t = \frac{3}{2} = 1 \frac{1}{2} h o u r \end{aligned}$ . The answer checks out.
::检查: 上流: t=128=112小时; 下游: t=32=112小时。检查答案。

Real-World Application: Biking
::Real- World 应用程序: 赌博

Peter rides his bicycle. When he pedals uphill he averages a speed of eight miles per hour, when he pedals downhill he averages 14 miles per hour. If the total distance he travels is 40 miles and the total time he rides is four hours, how long did he ride at each speed?
::彼得骑着自行车。当他踏上山坡时,平均速度为每小时8英里,当他踏下山时,平均速度为每小时14英里。如果他总行驶距离为40英里,总行驶时间为4小时,他每次行驶时长多少?

Define variables:
::定义变量 :

Let $\begin{aligned} d = \end{aligned}$ distance Peter bikes uphill at 8 miles per hour.
::让彼得的自行车以每小时8英里的速度向上行驶

Construct a table:
::构造表格 :

Direction	Distance (miles)	Rate (mph)	Time (hours)
Uphill	$\begin{aligned} d \end{aligned}$	8	$\begin{aligned} t_{1} \end{aligned}$
Downhill	$\begin{aligned} 40 - d \end{aligned}$	14	$\begin{aligned} t_{2} \end{aligned}$

Write an equation:
::写一个方程式 :

We know that $\begin{aligned} t i m e = \frac{d i s t a n c e}{r a t e} \end{aligned}$ .
::我们知道时间是距离

\begin{aligned} The time to go uphill is: & t_{1} = \frac{d}{8} \\ The time to go downhill is: & t_{2} = \frac{40 - d}{14} \\ We also know that the total time is 4 hours: & \frac{d}{8} + \frac{40 - d}{14} = 4 \end{aligned}

::上山时间是:t1=d8 下山时间是:t2=40-d14 我们还知道总时间是4小时:d8+40-d14=4

Solve the equation:
::解决方程式:

\begin{aligned} Find the lowest common denominator: & LCD = 56 \\ Multiply all terms by the common denominator: & 7 d + 160 - 4 d = 224 \\ Solve: & d = 21.3 m i \end{aligned}

::查找最小公分母 : LCD=56Multiply 所有条件, 以公分母: 7d+160-4d=224Solve: d=21.3 MI

Check: Uphill: $\begin{aligned} t = \frac{21.3}{8} = 2.67 h o u r s \end{aligned}$ ; downhill: $\begin{aligned} t = \frac{40 - 21.3}{14} = 1.33 h o u r s \end{aligned}$ . The answer checks out.
::检查: 向上: t=21.38=2.67小时; 向下: t=40-21.314=1.33小时。检查答案是否正确。

Real-World Application: Combining Money
::真实世界应用程序:货币组合

A group of friends decided to combine their money and buy a birthday gift that cost $200. Later 12 of the friends decided not to participate any more. This meant that each person paid $15 more than their original share. How many people were in the group to begin with?
::一群朋友决定合并他们的钱,买一份价值200美元的生日礼物。后来有12位朋友决定不参加。这意味着每人的薪酬比原来多15美元。有多少人从这个团体开始?

Define variables:
::定义变量 :

Let $\begin{aligned} x = \end{aligned}$ the number of friends in the original group.
::让 x = 原始组中的朋友数。

Make a table:
::绘制表格:

	Number of people	Gift price	Share amount
Original group	$\begin{aligned} x \end{aligned}$	200	$\begin{aligned} \frac{200}{x} \end{aligned}$
Later group	$\begin{aligned} x - 12 \end{aligned}$	200	$\begin{aligned} \frac{200}{x - 12} \end{aligned}$

Write an equation:
::写一个方程式 :

Since each person’s share went up by $15 after 12 people refused to pay, we write the equation $\begin{aligned} \frac{200}{x - 12} = \frac{200}{x} + 15 \end{aligned}$
::由于12人拒绝支付后每人的份额增加了15美元,所以我们写了200x-12=200x+15的方程。

Solve the equation:
::解决方程式:

\begin{aligned} Find the lowest common denominator: & LCD = x (x - 12) \\ Multiply all terms by the LCD: & x (x - 12) \cdot \frac{200}{x - 12} = x (x - 12) \cdot \frac{200}{x} + x (x - 12) \cdot 15 \\ Cancel common factors and simplify: & 200 x = 200 (x - 12) + 15 x (x - 12) \\ Eliminate parentheses: & 200 x = 200 x - 2400 + 15 x^{2} - 180 x \\ Get all terms on one side of the equation: & 0 = 15 x^{2} = 180 x - 2400 \\ Divide all terms by 15 : & 0 = x^{2} - 12 x - 160 \\ Factor: & 0 = (x - 20) (x + 8) \\ Solve: & x = 20, x = - 8 \end{aligned}

::查找最小公分母 : LCD=x( x- 12) =x( x-12) =200x- 12( 200x- 12) - 12=x( x- 12) =12x- 200x+x( x- 12) 15Cancel 常见因数并简化 : 200x=200( x- 12) +15x( x- 12) +15x( x- 12) ELI 括号: 200x=200x- 2400+15x2- 180x- 180xGeg 方程式的一边的所有条件 : 0=15x2=180x- 2400Divide 全部条件 15: 0=x2 - 12x- 160Facor: 0=( x-20) ( x+8) Solve:x=20, x_ 8

The answer that makes sense is $\begin{aligned} x = 20 \end{aligned}$ people.
::合理的答案是x=20人。

Check: Originally $200 shared among 20 people is $10 each. After 12 people leave, $200 shared among 8 people is $25 each. So each person pays $15 more. The answer checks out.
::检查 : 20 人之间最初共有200 美元, 每人10 美元。 12 人离开后, 8 人之间共有200 美元, 每人25 美元。所以每人多付 15 美元。答案会查清楚。

Example
::示例示例示例示例

Example 1
::例1

Carrie is a runner. When she runs uphill she averages a speed of 2 miles per hour, when she runs downhill she averages 5 miles per hour. If she runs through the hilly streets of San Francisco for 4.5 hours and the total distance she travels is 13.5 miles, how long did she run uphill and how long did she run downhill?
::Carrie是跑步者。当她上山时,平均速度为每小时2英里,当她下山时,平均速度为每小时5英里。如果她穿过旧金山山坡街道走4.5小时,总距离为13.5英里,她上山时,平均速度为每小时2英里,每小时5英里。如果她穿过旧金山山坡街道,总距离是13.5英里,她往山上跑多久,她往山下跑多久?

Define variables:
::定义变量 :

Let $\begin{aligned} t_{1} = \end{aligned}$ time Carrie runs uphill.
::让T1=时间 Carrie往上走

Let $\begin{aligned} t_{2} = \end{aligned}$ time Carrie runs downhill.
::让T2=时间 Carrie往下走

Let $\begin{aligned} d = \end{aligned}$ the distance Carrie runs uphill.
::让我们让卡莉的距离向上移动。

Construct a table:
::构造表格 :

Direction	Distance (miles)	Rate (mph)	Time (hours)
Uphill	$\begin{aligned} d \end{aligned}$	2	$\begin{aligned} t_{1} \end{aligned}$
Downhill	$\begin{aligned} 13.5 - d \end{aligned}$	5	$\begin{aligned} t_{2} \end{aligned}$

Write an equation:
::写一个方程式 :

We know that $\begin{aligned} t i m e = \frac{d i s t a n c e}{r a t e} \end{aligned}$ .
::我们知道时间是距离

\begin{aligned} The time to go uphill is: t_{1} = \frac{d}{2} \\ The time to go downhill is: t_{2} = \frac{13.5 - d}{5} \\ We also know that the total time is 4.5 hours: \frac{d}{2} + \frac{13.5 - d}{5} = 4.5 \end{aligned}

::上山时间是:t1=d2 下山时间是:t2=13.5-d5,我们也知道总时间是4.5小时:d2+13.5-d5=4.5。

Solve the equation:
::解决方程式:

\begin{aligned} Find the lowest common denominator: LCD = 2 \cdot 5 = 10 \\ Multiply all terms by the common denominator: 5 d + 27 - 2 d = 45 \\ Solve: d = 6 m i \end{aligned}

::查找最小公分母 : LCD=25=10Multiply 以公分母: 5d+27-2- 2d=45Solve: d=6 MI

Since $\begin{aligned} d = 6 \end{aligned}$ is the distance Carrie to ran uphill, then $\begin{aligned} 13.5 - d = 13.5 - 6 = 7.5 \end{aligned}$ is the distance Carrie ran downhill.
::由于d=6是Carrie上山的距离,那么13.5-d=13.5-6=7.5是Carrie下山的距离。

Check: Uphill: $\begin{aligned} t = \frac{6}{2} = 3 h o u r s \end{aligned}$ ; downhill: $\begin{aligned} t = \frac{13.5 - 6}{5} = 1.5 h o u r s \end{aligned}$ . The answer checks out.
::检查: 向上: t=62=3小时; 向下: t=13.5- 65=1.5小时。答案检查出来。

Review
::回顾

For 1-4, solve for $\begin{aligned} x \end{aligned}$ .
::1 -4 解答为 x 。

$\begin{aligned} \frac{3 x^{2} + 2 x - 1}{x^{2} - 1} = - 2 \end{aligned}$
::3x2+2x-1x2 - 1x2 - 12

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

$\begin{aligned} 3 + \frac{1}{x + 1} = \frac{2}{x} \end{aligned}$
::3+1x+1=2x

$\begin{aligned} \frac{1}{x} - \frac{x}{x - 2} = 2 \end{aligned}$
::1x-xx-2=2

For 5-10, solve the following applications.
::5-10, 解决以下应用程序。

Juan jogs a certain distance and then walks a certain distance. When he jogs he averages 7 miles/hour and when he walks he averages 3.5 miles per hour. If he walks and jogs a total of 6 miles in a total of 1.2 hours, how far does he jog and how far does he walk?
::胡安慢跑有一定的距离,然后走一定的距离。当他慢跑时,平均每小时7英里,行走时平均每小时3.5英里。如果他行走和慢跑总共6英里,总共1.2小时,他跑多远,行走多远?

A boat travels 60 miles downstream in the same time as it takes it to travel 40 miles upstream. The boat’s speed in still water is 20 miles per hour. Find the speed of the water.
::一艘船在下游60英里的时间内,在上游40英里的同一时间行驶在下游60英里。船在静水中的速度是每小时20英里。寻找水的速度。

Paul leaves San Diego driving at 50 miles per hour. Two hours later, his mother realizes that he forgot something and drives in the same direction at 70 miles per hour. How long does it take her to catch up to Paul?
::两小时后,他的母亲意识到他忘记了什么东西,开往同一方向,每小时70英里。她要多久才能赶上保罗?

On a trip, an airplane flies at a steady speed against the wind and on the return trip the airplane flies with the wind. The airplane takes the same amount of time to fly 300 miles against the wind as it takes to fly 420 miles with the wind. The wind is blowing at 30 miles per hour. What is the speed of the airplane when there is no wind?
::在一次旅行中,一架飞机以稳定的速度飞向风,在回程中,飞机随风飞行。飞机需要同样时间飞行300英里,与风飞行420英里相同。风以每小时30英里的速度吹。没有风,飞机的速度如何?

A debt of $420 is shared equally by a group of friends. When five of the friends decide not to pay, the share of the other friends goes up by $25. How many friends were in the group originally?
::420美元的债务由一群朋友平分。当五个朋友决定不支付时,其他朋友的份额增加了25美元。最初有多少朋友在小组中?

A non-profit organization collected $2250 in equal donations from their members to share the cost of improving a park. If there were thirty more members, then each member could contribute $20 less. How many members does this organization have?
::一个非营利组织从其成员收取了2 250美元的同等捐款,以分担改善公园的费用。如果还有30名成员,那么每个成员可以少捐款20美元。这个组织有多少成员?

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

Applications Using Rational Equations ::使用合理等量的应用

Real-World Application: Canoeing ::Real- World 应用程序: 独木舟

Real-World Application: Biking ::Real- World 应用程序: 赌博

Real-World Application: Combining Money ::真实世界应用程序:货币组合

Example ::示例示例示例示例

Example 1 ::例1

Review ::回顾

Review (Answers) ::回顾(答复)

Applications Using Rational Equations
::使用合理等量的应用

Real-World Application: Canoeing
::Real- World 应用程序: 独木舟

Real-World Application: Biking
::Real- World 应用程序: 赌博

Real-World Application: Combining Money
::真实世界应用程序:货币组合

Example
::示例示例示例示例

Example 1
::例1

Review
::回顾

Review (Answers)
::回顾(答复)