节: 解决合理等式 | 6.7 理衡等同-interactive

章节大纲

Lesson Objectives
::经验教训目标

Solve simple rational equations in one variable .
::在一个变量中解决简单的理性方程式。

Identify extraneous solutions.
::找出不相干的解决办法。

F ind the solutions using technology to graph the function .
::使用技术来绘制函数图解,找到解决方案。

Introduction: Going Swimming
::导言:游泳

It's almost summertime and Natasha is trying to fill her 6,000-gallon pool. She wants to use the hose , but it will take 16 hours to fill the pool. Feeling that this would take too long, Natasha also bought a water pump that can fill the pool in 6 hours. How long would it take if Natasha used the hose and the pump at the same time to fill up the pool?
::现在几乎是夏天了,娜塔莎正试图填充她的6000加仑游泳池。她想用水管,但是要花16小时才能填满水池。觉得这需要时间太长,娜塔莎还买了一个水泵,可以在6小时内填满水池。如果娜塔莎同时使用水管和水泵来填满水池,需要多长时间?

Activity 1: Solving Rational Equations Using Proportions
::活动1:利用比例解决合理等式

While there are many approaches to the problem above, this problem can be modeled using a rational equation . One approach to is through cross multiplication .
::虽然对上述问题有许多办法,但可以用一个合理的等式来模拟这一问题,一种办法是通过相互乘法。

Example
::示例示例示例示例

Solve for $\begin{aligned} x \end{aligned}$ in the equation $\begin{aligned} \frac{x}{2 x - 3} = \frac{3 x}{x + 11} \end{aligned}$
::在方程 x2x- 3=3xx+11 中为 xSolve for ex

You can use cross multiplication to solve a rational proportion . Cross multiplication refers to multiplying the numerator of one fraction by the denominator of the other across an equals sign. The image below demonstrates the process by color. The two terms in red are multiplied, as are the two terms in blue.
::您可以使用交叉乘法来解析一个合理比例。交叉乘法是指将一个分数的分子乘以一个等号的分母。下面的图像用颜色来显示过程。红色的两个词是乘法, 蓝色的两个词是乘法。

\begin{aligned} \frac{x}{2 x - 3} = \frac{3 x}{x + 11} \end{aligned}

::x2x-3=3xx+11

The result is an equation with no fractional terms:
::结果是一个没有分数术语的方程式 :

\begin{aligned} (x) (x + 11) = (2 x - 3) (3 x) \end{aligned}

x)(x+11)=(2x-3)(3x)

Once the new equation is set up, you can solve for $\begin{aligned} x . \end{aligned}$
::设置新方程式后,您可以解答 x。

\begin{aligned} (x) (x + 11) & = (2 x - 3) (3 x) \\ x^{2} + 11 x & = 6 x^{2} - 9 x \\ - x^{2} & - x^{2} \\ 11 x & = 5 x^{2} - 9 x \\ - 11 x & - 11 x \\ 0 & = 5 x^{2} - 20 x \end{aligned}

x)(xx+11) = (2x-3)(3x) = (2x-3)(3x) x2+11= 6x2-9x2-9x2-x2-x2x-211x=5x2-9x-9x-11x-11x-11xx0=5x2-20x)

Cross multiplication will result in the quadratic equation above which can be solved using factoring.
::交叉乘法将产生上述可使用乘数解析的二次方程。

\begin{aligned} 0 & = 5 x^{2} - 20 x \\ 0 & = 5 x (x - 4) \end{aligned}

::0=5x2 - 20x0=5x(x- 4)

T he factored quadratic will be used to find the solutions .
::系数二次曲线将用来寻找解决方案。

\begin{aligned} 5 x & = 0 \\ \div 5 & \div 5 \\ x & = 0 \end{aligned}

\begin{aligned} x - 4 & = 0 \\ + 4 & + 4 \\ x & = 4 \end{aligned}

Answer: 0 and 4
::答复:0和4

To make sense of this solution, look at it on a graph. Recall from Graphing Functions that any equation can be solved through finding the intersections of the functions determined by the left and right side of the equations. In this case, let $\begin{aligned} f (x) = \frac{x}{2 x - 3} \end{aligned}$ and $\begin{aligned} g (x) = \frac{3 x}{x + 11} . \end{aligned}$
::要理解这个解决方案, 请在图表中查看它。从图形函数中回顾, 任何方程式都可以通过找到由方程式左侧和右侧决定的函数的交叉点来解决。在此情况下, 请让 f( x) =x2x- 3 和 g( x) = 3xx+11 。

INTERACTIVE

Intersection of Two Rational Functions

Click the "+" or "-" buttons to zoom in or out.
::单击“+”或“-”按钮来缩放或剪切。

Then click on the red points to check the intersections of the functions.
::然后单击红点以检查函数的交叉点。

Activity 2: Going Swimming Continued
::活动2:继续游泳

The question posed in the introduction can be solved using the following rational equation where $\begin{aligned} t \end{aligned}$ is the time, in hours:
::在导言中提出的问题可以用以下合理公式解决,如果时间t为时,用小时:

\begin{aligned} \frac{1}{16} + \frac{1}{6} = \frac{1}{t} \end{aligned}

::116+16=1t

Each fraction represents the fraction of the task that can be completed in one hour.
::每个分数代表一个小时后完成的任务的分数。

Using the hose, $\begin{aligned} \frac{1}{16} \end{aligned}$ of the pool will be full after an hour because it takes 16 hours to fill the entire pool.
::用水管,116个游泳池一小时后会满, 因为它需要16小时才能填满整个游泳池。

Using the pump, $\begin{aligned} \frac{1}{6} \end{aligned}$ of the pool will be full after an hour because it takes 6 hours to fill the entire pool with the pump.
::使用泵,16个游泳池在一小时后会满满,因为用泵填满整个游泳池需要6个小时。

Using both hoses combined $\begin{aligned} \frac{1}{t} \end{aligned}$ of the pool will be full after an hour, assuming that it takes t hours to fill the entire pool.
::使用两个水龙头组合的1吨水龙头,一个小时后就会满满,假设整个水龙头需要2小时才能填满。

This expression $\begin{aligned} \frac{1}{16} + \frac{1}{6} \end{aligned}$ is equal to the amount the hose and the pump added together.
::此 116+16 表达式等于软管和泵加在一起的量。

There are several ways to solve a rational expression with three or more terms. One way is to combine any rational expressions on the same side using addition or subtraction , as defined by the problem. This approach will result in a rational proportion, which you already know how to solve.
::有几种方法可以用三个或三个以上的术语来解决理性表达方式。一种方法是使用问题定义的加法或减法将同一侧的任何合理表达方式结合起来。这种方法将产生合理的比例, 您已经知道如何解决。

\begin{aligned} \frac{1}{16} + \frac{1}{6} & = \frac{1}{t} \\ \frac{1}{16} \cdot \frac{3}{3} + \frac{1}{6} \cdot \frac{8}{8} & = \frac{1}{t} \\ \frac{3}{48} + \frac{8}{48} & = \frac{1}{t} \\ \frac{11}{48} & = \frac{1}{t} \end{aligned}

::116+16=1t11633+1688=1t348+848=1t1148=1t

Cross multiplying will result in the following:
::交叉乘法将导致以下结果:

\begin{aligned} \frac{11}{48} & = \frac{1}{t} \\ 11 t & = 48 \\ \div 11 & \div 11 \\ t & = \frac{48}{11} = 4. \bar{36} \end{aligned}

::1148=1t11t=48*11*11*11t=4811=4.36

Answer: It will take approximately 4.36 hours to fill the pool using both the hose and the pump.
::回答:用水管和水泵填充水池大约需要4.36小时。

Discussion Question : In the section Adding and Subtracting Rational Equations, you learned that the harmonic mean of two numbers, $\begin{aligned} a \end{aligned}$ and $\begin{aligned} b, \end{aligned}$ can be found using the formula $\begin{aligned} \frac{2}{\frac{1}{a} + \frac{1}{b}} \end{aligned}$ which simplifies to $\begin{aligned} \frac{2 a b}{a + b} . \end{aligned}$ Find the harmonic mean of the time it takes the hose and the pump to fill the pool and divide it by 2. Is the answer a coincidence? How can you explain it?
::讨论问题:在“增加和减少理性等式”一节中,你了解到两个数字(a和b)的调和平均值可以使用公式21a+1b找到,该公式简化为 2aba+b。找到用软管和泵填充水池和除以2的时间的调和平均值。答案是巧合吗?如何解释呢?

Here is the cross multiplication strategy applied to a more challenging problem:
::以下是适用于一个更具挑战性的问题的相互乘法:

Example
::示例示例示例示例

Solve to find $\begin{aligned} x \end{aligned}$ in the equation $\begin{aligned} \frac{2 x}{x - 3} = 2 + \frac{3 x}{x^{2} - 9} . \end{aligned}$
::在方程 2xx-3=2+3x22-9中找到 x。

To begin, add the two rational expressions on the right side of the equation.
::首先,在方程式右侧添加两个合理表达式。

\begin{aligned} 2 + \frac{3 x}{x^{2} - 9} & = \frac{2}{1} \cdot \frac{x^{2} - 9}{x^{2} - 9} + \frac{3 x}{x^{2} - 9} \\ = \frac{2 x^{2} - 18}{x^{2} - 9} + \frac{3 x}{x^{2} - 9} \\ = \frac{2 x^{2} + 3 x - 18}{x^{2} - 9} \end{aligned}

::2+3xx2-9=21x2-9x2-9x2-9+3x2-2-9=2x2-18x2-9+3x2-9=2x2-9=2x2-3x2-9=2x2-3x3x-18x2-9

The next step will be to cross multiply.
::下一步是交叉乘数。

$\begin{aligned} \frac{2 x}{x - 3} & = \frac{2 x^{2} + 3 x - 18}{x^{2} - 9} \\ 2 x (x^{2} - 9) & = (x - 3) (2 x^{2} + 3 x - 18) \\ 2 x^{3} - 18 x & = 2 x^{3} - 3 x^{2} - 27 x + 54 \\ 2 x^{3} - 18 x & = 2 x^{3} - 3 x^{2} - 27 x + 54 \\ + 18 x & + 18 x \\ 0 & = - 3 x^{2} - 9 x + 54 \\ 0 & = - 3 (x^{2} + 3 x - 18) \\ 0 & = - 3 (x + 6) (x - 3) \end{aligned}$

::2x3=2x2x2+3x3x18x2-92x(x2-9)=(x3)(2x2+3x-18)2x3-18x2x2x2x3-3x3x3-27x3x3-3x2x3-18x2-27x3x3-3x3x3-3x3x3-3x2-27x5x5x3-3x4+18x18x0*3x9x540}(x2+3x-180)3x2x3x3x6x(x-3)

The answers appear to be -6 and 3, but you can verify this by substituting -6 and 3 into the original equation.
::答案似乎是 -6和3 但你可以用原来的方程式替换 -6和3来验证

\begin{aligned} \frac{2 (- 6)}{(- 6) - 3} & = 2 + \frac{3 (- 6)}{(- 6)^{2} - 9} \\ \frac{- 12}{- 9} & = 2 + \frac{- 18}{27} \\ \frac{4}{3} & = 2 + \frac{- 2}{3} \\ \frac{4}{3} & = \frac{6}{3} + \frac{- 2}{3} \\ \frac{4}{3} & = \frac{4}{3} \end{aligned}

\begin{aligned} \frac{2 (3)}{(3) - 3} & = 2 + \frac{3 (3)}{(3)^{2} - 9} \\ \frac{6}{0} & = 2 + \frac{9}{0} \end{aligned}

O nly -6 is a solution to the equation, t he value 3 is an extraneous solution . An extraneous solution will satisfy the equation when cross multiplied and written without a denominator, but will involve dividing by zero in the equation's original form.
::只有 - 6 是方程的解决方案, 值 3 是一个不相干的解决办法。当交叉乘数和没有分母的书写时, 一个不相干的解决办法将满足方程, 但是在方程的原始形式中将包含以零除法。

Answer: $\begin{aligned} - 6 \end{aligned}$
::答复:-6

Activity 3: Solving Rational Equations By Canceling
::活动3:通过取消解决合理等等式

Another approach to solving a rational equation is to cancel the fractions. Th e fractions in a polynomial equation are canceled by multiplying every term with the least common denominator. T his same approach can be used to cancel the fractions in a rational equation.
::解决合理等式的另一种方法就是取消分数。多式方程式中的分数通过以最小的分母乘以每个术语来取消。同样的方法也可以用来在合理等式中取消分数。

INTERACTIVE

Solving Rational Equations

Click the button to see more examples of how to begin solving rational equations by finding their LCD.
::单击该按钮以查看更多实例,说明如何通过找到它们的液晶来开始解析理性方程式。

Example

::示例示例示例示例

Solve the rational equation for $\begin{aligned} x . \end{aligned}$
::解析 x 的合理方程式。

\begin{aligned} \frac{3}{x^{2} + 4 x + 4} + \frac{1}{x + 2} = \frac{2}{x^{2} - 4} \end{aligned}

::3x2+4x+4+4+1x+2=2x2-2-4

The LCD is $\begin{aligned} (x + 2) (x + 2) (x - 2) \end{aligned}$ . When multiplying each term by the LCD, any terms in the denominator of each rational expression will be canceled.
::LCD 是 (x+2)(x+2)(x-2)(x-2) 。当将每个词乘以 LCD 时, 每个理性表达式的分母中的任何词将被取消。

\begin{aligned} \frac{3}{x^{2} + 4 x + 4} + \frac{1}{x + 2} & = \frac{2}{x^{2} - 4} \\ (x + 2) (x + 2) (x - 2) \cdot \frac{3}{(x + 2) (x + 2)} + (x + 2) (x + 2) (x - 2) \cdot \frac{1}{x + 2} & = (x + 2) (x + 2) (x - 2) \cdot \frac{2}{(x - 2) (x + 2)} \\ 3 (x - 2) + (x - 2) (x + 2) & = 2 (x + 2) \\ 3 x - 6 + x^{2} - 4 & = 2 x + 4 \\ x^{2} + x - 14 & = 0 \end{aligned}

::3x2+4x+4+4+4+4+1x+2+2=2x2x2-2(x+2)(x-2)(x-2)(x+2)(x+2)(x+2)(x+2)(x+2)(x-2)(x+2)(x+2)(x-2)(x-2)(x-2)(x-2)(x-2)(x-2)(x-2)(x-2)(x-2)(x-2)(x-2)(x-2)(x+2)(x)(x2)(x+2)(x+2)(x+2)(x+2)(x+2(x)(x-2)(x-2)(x-2)(x-2)(x+2)(x+2)(x-2)(x+2)(x-2)(x+2)(x-2)(x-2)(x-2)(x-2)(x-2)(x-2)(x-2)(x-2)(x+2)(x-2)(x+2)(x+2)(x-2)(x+2)(x+2)(x+2)(x-2)(x+2)(x+2)(x)3x-2)(x+2)(x+2)(x+2)(x-2)(x+2)(x-2)(x+2)(x+2)(x+2)(x+2)(x+2)(x+2)(x+2)(x+2)(x+2)(x+2)(x+2)(x+2)(x+2)(x+2)(x+2)(x+2)(x+2)(x)3+2)(x+2)(x+2)(x+2)(x+2)(x+2)(x+2)(x)3)(x)3)(x+2)3x+2)(x+2)(x+2)3)(3)3)(3x+2)(3)(3)(3)(3)(x)3x)3x)(3x-2)(3)(3x-2)(3x)3x-2)(3x-2)(3x-2)(3)(x-2)(2)3)(3)(3)(3)(3)(3)(3)(3x-2)(3)(3)(3)(3)(3)(3)(3)(2)(2)(2)(2)(2)(2)(2)(2)(2)(x-2)(2)(2)(2)(2)(2)(x

This quadratic is not factorable, so you need to use the Quadratic Formula to solve for $\begin{aligned} x \end{aligned}$ .
::此二次方位是不可因数的, 所以您需要使用二次方位公式来解析 x 。

\begin{aligned} x = \frac{- (1) \pm \sqrt{(1)^{2} - 4 (1) (- 14)}}{2 (1)} = \frac{- 1 \pm \sqrt{57}}{2} \end{aligned}

::* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *

Answer: $\begin{aligned} x \approx 3.27 and - 4.27 \end{aligned}$
::答复:x_______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Use the interactive below to check both of these values to determine if they are extraneous.
::使用下面的交互效果来检查这两个值,以确定它们是否不相干。

INTERACTIVE

Solving Rational Equations By Canceling

Use the interactive to check the intersections of the two rational functions.

::使用交互功能检查两个理性函数的交叉点。

Click the "+" button to zoom in.
::单击“ +” 按钮以缩放。

Click the "-" button to zoom out.
::单击“ - ” 按钮缩放。

Click "Show Solution" to see the intersection.
::点击“ 显示解决方案” 查看十字路口。

Summary

To solve rational equations with proportions using cross multiplication:
- Multiply the numerator of one fraction by the denominator of the other across the equals sign. Do this for both sides.
  ::将一个分数的分子乘以对等符号之间的分母。为两边都这样做。
- Solve for x.
  ::解决x。
::使用交叉乘法用比例来解析合理方程式:将一个分数的分子乘以对等符号的分母。对两边都这样做。为 x 解决。

To solve rational equations using canceling:
- Find the LCD of all the rational expressions.
  ::找到所有理性表达方式的解码。
- Multiply every rational expression by the LCD, which will cancel the denominators.
  ::乘以液晶显示器的每个合理表达式, 这将取消分母。
- Solve for x.
  ::解决x。
::使用取消来解析理性方程式 : 查找所有理性表达式的液晶体。用液晶体乘以每个合理表达式, 它将取消分母。解决 x 。

章节大纲

Introduction: Going Swimming ::导言:游泳

Activity 1: Solving Rational Equations Using Proportions ::活动1:利用比例解决合理等式

Activity 2: Going Swimming Continued ::活动2:继续游泳

Activity 3: Solving Rational Equations By Canceling ::活动3:通过取消解决合理等等式

Wrap-Up: Review Questions ::总结:审查问题

Introduction: Going Swimming
::导言:游泳

Activity 1: Solving Rational Equations Using Proportions
::活动1:利用比例解决合理等式

Activity 2: Going Swimming Continued
::活动2:继续游泳

Activity 3: Solving Rational Equations By Canceling
::活动3:通过取消解决合理等等式

Wrap-Up: Review Questions
::总结:审查问题