2.3 解决双层两层

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  解决双层平方

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  Consider what happens when you get a cell phone bill that is more than you usually pay. You have a data plan that includes 2.5 gigabytes (GB) for $79.00 per month. For every 1 megabyte (MB) over that amount, the company charges an additional $3.00. (There are 1024 MB in 1 GB.) Last month, your cell phone bill was $115.00. How many megabytes of data did you use over the limit?
  ::想想当你收到超过通常支付额的手机账单时会发生什么。 您有一个数据计划, 包括每月279.00美元的2.5千兆字节(GB), 每1兆字节(MB), 公司要额外收费3.00美元。 上个月, 您的手机账单是115.00美元。 您在限额内使用了多少兆字节的数据?

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  There are two operations that we need to consider when trying to determine your total cell phone bill. We need to add the $79.00 you pay each month to the overage charges. The overage charges are $3.00 for every 1 MB—the operation between 3 and the number of megabytes is multiplication . How do we solve an equation with two operations in it? This is what we examine in this section. 
  ::在试图确定您的手机总帐单时,我们需要考虑两个操作。 我们需要在超龄收费中加上每月79美元。 超龄收费为每1 MB 3 和 兆字节之间的操作是乘法, 超龄收费为每1 MB 3 和 兆字节之间的操作是3. 00 。 我们如何用其中两个操作来解析等式 ? 这就是我们在本节中要检查的内容 。

  

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  Solving Two-Step Equations
  ::解决双层平方

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  A two-step equation is an algebraic equation with two operations in it that requires two steps to solve. To solve equations with multiple  operations, we need to first determine what order to do those operations in. Recall the that we use to evaluate an expression : " data-term="Parentheses" role="term" tabindex="0"> parentheses , exponents, multiplication and division , addition and subtraction . For example, if we wanted to evaluate  $\begin{array}{r}3x+2\end{array}$  for  $\begin{array}{r}x=5\end{array}$ , we would first multiply 5 by 3 to get 15. Then, we would add 15 and 2 to get 17. 
  ::两步方程是一个代数方程,其中有两个操作需要两个步骤才能解答。要用多个操作解析方程式,我们需要首先决定执行这些操作的顺序。 回顾我们用来评价表达式的表达式:括号、引言、乘法和除法、加法和减法。例如,如果我们想对 x=5 的 3x+2 值进行评估,我们先将5乘法乘法3,再乘法15,再加法15,再加法2,再加法17。

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  If we are given that $\begin{array}{r}3x+2=17\end{array}$  , how can we determine that ? We can think of this as unraveling the operations around  $\begin{array}{r}x.\end{array}$  We are going to perform the inverse of the last operation we did when evaluating first. The last operation we did was addition and its inverse is subtraction.
  ::如果给了我们 3x+2=17, 我们如何确定呢? 我们可以把它看作是在 x 周围拆解操作 。 我们将执行上次我们第一次评估时的反向操作 。 上次我们做的是附加操作, 其反面是减法 。

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  $$\begin{array}{r}3x+\cancel{2}=17\\ \underset{\_}{-\cancel{2}-2}\\ 3x=15\end{array}$$

  
  ::3x+2=17-2-2-2-2__3x=15

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  Now, we perform the inverse of the first operation we did when evaluating. That was multiplication and its inverse is division. 
  ::现在,我们执行我们第一次评估时做的操作的逆向。那是乘法,它的反向是分法。

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  $$\begin{array}{rl}\frac{\cancel{3}x}{\cancel{3}}& =\frac{15}{3}\\ x& =5\end{array}$$

  
  ::3x3=153x=5

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  In general, we want to perform addition or subtraction first, then perform multiplication or division. 
  ::一般说来,我们想先进行增减,然后进行乘法或除法。

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   by CK-12  provides an example of solving two-step equations.
  ::CK-12 提供了解决两步方程的例子。

  
   

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  Example 1
  ::例1

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  Solve $\begin{array}{r}4x-9=11.\end{array}$
  ::解决4x9=11

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  Solution:  We need to undo two operations. First, we do the inverse of subtraction and add 9 to both sides.
  ::解决方案: 我们需要撤销两个操作。 首先, 我们进行反减法, 并在两边加上 9 。

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  $$\begin{array}{rl}& 4x-\cancel{9}=11\\ & \underset{\_}{+\cancel{9}+9}& & \text{perform addition or subtraction}\\ & 4x=20\end{array}$$

  
  ::4 - 9=11 +9 +9 +9_ 性能增加或减 4x=20

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  The inverse of multiplication is division, so we divide both sides by 4 to solve for $\begin{array}{r}x\end{array}$ .
  ::乘法的反面是分裂, 因此我们把两边除以4 来解决x。

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  $$\begin{array}{rlrl}\frac{\cancel{4}x}{\cancel{4}}& =\frac{20}{4}& & \text{perform multiplication or division}\\ x& =5\end{array}$$

  
  ::4x4 = 204 功率乘数或分数x=5

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  Check your answer: $\begin{array}{r}4(5)-9=20-9=11.\end{array}$
  ::检查您的答案: 4(5)- 9=20- 9=11 。

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  Example 2
  ::例2

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  Let's consider the cell phone bill from the introduction. The bill  as an equation is  $\begin{array}{r}115=79+3m\end{array}$ , where  $\begin{array}{r}m\end{array}$  is the number of megabytes over the 2.5 GB limit. how many megabytes over the limit did you use?
  ::让我们来考虑一下导言中的手机帐单。 帐单的方程式是 115=79+3m, 其中 m 是超过 2.5 GB 限制的兆字节数 。 您使用多少兆字节 ?

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  Solution:  To determine how many megabytes over the limit you used, we solve for 
  ::解决方案: 要确定您使用的极限上方有多少兆字节, 我们解决

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  $\begin{array}{r}m.\end{array}$
  ::m. 小时

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  $$\begin{array}{r}\cancel{79}+3m=115\\ \underset{\_}{\cancel{-79}-79}& & \text{perform addition or subtraction}\\ 3m=36\\ \frac{\cancel{3}m}{\cancel{3}}=\frac{36}{3}& & \text{perform multiplication or division}\\ m=12\end{array}$$

  
  ::79+3m=115-79-79-79_工作效率增加或减值3m=363m3=363工作效率乘数或分数=12

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  You used 12 MB over your data limit for the month.
  ::您使用12 MB 超过月份的数据限值 。

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  Example 3
  ::例3

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  Solve $\begin{array}{r}\frac{3}{4}x-\frac{2}{3}=\frac{7}{3}\end{array}$ .
  ::解决34x-23=73

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  Solution: Even though this problem involves fractions, it is still solved the same way. 
  ::解决办法:尽管这个问题涉及分数,但还是以同样的方式解决。

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  $$\begin{array}{rl}& \frac{3}{4}x-\cancel{\frac{2}{3}}=\frac{7}{3}\\ & \underset{\_}{+\cancel{\frac{2}{3}}+\frac{2}{3}}& & \text{perform addition or subtraction}\\ & \frac{3}{4}x=\frac{9}{3}=3\\ & \frac{\cancel{4}}{\cancel{3}}\times \frac{\cancel{3}}{\cancel{4}}x=\cancel{3}\times \frac{4}{\cancel{3}}& & \text{perform multiplication or division}\\ & x=4\end{array}$$

  
  ::34 - 23=73 +23+23_ 完好增加或减34x=93=3 43x34x=3x3x3x43 完美乘数或分数x4

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  Check your answer: $\begin{array}{r}\frac{3}{4}(4)-\frac{2}{3}=3-\frac{2}{3}=\frac{9}{3}-\frac{2}{3}=\frac{7}{3}\end{array}$ .
  ::请检查您的回答: 34(4)-23=3-23=93-23=73。

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  See the Explore More section below for an alternative method to solve this problem.
  ::关于解决这一问题的替代方法,见下文“探索更多”一节。

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  Example 4
  ::例4

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  Solve  $\begin{array}{r}18=6-\frac{2}{5}y.\end{array}$
  ::解决18=6 -25y

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  Solution:  Even though the term with the variable is not first, we can still follow the same steps. 
  ::解决方案:即使变量的术语不是首先使用,我们仍然可以采取同样的步骤。

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  $$\begin{array}{rl}& 18=\cancel{6}-\frac{2}{5}y\\ & \underset{\_}{-6-\cancel{6}}\\ & 12=-\frac{2}{5}y\\ & -\frac{5}{2}\cdot 12=-\frac{\cancel{5}}{\cancel{2}}\cdot -\frac{\cancel{2}}{\cancel{5}}y\\ & -30=y\end{array}$$

  
  ::18=6-25y-6-6-6_1225y-52125225y-30=y

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  Check your answer: $\begin{array}{r}6-\frac{2}{5}(-30)=6+12=18\end{array}$
  ::检查您的答案: 6- 25(- 30)=6+12=18

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  by Mathispower4u demonstrates how to solve two-step equations that include fractions. 
  ::通过 Mathispower4u 演示如何解析包含分数的两步方程 。

   

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  Example 5
  ::例5

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  Solve  $\begin{array}{r}-\frac{t}{6}-23=\mathrm{42.7.}\end{array}$
  ::解决 - T6 - 23=42. 7。

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  Solution:  

  $$\begin{array}{r}-\frac{t}{6}-\cancel{23}=42.7\\ \underset{\_}{+\cancel{23}+23.0}\\ -\frac{t}{6}=65.7\\ -\cancel{6}\times -\frac{t}{\cancel{6}}=-6\times 65.7\\ t=-394.2\end{array}$$

  Check:  $\begin{array}{r}-\frac{-394.2}{6}-23=65.7-23=\mathrm{42.7.}\end{array}$
  ::解答:-t6-23=42.7+23+23.0t6=65.7-6666×65.7t394.2检查:394.26-23=65.7-23=42.7。

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  E xample 6
  ::例6

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  Solve  $\begin{array}{r}\frac{v}{9}+\sqrt{2}=5\sqrt{2}.\end{array}$
  ::解决 v9+2=52。

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  Solution:  

  $$\begin{array}{r}\frac{v}{9}+\cancel{\sqrt{2}}=5\sqrt{2}\\ \underset{\_}{-\cancel{\sqrt{2}}-\sqrt{2}}\\ \frac{v}{9}=4\sqrt{2}\\ \cancel{9}\times \frac{v}{\cancel{9}}=9\times 4\sqrt{2}\\ v=36\sqrt{2}\end{array}$$

  Check:  $\begin{array}{r}\frac{36\sqrt{2}}{9}+\sqrt{2}=4\sqrt{2}+\sqrt{2}=5\sqrt{2}.\end{array}$
  ::解答:v9+2=52-2-2_v9=429xv9=9x42v=362检查:3629+2=42+2=52。

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  While with linear equations in the form ax+b=c you should perform addition or subtraction first and then multiplication or division, you may not always have a linear equation in this form. Consider the following example.
  ::在以 Ax+b=c 形式出现的线性方程式中,您应该先执行增减,然后乘法或除法,但您不一定总有这种形式的线性方程式。请参考以下示例。

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  Example 7
  ::例7

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  Solve  $\begin{array}{r}\frac{x-2}{3}=5.\end{array}$
  ::解决 x- 23=5 。

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  Solution:  This is a linear equation. We can see this if we divide 3 into each of the terms in the numerator: $\begin{array}{r}\frac{1}{3}x-\frac{2}{3}=5\end{array}$ . However, instead of dividing 3 into each of the terms, it is helpful to multiply both sides of the equation by 3 to undo the division by 3 in the denominator. Next, we add 2 to both sides to undo the subtraction of 2.
  ::解决方案 : 这是一个线性方程式 。 如果我们在分子数中将 3 分为 3 : 13x- 23 = 5 。 我们可以看到这一点。 然而, 与其将 3 分为 3 = 5 , 不如将方程式的两边乘以 3 , 使分母的除法除以 3 。 其次, 我们向 双方增加 2 , 来抵消 2 的减法 。

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  $$\begin{array}{r}\frac{x-2}{3}=5\\ \\ \cancel{3}(\frac{x-2}{\cancel{3}})=3(5)\\ \\ x-2=15\\ \\ x\cancel{-2}=15\\ \underset{\_}{\cancel{+2}+2}\\ \\ x=17\end{array}$$

  Notice  we get the same answer if we proceed in the same way as we have done in this section.
  ::x-23=53(x-23)=3(5)x-2=15x-2=15+2+2+2_x=17 注意,如果我们按照本节的做法行事,我们会得到相同的答案。

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  $$\begin{array}{r}\frac{1}{3}x-\frac{2}{3}=5\\ \underset{\_}{+\frac{2}{3}+\frac{2}{3}}\\ \frac{1}{3}x=\frac{17}{3}\\ \cancel{3}\cdot \frac{1}{\cancel{3}}x=\frac{17}{\cancel{3}}\cdot \cancel{3}\\ x=17\end{array}$$

   
  ::13-23=5 +23+23_13x=1733/13x=1733/13x=173_3x=173_3x=177

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     WARNING
  ::警告

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  Do not multiply or divide only one term in $\begin{array}{r}ax\pm b.\end{array}$  The following is incorrect: 
  ::在 axb 中不乘或除以一个词。 以下不正确 :

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  $$\begin{array}{rlr}\frac{\cancel{2}x}{\cancel{2}}-3=\frac{8}{2}& & \text{OR}\cancel{2}\times \frac{1}{\cancel{2}}x-7=2\times 5\\ x-3=4& & x-7=10\end{array}$$

  
  ::2x2-3=82OR2x12x-7=2x5x-3=4x-7=10

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  Summary
  ::摘要

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  • In general, perform addition or subtraction first, then perform multiplication or division.
    ::一般说来,先执行增加或减法,然后执行乘法或除法。
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  • If the equation is in the form of a fraction equal to a number, multiply both sides by the denominator first.
    ::如果方程的形式是分数等于数,则双方首先乘以分母。

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  Review
  ::回顾

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  Solve the following equations and check your answer. Reduce any fractions.
  ::解决以下方程式并检查您的答案。 减少任何分数 。

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  1. $\begin{array}{r}2x-5=-17\end{array}$   
     ::2 - 517
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  2. $\begin{array}{r}-4b+3=-5\end{array}$  
     ::-4b+3+5
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  3. $\begin{array}{r}1.4p-5.6=2.8\end{array}$
     ::1.4p-5.6=2.8
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  4. $\begin{array}{r}\frac{3}{4}=\frac{1}{2}-\frac{1}{8}m\end{array}$
     ::34=12-18m
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  5. $\begin{array}{r}-1=\frac{s}{2}+6\end{array}$  
     ::-1=S2+6
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  6. $\begin{array}{r}-18=\frac{r}{-5}-3\end{array}$
     ::-18=r-5-3
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  7. $\begin{array}{r}\frac{4}{5}=-\frac{2}{5}+\frac{3}{2}x\end{array}$
     ::4525+32x
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  8. $\begin{array}{r}6-d=-22\end{array}$
     ::6-d22
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  9. True or False: 6 is a solution to  $\begin{array}{r}-\frac{5}{4}=\frac{1}{6}y+\frac{3}{12}\end{array}$
     ::真实或假: 6 是 ~ 54=16y+312 的解决方案
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  10. True or False: -6 is a solution to  $\begin{array}{r}14.4=-2.7n-1.8\end{array}$
      ::真实或假 : -6 是14.42. 7n- 1.8的解决方案

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  Explore More
  ::探索更多

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  1. An emergency plumber charges $65 as a call-out fee plus an additional $75 per hour. He arrives at a house at 9:30am and works to repair a water tank. If the total repair bill is $196.25, at what time was the repair completed?
  ::1. 紧急水管工收费65美元,每小时收费65美元,外加75美元,上午9时30分抵达一栋房屋,修理一个水箱,如果总修理费为196.25美元,修理何时完成?

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  2. The cost of a food processor is $52.74. If the sticker price is $49.99, what is the percent of sales tax?
  ::2. 食品加工员的费用为52.74美元,如果贴纸价格为49.99美元,销售税的百分率是多少?

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  3. If 20 less than t wo-thirds of the students in a college received some type of financial aid and 2,000 students received financial aid, how many students attend the college?
  ::3. 如果一个学院的20名不到三分之二的学生得到某种形式的财政援助,2 000名学生得到财政援助,有多少学生在学院上学?

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  4.  Three-fourths of the soccer team plus 3 more players pay in-state tuition. If 27 players pay in-state tuition, how many players are on the soccer team?
  ::4. 足球队四分之三加3个球员支付国家学费,如果27个球员支付国家学费,足球队有多少球员?

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  5. When dealing with fractions, one way to “get rid” of them is to multiply everything in the equation by the Least Common Denominator (LCD) for the entire equation. The LCD is the least common multiple of the denominators in the equation. Try this technique with the following equations. 
  ::5. 在处理分数时,一种 " 消除 " 的方法是将方程式中的所有部分由最小常见分数(LCD)乘以整个方程式中的所有部分,LCD是方程式中最小的分数的多个。用以下方程式尝试一下这一技术。

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  a.   $\begin{array}{r}\frac{3}{8}x-\frac{2}{5}=\frac{7}{4}\end{array}$ The LCD of 8, 5, and 4 is 40. $\begin{array}{r}40\left(\frac{3}{8}x-\frac{2}{5}=\frac{7}{4}\right)\end{array}$ . Distribute the 40 and solve.
  ::a. 38x-25=74 LCD为8、5和4,40(38x-25=74),40(38x-25=74)。

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  b.  $\begin{array}{r}\frac{10}{3}x+\frac{3}{4}=\frac{5}{2}\end{array}$
  ::b. 103x+34=52

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  Answers for Review and Explore More Problems
  ::回顾和探讨更多问题的答复

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  Please see the Appendix. 
  ::请参看附录。

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  PLIX
  ::PLIX

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  Try this interactive that reinforces the concepts explored in this section:
  ::尝试这一互动,强化本节所探讨的概念: