3.3 有理数的线性公式

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  有理数的线性线性公式

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  Using Decimals and Fractions in Equations 
  ::在等量中使用十进数和分数

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  Jenny Chang is the owner of a small shop, Empire Fudge, that sells one product: fudge. She is looking to lower her overall expenses by 5%. To do this, she plans to decrease her total material cost from $3.10 per pound to $2.83 per pound. The material cost is the cost of the ingredients that she uses to make one pound of fudge. While you possess the necessary skill to figure this out, in this chapter you are going to learn strategies to make solving problems involving decimals and fractions easier. This comes in especially handy when solving equations that involve money.
  ::Jenny Chang是一家小商店的老板,它卖的是一种产品:软糖。她希望将其总开支降低5%。为了做到这一点,她计划将其总材料成本从每磅3.10美元降低到每磅2.83美元。物质成本是她用来制造一磅软糖的原料成本。尽管你掌握了必要的技能来解决这个问题,但在本章中,你将学习如何使小数和分数问题更容易解决的战略。在解决涉及金钱的方程式时,这尤其有用。

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  Rational numbers are numbers that can be expressed as a fraction or the quotient of two integers . Examples of rational numbers are $\begin{array}{r}5,-16,\frac{1}{4},-9.12,33\frac{5}{6},\end{array}$  etc. When dealing with equations in business, all types of rational numbers are common. Money is often expressed rounded to the hundredths place to represents cents and percentages are often written as decimals or fractions. In this section, you will learn strategies for handling decimals and fractions in equations. 
  ::理性数字是指可以以小数或两个整数的商数表示的数字,理性数字的例子为5-16、14-9.12、3356等。在处理商业方程式时,所有类型的合理数字都是常见的。货币通常四舍五入到百位表示美分,百分比通常以小数或小数表示。在本节中,您将学习处理方程式中小数和小数的战略。

   

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  Clearing Decimals
  ::清结十进数

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  The idea behind  the strategy for dealing with fractions and decimals comes from the rule you learned in the previous section:
  ::处理分数和小数点的战略背后的想法来自你在上一节学到的规则:

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  Any operation done to one side of an equation must be done to the other side.
  ::对等方程式的一面进行的任何操作都必须对另一面进行。

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  Let’s look at an example of this. Below are two equations, every term in equation 1 was multiplied by 5 to get equation 2.
  ::让我们来看看这个例子。 下面是两个方程, 方程1中每个术语乘以5, 方程2。

  Equation 1	Equation 2
  播放段落 $\begin{array}{rl}x-3& =7\\ x& =10\end{array}$ ::x-3=7x=10 x-3=10	$\begin{array}{rl}5(x-3)& =5(7)\\ 5\cdot x-3\cdot 5& =5\cdot 7\\ 5x-15& =35\\ 5x& =50\\ x& =10\end{array}$

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  Both answers come out to 10. If you multiply every term in an equation by the same number, the answer to the equation will remain the same.  You can now use this idea to eliminate decimals and fractions.
  ::两个答案都显示为 10 。 如果您将方程式中的每个术语乘以相同数字, 方程式的答案将保持不变 。 您现在可以使用这个概念来消除小数数和分数 。

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  To clear a decimal you must multiply the number with the most decimal places by a power of 10 which will result in an integer. The powers of 10 are 10, 100, 1000, etc.
  ::要清除小数点, 您必须将数字乘以最小数位乘以10的功率, 得出整数。 10的功率是 10, 100, 1000 等 。

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  Use the interactive below to determine what the smallest power of ten that will give us a whole number is.
  ::使用下面的交互式来决定 10的最小功率 给我们一个完整的数字。

  

  INTERACTIVE

  Decimal Game

  

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  • Use the drop down menu to choose the smallest power of ten to clear the decimal.
    ::使用下拉菜单来选择十分之十的最小功率来清除小数。
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  • Click the New Number button to get more practice!
    ::单击新数字按钮以获得更多练习!

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  To clear all decimals you must multiply every term by the power of ten which will result in all decimals becoming integers.  This  approach will be the same as in the interactive above. Now that you know what to multiply by to clear the decimals, try an example.
  ::要清除所有十进制数, 您必须将每个词汇乘以十进制的功率, 从而导致所有十进制成整数。 此方法将与上面交互的相同 。 现在, 既然您知道如何乘以十进制数来清除十进制数, 请举一个例子 。

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  Example
  ::示例示例示例示例

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  $$\begin{array}{rl}0.13+1.6x& =1.25\\ 100(0.13+1.6x)& =100(1.25)\\ 0.13\cdot 100+1.6x\cdot 100& =1.25\cdot 100\\ 13+160x& =125\\ 160x& =112\\ x& =0.7\end{array}$$

   
  ::0.13+1.6x=1.25100(0.13+1.6x)=100(1. 255)0. 13_100+1.6x_100=1. 25/13+160x=125160x=112x=0. 7

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  Clearing the decimals allows you to solve the problem using whole numbers instead of decimals.
  ::清除小数点可使您使用整数而不是小数点来解决问题。

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  Discussion Questions
  ::讨论问题 讨论问题

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  1. What would have happened in the example if we had multiplied both sides of the equation by 10?
     ::如果我们将方程式两侧乘以10,这个例子会怎样?
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  2. Why should we multiply by 100 when 1,000 will work too?
     ::我们为什么要乘以100 当1000人也会工作的时候呢?

   

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  Clearing Fractions
  ::结算分分数

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  The  strategy for clearing fractions will be similar to  the strategy for clearing decimals. To clear a decimal you must multiply the denominator or denominators by the least common denominator which will result in an integer when reduced. The least common denominator is the smallest number that the denominator can divide into with no remainder.
  ::清除分数的战略将与清除小数点的战略相似。要清除小数点,您必须把分母或分母乘以最小的公分母,在减少时得出整数。最小的公分母是分母可以分得的最小数,没有剩余数。

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  Use the interactive below to determine what the least common denominator of the fractions is.
  ::使用下面的交互数据来确定分数中最小的公分母是什么。

  

  INTERACTIVE

  Find the LCD

  

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  • Enter the least common denominator in the box.
    ::输入框中的最小公分母。
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  • Click "Next" to try more fractions.
    ::单击“ 下一步” 来尝试更多分数 。

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  To clear all fractions,  you must multiply every term by the least common denominator of the fractions.  This  will result in all fractions becoming  whole numbers.  Now that you know what to multiply by to clear the fractions, try an example.
  ::要清除所有的分数, 您必须将每个值乘以分数中最小的共同分母。 这将导致所有分数成为整数。 现在, 既然您知道要乘以什么来清除分数, 请举一个例子 。

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  Example
  ::示例示例示例示例

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  $$\begin{array}{rl}\frac{3}{4}x+6\frac{1}{2}& =4\frac{2}{5}\\ \\ 20\left(\frac{3}{4}x+6\frac{1}{2}\right)& =20\left(4\frac{2}{5}\right)\\ \\ 20\cdot \frac{3}{4}x+20\cdot 6\frac{1}{2}& =20\cdot 4\frac{2}{5}\\ \\ 15x+130& =88\\ 15x& =\text{-}42\\ x& =2\frac{4}{5}\end{array}$$

  
  ::34x+612=42520(34x+612)=20(425)20(20)20(34x+612)=20(2425)20(20)20(34x+20_612)=20(425)42515x+130=8815x=-42x=245

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  Clearing the fractions allows  you to solve the problem using whole numbers instead of decimals, making it much easier.
  ::清除分数可以让您用整数而不是小数来解决问题, 使问题更容易解决 。

   

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  Cost Reduction
  ::减少费用

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  Earlier, you saw that the owner of Empire Fudge, Jenny Chang, wanted to reduce the cost of the ingredients used to make her fudge from $3.10/lb to $2.83/lb. Below are the costs per pound of the ingredients used to make the fudge.
  ::早些时候,你看到帝国软糖的主人Jenny Chang想将她软糖的原料成本从3.10/lb 降低到2.83/lb。 下面是用来做软糖的原料的每磅成本。

  Ingredient	Material Cost Per Pound of Fudge ($)
  Semisweet Chocolate Chips (2 cups)	1.92
  Sweetened Condensed Milk (14 ounces)	0.63
  Butter (1/4 cup)	0.55
  Total	3.10

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  Discussion Questions
  ::讨论问题 讨论问题

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  1. The material cost for each ingredient in the fudge at Empire fudge can be seen in the table. What are some ways to reduce the total material cost per pound from $3.10/lb to $2.83/lb?
     ::表格中可以看出帝国软糖中每个成分的物质成本。 将每磅总物质成本从3.10美元降低到2.83美元有什么办法?
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  2. Which material cost per pound do you think would be easiest to decrease? Which do you think would be the most difficult to decrease? Why?
     ::你认为降低每磅的材料成本最容易吗?你认为降低哪种材料成本最难吗?为什么?

     

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  Jenny  wants to decrease each material cost by the same amount to achieve her goal total material cost per pound of $2.83. How much would she need to decrease each material cost by? Jenny can use the following equation:
  ::珍妮希望将每件材料费用减少同样的数额,以实现她的目标,即每磅2.83美元的全部材料费用。

   

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  $$\begin{array}{r}\text{Current Total Material Cost - 3 }\cdot \text{Decrease For Each Material = Goal Total Material Cost}\end{array}$$

  
  ::当前材料总成本 - 每种材料减少3 = 目标 材料总成本

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  $\begin{array}{r}\bullet \end{array}$  Current Total material cost: $3.10
  ::总材料成本:3.10美元

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  $\begin{array}{r}\bullet \end{array}$  Goal Total Material Cost: $2.83
  ::* 材料费用总额:2.83美元

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  $\begin{array}{r}\bullet \end{array}$  Decrease For Each Material:  $\begin{array}{r}x\end{array}$
  ::* 每件材料减少:x

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  When you substitute these value in you obtain the following:
  ::当您用这些值替换这些值时,可获得以下数值:

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  $$\begin{array}{rl}3.1-3x& =2.83\\ 100(3.1-3x)& =100(2.83)\\ 100\cdot 3.1-100\cdot 3x=& =100\cdot 2.83\\ 310-300x& =283\\ \text{-}300x& =\text{-}27\\ x& =0.09\end{array}$$

  
  ::3.1-3x=2.83100(3.1-3x)=100(2.83)100(3.1-1003)-1003x=1003x=100_83310-300x=283-300x=27x=0.09

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  Jenny would need to decrease each material cost by $0.09.
  ::珍妮需要将每件材料费用减少0.09美元。

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     Summary
  ::摘要

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  To make solving equations with rational numbers easier :
  ::使用合理数字解析方程式更加容易:

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  • Multiply equations by powers of ten to clear the decimals.
    ::乘以十位数的功率乘以十位数来清除小数。
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  • Multiply equations by the least common denominator to  change any fractions into whole numbers.
    ::乘以最小公分母的方程式,将任何分数转换成整数。