节: 单步等量和反向操作 | 3.1 单级等量和反向作业

章节大纲

One-Step Equations and Inverse Operations
::单步等量和反向操作

Nadia is buying a new mp3 player. Peter watches her pay for the player with a $100 bill. She receives $22.00 in change, and from only this information, Peter works out how much the player cost. How much was the player?
::Nadia 正在购买一个新的 mp3 播放器。 Peter 拿着100美元钞票看球员的薪水。她收到22美元零零的零钱, 仅从这个信息, Peter 就能知道玩家要花多少钱。玩家花了多少钱?

In algebra, we can solve problems like this using an equation . An equation is an algebraic expression that involves an equals sign. If we use the letter $\begin{aligned} x \end{aligned}$ to represent the cost of the mp3 player, we can write the equation $\begin{aligned} x + 22 = 100 \end{aligned}$ . This tells us that the value of the player plus the value of the change received is equal to the $100 that Nadia paid.
::在代数中, 我们可以用方程式来解决这样的问题。方程式是一个代数表达式, 包含等值符号。如果我们使用字母 x 来代表 mp3 播放器的成本, 我们可以写公式 x+22=100 。这告诉我们, 玩家的值加上所收到更改的值等于 Nadia 支付的 $100 。

Another way we could write the equation would be $\begin{aligned} x = 100 - 22 \end{aligned}$ . This tells us that the value of the player is equal to the total amount of money Nadia paid $\begin{aligned} (100 - 22) \end{aligned}$ . This equation is mathematically equivalent to the first one, but it is easier to solve.
::另一种我们可以写公式的方式是 x=100- 22。这告诉我们玩家的价值等于Nadia支付的钱总额( 100- 22) 。这个公式在数学上相当于第一个, 但比较容易解决。

In this chapter, we will learn how to solve for the variable in a one-variable linear equation . Linear equations are equations in which each term is either a constant , or a constant times a single variable (raised to the first power). The term linear comes from the word line, because the graph of a linear equation is always a line.
::在本章中, 我们将学习如何用一个可变线性方程式解决变量。线性方程式是方程式, 其中每个词都是一个常数, 或者一个常数乘以一个变量( 提升到第一个变数 ) 。线性词来自单词行, 因为线性方程式的图形总是一条线。

Solving Equations Using Addition and Subtraction
::利用增加和减法解决等量

When we work with an algebraic equation , it’s important to remember that the two sides have to stay equal for the equation to stay true. We can change the equation around however we want, but whatever we do to one side of the equation, we have to do to the other side. In the introduction above, for example, we could get from the first equation to the second equation by subtracting 22 from both sides:
::当我们用代数方程式工作时,重要的是要记住,双方必须保持平等,方程式才能保持真实。我们可以随心所欲地改变方程式,但不管我们如何对待方程式的一边,我们必须对另一方采取行动。例如,在上文的导言中,我们可以从第一个方程式到第二个方程式,从双方减去22个方程式:

$\begin{aligned} x + 22 & = 100 \\ x + 22 - 22 & = 100 - 22 \\ x & = 100 - 22 \end{aligned}$

::x+22=100x-22=100-22=100-22x=100-22x=100-22x=100-22

Similarly, we can add numbers to each side of an equation to help solve for our unknown.
::同样,我们可以在等式的每侧增加数字,以帮助解决我们未知的问题。

1. Solve $\begin{aligned} x - 3 = 9 \end{aligned}$ .
::1. 解决 x-3=9。

To solve an equation for $\begin{aligned} x \end{aligned}$ , we need to isolate $\begin{aligned} x - \end{aligned}$ that is, we need to get it by itself on one side of the equals sign. Right now our $\begin{aligned} x \end{aligned}$ has a 3 subtracted from it. To reverse this, we’ll add 3—but we must add 3 to both sides.
::为了解决 x 的方程, 我们需要孤立x , 也就是说, 我们需要将它自己放在等号的一边。现在, 我们的 x 已经从中扣除了 3 个。为了扭转这种情况, 我们会增加 3 个 — — 但我们必须在两边加 3 个。

$\begin{aligned} x - 3 & = 9 \\ x - 3 + 3 & = 9 + 3 \\ x + 0 & = 9 + 3 \\ x & = 12 \end{aligned}$

::x-3=9x-3+3=9+3x+0=9+3x=12

2. Solve $\begin{aligned} z - 9.7 = - 1.026 \end{aligned}$
::2. 解决z-9.71.026

It doesn’t matter what the variable is—the solving process is the same.
::不管变量是什么——解决过程是一样的。

$\begin{aligned} z - 9.7 & = - 1.026 \\ z - 9.7 + 9.7 & = - 1.026 + 9.7 \\ z & = 8.674 \end{aligned}$

::-9.71.026z-9.7+9.71.026+9.7z=8.674

Make sure you understand the addition of decimals in this example!
::请确保您了解在此示例中增加的小数点 !

3. Solve $\begin{aligned} x + \frac{4}{7} = \frac{9}{5} \end{aligned}$ .
::3. 解决x+47=95。

To isolate $\begin{aligned} x \end{aligned}$ , we need to subtract $\begin{aligned} \frac{4}{7} \end{aligned}$ from both sides.
::为了分离x,我们需要从两侧减去47。

$\begin{aligned} x + \frac{4}{7} & = \frac{9}{5} \\ x + \frac{4}{7} - \frac{4}{7} & = \frac{9}{5} - \frac{4}{7} \\ x & = \frac{9}{5} - \frac{4}{7} \end{aligned}$

::x+47=95x+47-47=95-47=95-47=95-47

Now we have to subtract fractions, which means we need to find the LCD. Since 5 and 7 are both prime, their lowest common multiple is just their product, 35.
::现在,我们必须减去分数,这意味着我们需要找到液晶体晶体。由于5和7都是初级的,它们最小的公倍数就是它们35的产物。

$\begin{aligned} x & = \frac{9}{5} - \frac{4}{7} \\ x & = \frac{7 \cdot 9}{7 \cdot 5} - \frac{4 \cdot 5}{7 \cdot 5} \\ x & = \frac{63}{35} - \frac{20}{35} \\ x & = \frac{63 - 20}{35} \\ x & = \frac{43}{35} \end{aligned}$

::x=95-47x=7975-4575x=6335-2035x63-2035x63-2035x=4335

Make sure you’re comfortable with decimals and fractions! To master algebra, you’ll need to work with them frequently.
::确保您对小数数和分数感到自在! 要掌握代数, 您需要经常与代数合作。

Example
::示例示例示例示例

Example 1
::例1

Solve $\begin{aligned} x + 10 = 17 \end{aligned}$ .
::解决 x+10=17。

To solve an equation for $\begin{aligned} x \end{aligned}$ , we need to isolate $\begin{aligned} x - \end{aligned}$ that is, we need to get it by itself on one side of the equals sign. Right now our $\begin{aligned} x \end{aligned}$ has 10 added to it. To reverse this, we’ll subtract 10—but we must subtract 10 to both sides.
::为了解决 x 的方程, 我们需要孤立x , 也就是说, 我们需要将它自己放在等号的一边。现在我们的 x 增加了 10 个。要扭转这一局面, 我们要减 10 个 — — 但是我们必须在两边都减 10 个。

$\begin{aligned} x + 10 & = 17 \\ x + 10 - 10 & = 17 - 10 \\ x + 0 & = 17 - 10 \\ x & = 7 \end{aligned}$

::x+10=17x+10-10=17-10x+0=17-10x=7

Review
::回顾

For 1-5, solve the following equations for $\begin{aligned} x \end{aligned}$ .
::对于 1-5, 解答 x 的以下方程式。

$\begin{aligned} x - 11 = 7 \end{aligned}$
::x- 11=7

$\begin{aligned} x - 1.1 = 3.2 \end{aligned}$
::x- 1. 1=3.2

$\begin{aligned} x + 0.257 = 1 \end{aligned}$
::x+0.257=1

$\begin{aligned} x + \frac{5}{2} = \frac{2}{3} \end{aligned}$
::x+52=23

$\begin{aligned} x - \frac{5}{6} = \frac{3}{8} \end{aligned}$
::x-56=38

For 6-10, solve the following equations for the unknown variable.
::对于 6-10, 解析未知变量的下列方程式。

$\begin{aligned} q - 13 = - 13 \end{aligned}$
::q-13 13

$\begin{aligned} z + 1.1 = 3.0001 \end{aligned}$
::zz+1.1=3.0001

$\begin{aligned} r + 1 = \frac{2}{5} \end{aligned}$
::r+1=25

$\begin{aligned} t + \frac{1}{2} = \frac{1}{3} \end{aligned}$
::t+12=13

$\begin{aligned} \frac{3}{4} = - \frac{1}{2} - y \end{aligned}$
::3412-y

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

One-Step Equations and Inverse Operations ::单步等量和反向操作

Solving Equations Using Addition and Subtraction ::利用增加和减法解决等量

Example ::示例示例示例示例

Example 1 ::例1

Review ::回顾

Review (Answers) ::回顾(答复)

One-Step Equations and Inverse Operations
::单步等量和反向操作

Solving Equations Using Addition and Subtraction
::利用增加和减法解决等量

Example
::示例示例示例示例

Example 1
::例1

Review
::回顾

Review (Answers)
::回顾(答复)