课程： 7.10 单一变数等量

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The incoming class at tennis camp is large this year. This is a special program that will have 6 people in each class. However, there is a maximum number of 42 classes. The incoming people need to be divided into groups of 6, so that the number of groups is 42. What is the maximum capacity, $\begin{aligned} c \end{aligned}$ , of the incoming class? How can you write an equation for $\begin{aligned} c \end{aligned}$ , and then solve this equation?
::今年网球营的入门班规模很大。这是一个特殊课程, 每班有6人。然而, 最多有42个班。入门班需要分为6人组, 这样入门班最多能容纳42人, c 最多能容纳42人组? 您如何为C 写一个方程, 然后解开这个方程 ?

In this concept, you will learn to solve single variable division equations.
::在此概念中,您将学会解析单一变量分割方程式。

Solving Single Variable Division Equations
::解决单一变数等量

To solve an equation in which a variable is divided by a number, you use the inverse of division, multiplication , to isolate the variable and solve the equation.
::要解析变量除以数字的方程式,您要使用除法、乘法的反函数来分隔变量并解析方程式。

You can multiply both sides of an equation by a number because of the Multiplication Property of Equality , which states:
::您可以将等式的两边乘以数,因为“平等乘法属性”规定:

if $\begin{aligned} a = b \end{aligned}$ , then $\begin{aligned} a \times c = b \times c \end{aligned}$ .
::a=b,然后axc=bxc。

This means that if you multiply one side of an equation by a number $\begin{aligned} c \end{aligned}$ , you must multiply the other side of the equation by that same number $\begin{aligned} c \end{aligned}$ , to keep the values on both sides of the equation equal.
::这意味着,如果将方程式的一面乘以一个数字 c,则必须将方程式的另一面乘以同一数字 c,以使方程式两侧的数值相等。

Here is an example.
::举一个例子。

Solve the equation for $\begin{aligned} k \end{aligned}$ .
::解决K的方程

$\begin{aligned} \frac{k}{- 4} = 12 \end{aligned}$
::k-4=12 (k-4=12)

First, use the multiplication property of equality, and multiply both sides of the equation by -4 to isolate the variable $\begin{aligned} k \end{aligned}$ .
::首先,使用平等的乘法属性,并将方程的两边乘以-4,以孤立变量 k。

$\begin{aligned} \begin{array}{rcl} \frac{k}{- 4} & = & 12 \\ - 4 \times \frac{k}{- 4} & = & - 4 \times 12 \\ \frac{- 4 k}{- 4} & = & - 48 \end{array} \end{aligned}$

::k-4=12 - 4xk- 4}4x12 - 4k-448

Next, separate the fraction and simplify.
::接下来,将分数分离并简化。

$\begin{aligned} \begin{array}{rcl} \frac{- 4}{- 4} k & = & - 48 \\ 1 k & = & - 48 \\ k & = & - 48 \end{array} \end{aligned}$

::-4 -4k481k48k48

The answer is $\begin{aligned} k = - 48 \end{aligned}$ .
::答案是k48。

Here is another example.
::下面是另一个例子。

Solve for $\begin{aligned} n \end{aligned}$ in the equation $\begin{aligned} \frac{n}{1.5} = 10 \end{aligned}$ .
::在 n1.5=10 方程式中解决 n。

First, use the multiplication property of equality to multiply both sides of the equation by 1.5.
::首先,使用等式的乘法属性将等式的两侧乘以1.5。

$\begin{aligned} 1.5 \times \frac{n}{1.5} = 10 \times 1.5 \end{aligned}$
Next, since $\begin{aligned} 1.5 \frac{n}{1.5} = \frac{1.5}{1} \times \frac{n}{1.5} \end{aligned}$ , you can rewrite that multiplication as one fraction.
::1.5xn1.5=10x1.5 下一步,从1.5n1.5=1.551xn1.5开始,您可以将该乘数重写为一个分数。

$\begin{aligned} \frac{1.5 n}{1.5} = 10 \times 1.5 \end{aligned}$

::1.5n1.5=10x1.5

Next, you separate the fraction and simplify.
::接下来,你把分数分开,简化。

$\begin{aligned} \begin{array}{rcl} \frac{1.5}{1.5} n & = & 15 \\ 1 n & = & 15 \\ n & = & 15 \end{array} \end{aligned}$
The answer is $\begin{aligned} n = 15 \end{aligned}$ .
::1.51.5n=151n=15n=15 答案是n=15。

Examples
::实例

Example 1
::例1

Earlier, you were given a problem about the tennis camp.
::早些时候,有人给了你一个网球营的问题。

The incoming students need to be divided into groups of 6, but there can only be 42 classes in total. Can you write a division equation, where $\begin{aligned} c \end{aligned}$ , is the maximum number of people in the incoming class so that there are 6 people in each group, and then solve it?
::新生学生需要分为6组,但总共只能有42个班级。你能写出一个分裂方程式吗? C是新进班的最大人数,每个班有6人,然后解决吗?

First, translate the language into an equation. Let $\begin{aligned} c \end{aligned}$ , be the maximum number of people in the incoming class. This number, divided by 6, should equal 42 classes.
::首先,将语言翻译为等式。让 c, 成为新进班级的最大人数。这个数字除以 6 , 应该等于 42 个班级。

$\begin{aligned} \frac{c}{6} = 42 \end{aligned}$

::c6=42

Next, use the multiplication property of equality and multiply both sides of the equation by 6.
::其次,使用平等的乘法属性,将等式的两边乘以6。

$\begin{aligned} 6 \times \frac{c}{6} = 6 \times 42 \end{aligned}$

::6xc6=6x42

Then, re-write the multiplication by a fraction and simplify.
::然后,将乘法重新写成一个分数并简化。

$\begin{aligned} \begin{array}{rcl} \frac{6}{1} \times \frac{c}{6} & = & 252 \\ \frac{6}{6} \times \frac{c}{1} & = & 252 \\ 1 c & = & 252 \\ c & = & 252 \end{array} \end{aligned}$

::61xc6=25266xc1=2521c=252c=252c=252

The answer is that there can be a maximum number of 252 students in the incoming class.
::答案是,进入班级的学生人数最多可达252人。

Example 2
::例2

Three friends evenly split the total cost of the bill for their lunch. The amount each friend paid was $4.25.
::三位朋友平分午餐总费用,每人支付4.25美元。

Write a division equation to represent $\begin{aligned} c \end{aligned}$ , the total cost, in dollars, of the bill for lunch.
::写一个分部方程式代表C, 午餐费用总额,以美元计。

Solve the equation to solve for the total cost of the bill.
::解决这个方程式解决账单的总成本问题

Consider part a first.
::考虑第一部分。

First, rephrase the question to help you solve the problem: The total cost, $\begin{aligned} c \end{aligned}$ , divided by three equals 4.25, the amount each person paid.
::首先,重写问题来帮助你解决问题:总成本,c,除以3等于4.25,每人支付的数额。

Then, express this as an equation.
::然后,用方程式来表达这一点。

$\begin{aligned} \frac{c}{3} = 4.25 \end{aligned}$

::c3=4.25 (千分之3=4.25)

Now consider part b .
::现在审议B部分。

Solve the equation by using the multiplication property of equality. Multiply both sides of the equation by 3.
::使用平等乘法属性来解析等式。将等式的两边乘以 3 。

$\begin{aligned} \begin{array}{rcl} \frac{c}{3} & = & 4.25 \\ 3 \times \frac{c}{3} & = & 3 \times 4.25 \end{array} \end{aligned}$

::c3=4.253×c3=3×4.25

Next, rearrange the multiplication of fractions.
::下一步,重新排列分数的乘法。

$\begin{aligned} \frac{3}{3} c = 12.75 \end{aligned}$

::33c=12.75

Now, simplify and solve.
::现在,简化和解决。

$\begin{aligned} \begin{array}{rcl} 1 c & = & 12.75 \\ c & = & 12.75 \end{array} \end{aligned}$

::1c=12.75c=12.75

The answer is that the bill was $12.75.
::答案是账单是12.75美元

Solve each equation.
::解决每个方程式

Example 3
::例3

$\begin{aligned} \frac{x}{- 2} = 5 \end{aligned}$
::x-2=5

First, use the multiplication property of equality and multiply both sides of the equation by -2.
::首先,使用平等的乘法属性,将等式的两侧乘以-2。

$\begin{aligned} - 2 \frac{x}{- 2} = - 2 \times 5 \end{aligned}$

::− 2x-2=%2×5

Next, simplify and solve for $\begin{aligned} x \end{aligned}$ .
::下一步,简化和解决 x。

$\begin{aligned} \begin{array}{rcl} \frac{- 2}{- 2} x & = & - 10 \\ 1 x & = & - 10 \\ x & = & - 10 \end{array} \end{aligned}$

::-2 -2x 101x10x10

The answer is $\begin{aligned} x = - 10 \end{aligned}$ .
::答案是 x10。

Example 4
::例4

$\begin{aligned} \frac{y}{5} = 6 \end{aligned}$
::y5=6 y5=6

First, use the multiplication property of equality and multiply both sides of the equation by 5.
::首先,使用平等的乘法属性,将等式的两侧乘以5。

$\begin{aligned} 5 \frac{y}{5} = 5 \times 6 \end{aligned}$

::5y5=5x6

Next, simplify and solve for $\begin{aligned} y \end{aligned}$ .
::接下来,简化和解决y。

$\begin{aligned} \begin{array}{rcl} \frac{5}{5} y & = & 30 \\ 1 y & = & 30 \\ y & = & 30 \end{array} \end{aligned}$

::55y=301y=30y=30y=30

The answer is $\begin{aligned} y = 30 \end{aligned}$ .
::答案是y=30。

Example 5
::例5

$\begin{aligned} \frac{b}{- 4} = - 3 \end{aligned}$
::b-43

First, use the multiplication property of equality and multiply both sides of the equation by -4.
::首先,使用平等的乘法属性,将等式的两侧乘以-4。

$\begin{aligned} - 4 \frac{b}{- 4} = - 4 \times - 3 \end{aligned}$

::-4b -4443

Next, simplify and solve for $\begin{aligned} b \end{aligned}$ .
::接下来,简化和解决b.

$\begin{aligned} \begin{array}{rcl} \frac{- 4}{- 4} b & = & 12 \\ 1 b & = & 12 \\ b & = & 12 \end{array} \end{aligned}$

::-4-4b=121b=12b=12

The answer is $\begin{aligned} b = 12 \end{aligned}$ .
::答案是b=12。

Review
::回顾

Solve each single variable division equation for the missing value.
::为缺失值解决每个单个变量分割方程式。

$\begin{aligned} \frac{x}{5} = 2 \end{aligned}$
::x5=2

$\begin{aligned} \frac{y}{7} = 3 \end{aligned}$
::y7=3 y7=3

$\begin{aligned} \frac{b}{9} = - 4 \end{aligned}$
::b94

$\begin{aligned} \frac{b}{8} = - 10 \end{aligned}$
::b810

$\begin{aligned} \frac{b}{8} = 20 \end{aligned}$
::b8=20 (b8=20)

$\begin{aligned} \frac{x}{- 3} = 10 \end{aligned}$
::x-3=10 x-3=10

$\begin{aligned} \frac{y}{18} = - 20 \end{aligned}$
::y 18 20

$\begin{aligned} \frac{a}{- 9} = - 9 \end{aligned}$
:a-99)

$\begin{aligned} \frac{x}{11} = - 12 \end{aligned}$
::x1112

$\begin{aligned} \frac{x}{3} = - 3 \end{aligned}$
::x33

$\begin{aligned} \frac{x}{5} = - 8 \end{aligned}$
::x58

$\begin{aligned} \frac{x}{1.3} = 3 \end{aligned}$
::x1.3=3 x1.3=3

$\begin{aligned} \frac{x}{2.4} = 4 \end{aligned}$
::x2.4=4

$\begin{aligned} \frac{x}{6} = 1.2 \end{aligned}$
::x6=1.2x6=1.2

$\begin{aligned} \frac{y}{1.5} = 3 \end{aligned}$
::y1.5=3

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。

Resources
::资源

章节大纲

常规

单变数分区

Solving Single Variable Division Equations ::解决单一变数等量

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Review ::回顾

Review (Answers) ::回顾(答复)

Resources ::资源

Solving Single Variable Division Equations
::解决单一变数等量

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Example 5
::例5

Review
::回顾

Review (Answers)
::回顾(答复)

Resources
::资源