Course: 7.8 单一变数减法

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单变数减法
The baking club realized they only have 2.7 kg of flour left from their original order . The only person to use the flour since they bought it was Sal who used .9 kg of flour to make cakes. Can you write an equation to solve for how much flour they had before Sal took some, and then solve that equation?
::烘烤俱乐部意识到他们只有2.7公斤的面粉,从最初的订单上就剩下了。唯一使用面粉的人,自从他们购买面粉以来,是Sal用9公斤的面粉做蛋糕。你能写一个方程来解答在Sal拿一些面粉之前他们有多少面粉,然后解开这个方程吗?

In this concept, you will learn to solve single variable subtraction equations.
::在此概念中,您将学会解答单变量减法方程式。

Solving Single Variable Subtraction Equations
::解决单变量减法等量

To solve an equation in which a number is subtracted from a variable, you can use addition to isolate the variable .
::要解析从变量中减去数字的方程式, 您可以使用添加来分隔变量。

You can add the same number to both sides of the equation because of the Addition Property of Equality , which states: if $\begin{aligned} a = b \end{aligned}$ , then $\begin{aligned} a + c = b + c \end{aligned}$ .
::您可以在等式的两侧加上相同的数字,因为“平等加属性”规定:如果a=b,那么a+c=b+c。

This means that if you add a number, $\begin{aligned} c \end{aligned}$ , to one side of an equation, you must add that same number, $\begin{aligned} c \end{aligned}$ , to the other side, in order to keep both sides of the equation equal.
::这意味着,如果在等式的一面添加一个数字,c,则必须在另一面添加相同的数字,c,以使等式的两面保持相等。

Let’s look at an example.
::让我们举个例子。

Solve for $\begin{aligned} a \end{aligned}$ .
::解决一个。

$\begin{aligned} a - 15 = 18 \end{aligned}$

::a-15=18

In the equation, 15 is subtracted from $\begin{aligned} a \end{aligned}$ . So, you use the addition property of equality to add 15 to both sides of the equation. This will isolate the variable, $\begin{aligned} a \end{aligned}$ .
::在等式中, 从 a 中减去 15 个。所以, 您使用等式的附加属性在等式的两侧添加 15 个。这将分离变量 a 。

$\begin{aligned} \begin{array}{rcl} a - 15 & = & 18 \\ a - 15 + 15 & = & 18 + 15 \\ a + - 15 + 15 & = & 33 \\ a + 0 & = & 33 \\ a & = & 33 \end{array} \end{aligned}$

::a-15=18a-15+15=18+15a=15+1515=1533a+0=33a=33

Notice that you can rewrite $\begin{aligned} (- 15) \end{aligned}$ as $\begin{aligned} (+ - 15) \end{aligned}$ . This is a very useful strategy in solving equations . You can rewrite subtraction as adding a negative number.
::注意您可以将 (- 15) 重写为 (\\\\ 15) 。这是解决方程式的非常有用的战略。您可以将减法重写为负数。

The answer is $\begin{aligned} a = 33 \end{aligned}$ .
::答案是a=33。

Here is another example.
::下面是另一个例子。

Solve for $\begin{aligned} k \end{aligned}$ .
::解决k。

$\begin{aligned} k - \frac{1}{3} = \frac{2}{3} \end{aligned}$

::k- 13=23

In the equation, $\begin{aligned} \frac{1}{3} \end{aligned}$ is subtracted from $\begin{aligned} k \end{aligned}$ . So, you use the addition property of equality, and add $\begin{aligned} \frac{1}{3} \end{aligned}$ to both sides of the equation. This isolates the variable $\begin{aligned} k \end{aligned}$ .
::在方程中,从 k 中减去 13 。所以, 您使用平等的附加属性, 并在方程的两侧添加 13 。这分隔了变量 k 。

First, add $\begin{aligned} \frac{1}{3} \end{aligned}$ to both sides of the equation.
::首先,在等式的两边加13。

$\begin{aligned} \begin{array}{rcl} k - \frac{1}{3} & = & \frac{2}{3} \\ k - \frac{1}{3} + \frac{1}{3} & = & \frac{2}{3} + \frac{1}{3} \end{array} \end{aligned}$

::k- 13=23k- 13+13=23+13

Since the fractions have the same denominators you can add them together.
::由于分数具有相同的分母, 您可以将它们加在一起。

$\begin{aligned} \begin{array}{rcl} k - \frac{1}{3} + \frac{1}{3} & = & \frac{2}{3} + \frac{1}{3} \\ k & = & \frac{3}{3} \\ k & = & 1 \end{array} \end{aligned}$

::k- 13+13=23+13k=33k=1

The answer is $\begin{aligned} k = 1 \end{aligned}$ .
::答案是 k=1 。

Examples
::实例

Example 1
::例1

Earlier, you were given a problem about the baking club and Sal, who used their flour to make cakes.
::早些时候,你得到一个问题关于烘烤俱乐部和萨尔, 谁用他们的面粉做蛋糕。

They need to know how much flour they originally had so they can pay their supplier.
::他们需要知道他们最初有多少面粉,以便支付供应商。

The club started with some flour, but Sal took .9 kg of this flour to make cake. They had 2.7 kg left. You have to find how much they had to begin with.
::俱乐部开始时是面粉,但Sal拿了9公斤的面粉做蛋糕,剩下2.7公斤。你必须找到他们必须做的。

First, you need to write an equation that represents this information. Let $\begin{aligned} x \end{aligned}$ , represent the amount of flour the group originally had. This amount, minus the .9 kg Sal took equals 2.7 kg.
::首先,您需要写一个代表此信息的方程式。 Let x, 代表该组原先的面粉数量。减去9千克 Sal 等于 2.7千克的面粉数量。

$\begin{aligned} x - .9 = 2.7 \end{aligned}$

::-9=2.7x-9.=2.7

Use the addition property of equality and add .9 to both sides of the equation and solve.
::使用平等的附加财产,并在等式和解决办法的两侧增加9。

$\begin{aligned} \begin{array}{rcl} x - .9 + .9 & = & 2.7 + .9 \\ x & = & 3.6 \end{array} \end{aligned}$

::-9+9=2.7+.9x=3.6 -9+9=2.7+9x=3.6

The answer is they originally had 3.6 kg of flour.
::答案是,他们最初有3.6公斤面粉。

Example 2
::例2

Harry earned $19.50 this week. That is $6.50 less than he earned last week.
::哈利这周挣了19.50美元比上星期少了6.50美元

Write an equation to represent $\begin{aligned} m \end{aligned}$ , the amount of money, in dollars, that he earned last week. Then solve for $\begin{aligned} m \end{aligned}$ .
::写一个方程式来代表m, 他上周赚到的钱, 以美元计的金额。然后解决m。

Let $\begin{aligned} m \end{aligned}$ be the amount of money Harry earned last week. Then you can write an equation.
::让哈利上周赚的钱多一点然后你就可以写一个方程式了

$\begin{aligned} 19.50 = m - 6.50. \end{aligned}$

::19.50=m-6.50。

Next, solve the equation. Use the addition property of equality to add 6.50 to each side of the equation.
::接下来,解决方程式。使用等式的附加属性来为方程式的每侧增加6.50。

$\begin{aligned} \begin{array}{rcl} 19.50 & = & m - 6.50 \\ 19.50 + 6.50 & = & m - 6.50 + 6.50 \\ 26.00 & = & m + (- 6.50 + 6.50) \\ 26 & = & m + 0 \\ 26 & = & m \end{array} \end{aligned}$

::19.50=m-6.5019.50+6.50=m-6.50+6.5026.00=m+(-6.50+6.5026)26=m+0.026=m

The answer is Harry earned $26.00 last week.
::答案是哈利上周挣了26美元

Solve each equation.
::解决每个方程式

Example 3
::例3

Solve for $\begin{aligned} x \end{aligned}$ .
::解决x。

$\begin{aligned} x - 44 = 22 \end{aligned}$

::x-44=22

Use the addition property of equality and add 44 to both sides of the equation.
::增加平等财产,在等式的两侧增加44个。

$\begin{aligned} \begin{array}{rcl} x - 44 & = & 22 \\ x - 44 + 44 & = & 22 + 44 \\ x & = & 66 \end{array} \end{aligned}$

::x-44=22x-44+44=22+44x=66

The answer is $\begin{aligned} x = 66 \end{aligned}$ .
::答案是 x=66。

Example 4
::例4

Solve for $\begin{aligned} x \end{aligned}$ .
::解决x。

$\begin{aligned} x - 1.3 = 5.6 \end{aligned}$

::x-1.3=5.6

Use the addition property of equality and add 1.3 to both sides of the equation.
::增加平等财产,在等式的两侧增加1.3。

$\begin{aligned} \begin{array}{rcl} x - 1.3 & = & 5.6 \\ x - 1.3 + 1.3 & = & 5.6 + 1.3 \\ x & = & 6.9 \end{array} \end{aligned}$

::x-1.3=5.6x-1.3+1.3=5.6+1.3x=6.9

The answer is $\begin{aligned} x = 6.9 \end{aligned}$ .
::答案是 x=6. 9 。

Example 5
::例5

Solve for $\begin{aligned} y \end{aligned}$ .
::解决你。

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

::y- 14=12

Use the addition property of equality and add $\begin{aligned} \frac{1}{4} \end{aligned}$ to both sides of the equation.
::增加平等财产,在等式的两侧增加14个。

$\begin{aligned} \begin{array}{rcl} y - \frac{1}{4} & = & \frac{1}{2} \\ y - \frac{1}{4} + \frac{1}{4} & = & \frac{1}{2} + \frac{1}{4} \\ y & = & \frac{1}{2} + \frac{1}{4} \end{array} \end{aligned}$

::y- 14= 12- 14+14= 12+14y= 12+14

To add $\begin{aligned} \frac{1}{2} + \frac{1}{4} \end{aligned}$ you need common denominators. Write $\begin{aligned} \frac{1}{2} \end{aligned}$ as the equivalent fraction $\begin{aligned} \frac{2}{4} \end{aligned}$ and then add.
::要添加 12+14, 您需要共同的分母。将 12 写入等效分数 24, 然后添加。

$\begin{aligned} \begin{array}{rcl} y & = & \frac{2}{4} + \frac{1}{4} \\ y & = & \frac{3}{4} \end{array} \end{aligned}$

::y=24+14y=34

The answer is $\begin{aligned} y = \frac{3}{4} \end{aligned}$ .
::答案是y=34

Review
::回顾

Solve each single-variable subtraction equation.
::解决每个单变量减法方程式。

$\begin{aligned} x - 8 = 9 \end{aligned}$
::x-8=9

$\begin{aligned} x - 18 = 29 \end{aligned}$
::x-18=29

$\begin{aligned} a - 9 = 29 \end{aligned}$
::a-9=29

$\begin{aligned} a - 4 = 30 \end{aligned}$
::a-4=30

$\begin{aligned} b - 14 = 27 \end{aligned}$
::b-14=27

$\begin{aligned} b - 13 = 50 \end{aligned}$
::b-13=50

$\begin{aligned} y - 23 = 57 \end{aligned}$
::y - 23=57

$\begin{aligned} y - 15 = 27 \end{aligned}$
::y - 15=27

$\begin{aligned} x - 9 = 32 \end{aligned}$
::x- 9=32 x- 9=32

$\begin{aligned} c - 19 = 32 \end{aligned}$
::c-19=32

$\begin{aligned} x - 1.9 = 3.2 \end{aligned}$
::x-1.9=3.2

$\begin{aligned} y - 2.9 = 4.5 \end{aligned}$
::y-2.9=4.5

$\begin{aligned} c - 6.7 = 8.9 \end{aligned}$
::c-6.7=8.9

$\begin{aligned} c - 1.23 = 3.54 \end{aligned}$
::c-1.23=3.54

$\begin{aligned} c - 5.67 = 8.97 \end{aligned}$
::c-5.67=8.97

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

Section outline

General

单变数减法

Solving Single Variable Subtraction Equations ::解决单变量减法等量

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 Subtraction Equations
::解决单变量减法等量

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