节: 解决单步方位 | 2.2 溶解单步方位

章节大纲

You are organizing a fundraiser for a local youth group and you want to raise $10,000. Currently, you have raised $3,500 from community events and $4,800 from various donors. How much more needs to be raised?
::您正在组织一个当地青年团体的筹款活动,您想要筹集10,000美元。目前,您已经从社区活动筹集了3,500美元,从各捐助方筹集了4,800美元。您还需要筹集多少资金?

In this section, we will find the missing value by doing one arithmetic operation— addition , subtraction , multiplication , or division .
::在本节中,我们将通过一个算术操作——增加、减法、乘法或除法——找到缺失的价值。

lesson content

Solving One-Step Equations
::解决单步方位

An equation is a statement in which both sides are the same or equal . In the next few sections, we will discuss solving linear equations.
::等式是双方相同或平等的声明。在接下来的几节中,我们将讨论解决线性等式。

Linear Equations

::线性等式

A linear equation is an equation that can be written in the form

\begin{aligned} a x + b = c \end{aligned}

$\begin{align*}ax+b=c\end{align*}$ , where

\begin{aligned} a \end{aligned}

$\begin{align*}a\end{align*}$ ,

\begin{aligned} b \end{aligned}

$\begin{align*}b\end{align*}$ , and

\begin{aligned} c \end{aligned}

$\begin{align*}c\end{align*}$ are real numbers. (Another way to recognize a linear equation is to note that the exponent of

\begin{aligned} x \end{aligned}

$\begin{align*}x\end{align*}$ , or in general the variable , is 1).

The table below shows some examples of linear and nonlinear equations.
::下表列出了线性方程式和非线性方程式的一些实例。

Linear Equations	Nonlinear Equations
$\begin{align}2x+3=4\end{align}$	$\begin{align}x^2-2x=0\end{align}$
$\begin{align}-5(3x+2) -4x=15 \longrightarrow -19x - 10 = 15\end{align}$	$\begin{align}\sqrt{b}-5=0\end{align}$
$\begin{align}0.5n+2=3n-7 \longrightarrow -2.5x + 2 = -7\end{align}$	$\begin{align}\frac{2}{x} -4 = 8\end{align}$

Sometimes when we have an algebraic expression in an equation, we would like to find a number or set of numbers that make the equation a true statement, that is, a number (or numbers) for which both sides of the equation are equal. This number is called a solution to the equation and the process of finding the number is called solving (a set of numbers is called the solution set and the elements of it are called solutions ).
::有时当我们在一个方程中有一个代数表达式时,我们希望找到一个数字或一组数字,使方程成为真实的语句,也就是一个数字(或数字),而该方程的两边是相等的。这个数字被称为方程的解决方案,找到数字的过程被称为解决(一组数字被称为套数,其元素被称为解决方案)。

When solving an equation for a variable, you must get the variable by itself or isolate the variable on one side . In this and the following sections of this chapter, we will discuss the properties that you can use to do this. Here, we will only discuss solving one-step equations , equations that can be solved by performing one operation : addition, subtraction, multiplication, or division.
::当解析变量的方程时,您必须自己获得变量,或者将变量孤立在一边。在本章的这一节和以下各节中,我们将讨论您可用于此目的的属性。在这里,我们将只讨论解决单步方程,通过执行一个操作可以解决的方程:加法、减法、乘法或除法。

Properties of Equality
::平等财产

One way to think of an equation is as a scale where both sides of the scale are at the same height. As in the picture below, we can change the number of ovals on each side of the s cale and it still remains balanced.
::计算等式的一个方法就是将比例尺两侧高度都放在同一高度的尺度。和下面的图景一样,我们可以改变比例尺两侧的奥瓦尔数量,并且它仍然保持平衡。

We can think of what we need to do to solve equations in the same way. We want to keep the equations balanced. Formally, we will do this by using the properties of equality.
::我们可以想出如何以同样的方式解决方程式问题。我们希望平衡方程式。形式上,我们将通过使用平等性来做到这一点。

The Properties of Equality

Addition Property of Equality
::增加平等财产

If $\begin{align*}a=b\end{align*}$ , then $\begin{align*}a+c=b+c\end{align*}$ for real numbers $\begin{align*}a\end{align*}$ , $\begin{align*}b\end{align*}$ , and $\begin{align*}c\end{align*}$ .
::如果 a=b,则a+c=b+c,实际数字a、b和c为a、b和c。

Multiplication Property of Equality
::平等乘以财产

If $\begin{align*}a=b\end{align*}$ , then $\begin{align*}a \times c = b \times c\end{align*}$ for real numbers $\begin{align*}a\end{align*}$ , $\begin{align*}b\end{align*}$ , and $\begin{align*}c\end{align*}$ .
::如果 a=b, 则a×c=bxc, 实际数字a、b和c为a、b和c。

Subtraction follows from addition:
::减法如下:

If $\begin{align*}a=b\end{align*}$ , then $\begin{align*}a +(\text{-}c)= b+(\text{-}c)\end{align*}$ or $\begin{align*}a-c = b-c\end{align*}$ .
::如果 a=b,则a+(-c)=b+(-c)或a-c=b-c。

Division follows from multiplication:
::乘法除法如下:

If $\begin{align*}a=b\end{align*}$ , then $\begin{align*}a\times \frac{1}{c}=b\times \frac{1}{c}\end{align*}$ or $\begin{align*}\frac{a}{c}=\frac{b}{c}\end{align*}$ for $\begin{align*}c\neq 0.\end{align*}$
::如果 a=b, 那么 ax1c=bx1c 或 ac=bc 表示 c=0 。

Solving Equations by Addition or Subtraction
::以增加或减法解决的等号

The above properties show that if we add or subtract the same number on both sides of an equation, then the resulting sides of the equation will be equal. So , the question is: Which number do you choose to add or subtract on both sides to get the variable by itself? Let's consider an example.
::上述属性显示,如果我们在等式两侧加或减相同数字,则该等式的后半边将是相等的。因此,问题在于:您选择在两侧加或减哪个数字来获得变量本身? 让我们来举一个例子。

Example 1
::例1

Solve $\begin{align*}j-5=17\end{align*}$
::解决 j- 5= 17

Solution: We want to get j by itself on one side of the equation. To do this, we can use two properties of that we discussed in Chapter 1—the additive inverse property and the additive identity property .
::解答: 我们想自己在方程的一面得到j。为了做到这一点, 我们可以使用我们在第一章中讨论过的两种属性—— 添加反向属性和添加身份属性。

Recall that the additive inverse property states that the sum of a number and its opposite is zero or $\begin{align*}a+(-a)=0\end{align*}$ (the additive inverse of a number is another way to describe the opposite of a number). The a dditive identity property states that the sum of a number and zero is the number or $\begin{align*}a+0=a\end{align*}$ .
::回想起来,该添加物的反面属性规定,数字之和与数字之和为零或a+(a)=0(乘数之反数是描述数字之反的另一种方式)。

We only want j on one side of the equation; we do not want to subtract 5. The number we want to add to j is 0 and the way to get 0 is to add the opposite of -5 or +5. But, if we add 5 to one side of the equation, we have to add 5 to the other side of the equation to keep it balanced.
::我们只想在等式的一面加上j; 我们不想减5。我们想要加到 j 的数是 0, 获得 0 的方法是增加 5 或 + 5 的反面。但是,如果我们在等式的一面加上 5, 我们必须在等式的另一面加上 5 来保持平衡。

\begin{aligned} j - 5 = 17 \\ \underline{+ 5 + 5} & addition property of equality \\ j = 22 & additive inverse property, - 5 + 5 = 0 \\ additive identity property, j + 0 = j \end{aligned}

$\begin{align*}& \ \ \ j\ -\cancel{5} = 17 \\ & \underline{\ \quad \ +\cancel{5} \ \ +5 \; } && \color{gray}{\text{addition property of equality}}\\ & \ \qquad \ \ j = 22 && \color{gray}{\text{additive inverse property,} \ −5+5 = 0}\\ & \qquad && \color{gray}{\text{additive identity property,} \ j +0 = j}\end{align*}$
::j-5=17+5+5+5_平等附加财产j=22additive 反向财产,-5+5=0additive 身份财产,j+0=j

To determine if 22 makes the equation true or to check the solution , we substitute 22 into the original equation: $\begin{align*}22-5 = 17.\end{align*}$
::为了确定22方程是否正确,或检查解决办法,我们用22方程替换原方程:22-5=17。

Example 2
::例2

Solve $\begin{align*}7+y=16\end{align*}$
::解决 7+y=16

Solution: To isolate y , we add the opposite of 7 or -7 to both sides of the equation. This is the same as subtracting 7 on both sides. We say that addition and subtraction are inverse operations , because addition undoes subtraction and vice versa.
::解决方案: 要孤立 y , 我们将在等式的两侧加上 7 或 - 7 的反面。这与在两侧减去 7 相同。我们说, 增减是反向操作, 因为加减可以取消减值, 反之亦然。

\begin{aligned} 7 + y = 16 \\ \underline{- 7 - 7} & addition property of equality \\ y = 9 & additive inverse property, 7 + (- 7) = 0 \\ additive identity property, 0 + y = y \end{aligned}

$\begin{align*}& \ \ \ \cancel{7}+y = 16 \\ & \underline{-\cancel{7} \quad \quad -7 \; \;} && \color{gray}{\text{addition property of equality}}\\ & \ \qquad \ \ y = 9 && {\color{gray}{\text{additive inverse property,} \ 7+ (−7) = 0}} \\ & \qquad && \color{gray}{\text{additive identity property,} \ 0+y = y } \end{align*}$
::7+y=16-7-7-7_增加平等属性y=9附加反向财产,7+(-7)=0附加身份属性,0+y=y

You can check that y = 9 is correct by plugging 9 back into the original equation. Since $\begin{align*}7+9=16\end{align*}$ is a true statement, 9 is the solution.
::您可以在原始方程中将 9 插入到 7+9=16 是一个真实的语句, 来检查 Y = 9 是否正确。由于 7+9=16 是真实的语句, 9 是解决方案。

by CK-12 provides an explanation of how to solve equations using addition and subtraction. ( Note: The example labels do not align with this text ).
::在 CK-12 中,对如何使用添加和减法解析方程式作了解释。 (注:示例标签与本文本不一致)。

Example 3
::例3

Solution: First, let x be how much money you still need to raise.
::解决方案:首先,让我们先看看您还需要筹集多少钱。

Next, you need to translate the language into an equation. On one side of the equation we have the sum of all of the monies raised plus what they still need to raise, x , and on the other side we have the total amount needed.
::接下来,你需要将语言转换成方程式。在方程式的一面,我们有所筹到的所有资金加他们仍需要筹集的资金的总和, x, 而另一方面,我们有需要的总额。

\begin{aligned} 3, 500 + 4, 800 + x = 10, 000 \end{aligned}

$\begin{align*}3,500+4,800+x=10,000\end{align*}$
::3 500+4 800+x=10 000

Next, add the numbers together.
::下一步,将数字加在一起。

\begin{aligned} 8, 300 + x = 10, 000 \end{aligned}

$\begin{align*}8,300 + x = 10,000\end{align*}$
::8 300+x=10 000

Now, subtract 8,300 from both sides of the equation. This isolates the variable x .
::现在,从方程两侧减去8,300美元。此选项将变量 x 分离出来。

\begin{aligned} 8, 300 + x = 10, 000 \\ \underline{- 8, 300 - 8, 300} & addition property of equality \\ x = 1, 700 & additive inverse property, 8, 300 + (- 8, 300) = 0 \\ additive identity property, 0 + x = x \end{aligned}

$\begin{align*}& \ \ \ \cancel{8,300}+x = 10,000 \\ & \underline{-\cancel{8,300} \quad \quad -8,300 \; \;} && \color{gray}{\text{addition property of equality}}\\ & \ \qquad \ \ x = 1,700 && {\color{gray}{\text{additive inverse property,} \ 8,300+ (−8,300) = 0}} \\ & \qquad && \color{gray}{\text{additive identity property,} \ 0+x = x }\end{align*}$
::8 300+x=10 000-8 300-8 300-8 300_增加平等财产x=1 700 附加反向财产,8 300+(-8 300)=0额外身份财产,0+x=x

You still need to raise $1,700.
::你还需要筹到1700美元

WARNING
::警告

1. Do not perform the operation on only one side of the equation. The following is incorrect:
::1. 不只对等式的一面执行操作,以下不正确:

\begin{aligned} x - 3 = 5 & OR & x - 3 = 5 \\ \underline{+ 3} & \underline{+ 3} \\ x - 3 \neq 8 & x \neq 5 \end{aligned}

$\begin{align*}x-3=5 && \text{OR} & \qquad \ x-3=5 \\ \underline{\qquad \ +3} && & \qquad \ \underline{\ \ +3 \ \ \qquad }\\ x-3\neq8 && &\qquad \ x\neq5 \\\end{align*}$
::x-3=5OR x-3=5+3_3_3_x-3_3_8 x%5

2. Do not perform the same operation as in the equation on both sides. The following is incorrect:
::2. 不执行与双方方程式相同的行动,以下不正确:

\begin{aligned} x + 3 = 5 \\ \underline{+ 3 + 3} \\ x = 8 \end{aligned}

$\begin{align*}x+3 = 5\\ \underline {+3 +3} \\ x = 8\end{align*}$
::x+3=5+3+3_x=8 x3=8

Solving Equations by Multiplication or Division
::按乘数或除法解决等号

The process for solving equations with multiplication and division is similar to the process of solving equations with addition and subtraction. We need to determine which number to choose to multiply or divide by to isolate the variable.
::用乘法和除法解析方程式与加法和减法解析方程式相似。我们需要确定选择乘法或除法的乘法或除法, 以孤立变量。

Recall that the multiplicative inverse property states that the product of any number and its reciprocal is 1 or $\begin{align*}a\times \frac{1}{a}=1\end{align*}$ (the multiplicative inverse of a number is another way to say the reciprocal of a number). There is also a multiplicative identity property that says the product of a number and one is the number or $\begin{align*}a\times 1 = a\end{align*}$ .
::回顾多重反向财产规定,任何数字的产物及其对等性为1或ax1a=1(乘以数的反差是表示数字对等的另一种方式)。还存在一种多重复身份财产,表示数字的产物和数字的一个是数字或ax1=a。

Let's see how we can use these properties to solve equations as we did with the properties of addition.
::让我们看看我们如何用这些属性解析方程式, 就像我们用添加的属性解析等式一样。

Example 4
::例4

Solve $\begin{align*}\frac{h}{6}= -11\end{align*}$
::解决 h6 @% 11

Solution: We want to get h by itself on one side of the equation. We can use the multiplicative inverse property and multiply $\begin{align*}\frac{1}{6}\end{align*}$ by 6 to get 1. If we multiply by 6 on one side, we have to multiply by 6 on the other side of the equation. Then, by the multiplicative identity property 1 times h is h .
::解答: 我们想要自己在方程的一面得到 h。我们可以使用多倍反向属性, 乘以 16 乘以 6 来获得 1 。如果我们在一面乘以 6, 我们必须在方程的另一面乘以 6 。然后, 通过多倍身份属性, 乘以 1 h 。

\begin{aligned} 6 \cdot \frac{h}{6} & = - 11.6 & multiplication property of equality \\ multiplicative inverse property, 6 \times \frac{1}{6} = 1 \\ h & = - 66 & multiplicative identity property, 1 \times h = h \end{aligned}

$\begin{align*}\cancel{6} \cdot \frac{h}{\cancel{6}} &= -11.6 && \color{gray}{\text{multiplication property of equality}} \\ & \quad && \color{gray}{\text{multiplicative inverse property,} \ 6 \times \frac{1}{6}= 1} \\ h &= -66 && \color{gray}{\text{multiplicative identity property}, \ 1 \times h = h} \end{align*}$
::6h611.6 等式重复性反向财产的倍增属性,6x16=1h=1h=66倍倍增性身份属性,1xh=h

To check the answer, we can substitute h = -66 into the original equation: $\begin{align*}\frac{-66}{6}=-11\end{align*}$ . Since -11 = -11, h = -66 is the solution.
::要检查答案, 我们可以将 h = - 66 替换为原始方程式 : - 666\\\\ 11。因为 - 11 = - 11, h = - 66 是解决方案。

Example 5
::例5

Solve $\begin{align*}-7b=84\end{align*}$
::解决 - 7b=84

Solution: The reciprocal of -7 is $\begin{align*}-\frac{1}{7}\end{align*}$ , so we should multiply both sides of the equation by $\begin{align*}-\frac{1}{7}\end{align*}$ . But, multiplication by $\begin{align*}−\frac{1}{7}\end{align*}$ is the same as division by -7. Note multiplication and division are inverse operations for this reason. We divide both sides by -7 to solve for b .
::解决方案: - 7 的对等为 - 17, 因此我们应该将等式的两边乘以 17。但是, 17 的乘法与除法乘以 - 相同。注意乘法和除法是反向的操作。我们将两边除以 - 7 来解决 b 。

\begin{aligned} \frac{- 7 b}{- 7} & = \frac{84}{- 7} & multiplication property of equality \\ multiplicative inverse property, - 7 \times - \frac{1}{7} = 1 \\ b & = - 12 & multiplicative identity property, 1 \times b = b \end{aligned}

$\begin{align*}\frac{-\cancel{7}b}{-\cancel{7}} &= \frac{84}{-7} && \color{gray}{\text{multiplication property of equality}}\\ &\quad && \color{gray}{\text{multiplicative inverse property,} \ −7 \times −\frac{1}{7}= 1} \\ b &= -12 && \color{gray}{\text{multiplicative identity property,} \ 1 \times b = b} \end{align*}$ To check the answer, we can substitute b = -12 into the original equation:

\begin{aligned} - 7 \times - 12 = 84. \end{aligned}

$\begin{align*}-7\times -12=84.\end{align*}$ Since both sides are equal, b = -12 is the solution to the equation.
::7b-7=84-7 平等重复性反向财产的倍数属性, - 717=1b12 多倍性身份属性, 1xb=b 为检查答案, 我们可以将 b= - 12 替换为原始方程 : - 712=84。由于两面相等, b= - 12 是方程的解决方案。

Example 6
::例6

Solve $\begin{align*}\frac{3}{8}x = \frac{3}{2}\end{align*}$
::解决38x=32

Solution: The variable is being multiplied by a fraction . Instead of dividing by a fraction, we multiply by the reciprocal of $\begin{align*}\frac{3}{8}\end{align*}$ , which is $\begin{align*}\frac{8}{3}\end{align*}$ . These fractions cancel each other out, which results in 1 x, or just x on the left side of the equation.
::解答: 变量正被乘以一个分数。我们不是以一个分数除以一个分数, 而是以38的对等数乘以38, 也就是83。这些分数相互取消, 结果是在方程的左侧有1x, 或者仅x。

\begin{aligned} \frac{8}{3} \cdot \frac{3}{8} x & = \frac{3}{2} \cdot \frac{8}{3} \\ x & = \frac{8}{2} = 4 \end{aligned}

$\begin{align*}\frac{\cancel{8}}{\cancel{3}} \cdot \frac{\cancel{3}}{\cancel{8}} x &= \frac{\cancel{3}}{2} \cdot \frac{8}{\cancel{3}}\\ x &= \frac{8}{2}=4\end{align*}$
::83_38x=32_83x=82=4

Check the answer: $\begin{align*}\frac{3}{8} \cdot {4}=\frac{3}{2} \end{align*}$ . Both sides of the equation are equal, so we know that $\begin{align*}x = 4\end{align*}$ is the solution.
::检查答案: 384=32。等式的两边相等, 所以我们知道 x=4 是解决方案。

WARNING
::警告

1. Multiply by the reciprocal. The following is incorrect:

\begin{aligned} \frac{2}{3} x \cdot \frac{2}{3} x = 5 \cdot \frac{2}{3} \\ x = \frac{10}{3} \end{aligned}

$\begin{align*}\frac{2}{3}x\cdot \frac{2}{3}x = 5\cdot \frac{2}{3}\\\\ x = \frac{10}{3}\end{align*}$ 2. Multiply or divide by a negative number when you need to move the negative number to the other side of the equation. The following is incorrect:
::2. 当您需要将负数移到方程式的另一侧时,乘或除以负数。以下不正确:

\begin{aligned} \frac{- 6 x}{6} = \frac{48}{6} & O R & 7 \cdot \frac{- x}{7} = 8 \cdot 7 \\ x = 8 & x = 56 \end{aligned}

$\begin{align*}\frac{-6x}{6} = \frac{48}{6} && \text OR && 7\cdot \frac{-x}{7}= 8\cdot7\\\\ x = 8 && && x = 56\end{align*}$ 3. When we do not write the coefficient , it is 1 since 1 · x = x. This is true for -1 as well since −x = −1 · x. Do not divide by the negative sign. The following is incorrect:
::3. 当我们不写系数时,系数为1, 因为1 =x=xxxxxxxxxxxxxxxxxx。对于-1,这是事实,因为-x=-1=-1xxxxxxxxxxxxxxxxxx。不要除以负符号。以下不正确:

\begin{aligned} \frac{- y}{-} = \frac{4}{-} \end{aligned}

$\begin{align*}\frac{\cancel{-y}}{\cancel{-}}=\frac{4}{-}\end{align*}$
::~ ~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

by CK-12 demonstrates how to solve equations using multiplication and division.
::通过 CK-12 演示如何使用乘法和除法解析方程式。

Example 7
::例7

Sarvenaz earns $8 for each hour she works. She earned a total of $168 last week. Determine how many hours Sarvenaz worked last week.
::Sarvenaz每工作一个小时挣8美元,上周总共挣了168美元,确定Sarvenaz上星期工作了多少小时。

Solution: Let h be the number of hours Sarvenaz worked. She earns $8 for each hour she works, so you multiply the number of hours she worked by $8 to find the total amount she earned.
::解答:让Sarvenaz的工作小时数算一算。她每工作小时挣8美元,所以你将她的工作小时数乘以8美元,以找到她所得的总金额。

We solve the equation $\begin{align*}8h=168\end{align*}$ to find h , the number of hours she worked last week. We can use the inverse operation of multiplication. We divide both sides of the equation by 8.
::我们解决了8小时=168的方程, 找到她上星期的工作小时数。我们可以使用乘法的反向操作。我们将方程的两边除以 8 。

\begin{aligned} 8 h = 168 \\ \frac{8 h}{8} = \frac{168}{8} \\ h = 21 \end{aligned}

$\begin{align*}8h = 168\\\\ \frac{\cancel{8}h}{\cancel{8}} = \frac{168}{8}\\\\ h = 21\end{align*}$ Sarvenaz worked 21 hours last week.
::8h=1688h8=1688h=21 Sarvenaz上周工作21小时。

How to Solve Linear Equations With Desmos
::如何用脱mos 来解析线性等式

Enter each side of the equation into Desmos in the form y=____. Then find the value of x where the two lines intersect.
::将方程式的每个侧面输入成以 y 为形式的 Desmos。然后在两条线交叉处找到 x 的值。

Consider, $\begin{align*}x-3=2.\end{align*}$
::考虑, x-3=2。

Graph $\begin{align*}y=x-3 &&\text{and}&& y = 2.\end{align*}$
::图y=x-3andy=2。

The two lines intersect when $\begin{align*}x=5\end{align*}$ , so that is the solution.
::两条线在 x=5 时交叉, 所以这就是解决方案。

How to Solve Linear Equation With a TI-83/84
::如何用 TI-83/84 解决线性赤道

1. Press the \fbox{Y=} button to enter each side of the equation in the form Y=____. To enter the variable, use the \fbox{X} button to use x as a variable.
::1. 按下\fbox{Y} 按钮以输入以 Y为形式的方程式的每侧。要输入变量,请使用\fbox{X} 按钮将 x 用作变量。

2. Press the \fbox{GRAPH} button to graph the two lines.
::2. 按下\fbox{GRAPH} 按钮以图形显示两行。

3. To find the point where the two lines intersect, press \fbox{2nd} \ \fbox{CALC} \ \fbox{5}, which is the intersection command under the calculate menu.
::3. 要找到两行交叉的点, 请按\ fbox{ 2nd}\\\ fbox{ CALC}\\\\ fbox{ 5}, 这是计算菜单下的交叉命令。

4.Press \fbox{ENTER} on both lines. The up and down arrows can help you select the lines.
::4. 在两条线上都显示\fbox{ENTER}。向上和向下箭头可以帮助您选择线条。

5. Press \fbox{ENTER} for the guess. The intersection will display at the bottom of the window. The x-coordinate is the solution.

Feature: Where the Cassowary Roams
::Cassowary Roams 的地方

by Deirdre Mundy
::由Deirdre Mundy 编辑

It is one of the oddest birds on Earth. It is tall. It does not fly. It eats mainly rain forest fruits. It is the cassowary, and it is endangered.
::它是地球上最奇特的鸟类之一,是高的,不飞,主要吃雨林果实,是开胃菜,是濒临灭绝的。

Each male cassowary needs about 7 square miles of rain forest to survive. Females share territory with males. Males raise the children, while females mate with multiple males. About $\begin{align*}\frac{2}{3}\end{align*}$ of all cassowaries are male.
::女性与男性共享领地,男性抚养子女,而女性与多个男性为伴侣,大约23个大屠宰场为男性。

The amount of rain forest required to support a cassowary population of size p is approximately $\begin{align*}7(\frac{2}{3}p)\end{align*}$ . The cassowary population of Australia has been shrinking because the rain forest is being destroyed. Subdivisions, farms, and roads carve up the rain forest into tiny patches. The surviving cassowaries have trouble finding enough food to support themselves. They start wandering across roads and into neighborhoods. Cars hit them, dogs attack them, and these rare, bizarre birds die.
::养活一个面积约7(23p)的屠宰场所需的雨林面积约为7(23p)。澳大利亚的屠宰场人口由于雨林正在被摧毁而不断缩小。分块、农场和道路将雨林分割成小块地。幸存的屠宰场难以找到足够的食物来养活自己。它们开始横跨公路和邻里。汽车撞到他们,狗袭击他们,这些稀有的怪鸟死亡。

This video by National Geographic describes a cassowary research project.
::国家地理杂志的这段影片描述了一个授礼研究项目。

If Australia is home to 2,000 cassowaries, how many square miles of contiguous rain forest does it need to support them?
::如果澳大利亚是2 000个天文区的家园,那么需要多少平方英里的毗连雨林来支撑这些天文?

We could determine the square miles needed by multiplying the number of male cassowaries by 7. We first need to approximate the number of male cassowaries in a population of 2,000 cassowaries. We can multiply 2,000 by $\begin{align*}7(\frac{2}{3})\end{align*}$ to determine about how many of the cassowaries are male. Therefore , the solution is $\begin{align*}\frac{28000}{3}\end{align*}$ square miles.
::我们可以确定所需的平方英里,方法是将男性屠宰场数量乘以7。我们首先需要接近2 000个屠宰场人口的男性屠宰场数量。我们可以乘以2 000,乘以7(23),以确定有多少男性屠宰场。因此,解决方案是280003平方英里。

For more information about the cassowary, see the Resources tab.
::欲了解更多关于食堂的信息,请见资源标签。

Summary
::摘要

To isolate the variable on one side of the equation, perform the inverse operation on both sides of the equation.
::要分离方程式一边的变量, 在方程式两侧执行反向操作。

Addition and subtraction are inverse operations, as are multiplication and division.
::增减是反向操作,乘法和除法也是反向操作。

Review
::回顾

Solve each equation below and check your answer. Reduce any fractions.
::在下面解决每个方程式并检查您的答案。减少任何分数。

$\begin{align*}-3+x=-1\end{align*}$
::- 3+x%1

$\begin{align*}\frac{7}{8}= p + \frac{1}{2}\end{align*}$
::78=p+12

$\begin{align*}r-36.23=2.8\end{align*}$
::-36.23=2.8

$\begin{align*}7\sqrt{5}= b + 3\sqrt{5}\end{align*}$
::75=b+35

$\begin{align*}7=\frac{g}{-8}\end{align*}$
::7=g-8

$\begin{align*}35.75 = 3.25 n\end{align*}$
::35.75=3.25n 35.75=3.25n

$\begin{align*}\frac{6}{7}a = 36\end{align*}$
::67a=36

$\begin{align*}-8k=-64\end{align*}$
::-8k64

True or False: -21 is the solution of $\begin{align*}-16=y-5\end{align*}$ .
::真实或假 : - 21 是 - 16=y-5 的解决方案。

True or False: $\begin{align*}-\frac{20}{11}\end{align*}$ is the solution of $\begin{align*}\frac{3}{5}q = -\frac{12}{11}\end{align*}$ .
::真实或假 : - 2011 是35Q1211的解决方案。

Explore More
::探索更多

You received a gift card for your birthday. You bought an app for $10 and had $25 still on the card. What was the original amount of your gift card?
::你买了一个10美元的应用程序,卡上还有25美元。

Malcolm wrote 165 more lines of code than Sheila. Sheila wrote 1,595 lines of code. How many lines of code did Malcolm write?
::Malcolm比Sheila写了165行代码。Sheila写了1 595行代码。Malcolm写了多少行代码?

Your yearly salary was divided into 12 equal amounts and paid to you monthly. Each month you made $2,400. What was your yearly salary?
::您的年薪分为12个等额,并按月发放。每个月您挣了2,400美元。您的年薪是多少?您的年薪是多少?

One serving of fries contains 30% of the daily recommended amount of sodium. The amount of sodium in one serving of fries is 760 milligrams. What is the daily recommended amount of sodium?
::一种薯条中含有30%的每日推荐钠,一种薯条中含有760毫克的钠。

Identify the error: $\begin{align*}\frac{4}{5} \times -\frac{5}{4}y = -\frac{2}{3}\end{align*}$ . Therefore, $\begin{align*}y = -\frac{2}{3}\times \frac{4}{5}= -\frac{8}{15}\end{align*}$ .
::识别错误: 4554y23。因此,y23×45815。

To solve this problem, a proportion, we can use a technique called cross multiplication. In cross multiplication, we multiply the numerator of the fraction on one side of the equation with the denominator of the fraction on the opposite side of the equation and set the results equal to each other.
$\begin{aligned} \frac{a}{b} = \frac{c}{d} \end{aligned}$
$\begin{align*}\frac{a}{b} = \frac {c}{d}\end{align*}$
$\begin{aligned} a d = b c \end{aligned}$
$\begin{align*}ad=bc\end{align*}$ The reason this technique works is you can think of multiplying both sides of the equation by $\begin{align*}b\end{align*}$ and then by $\begin{align*}d\end{align*}$ .

Let's try a problem that uses this technique.

A nurse wants to find the number of milliliters of solution to administer to a patient. If the drug is available in 20 milligrams per 5 milliliters solution, how many milliliters of the solution does the nurse need to administer so the patient gets 70 mg of the drug?
::为了解决这个问题, 比例, 我们可以使用一种叫做交叉乘法的技术。在交叉乘法中, 我们将方程式一边的分数数数乘以方程式对面的分数分数分母, 并将结果设置为相等。 ab=cd ad=bcc 这个方法的原理是, 您可以将方程式的两边乘以 b 乘以。让我们尝试一个使用此技术的问题。护士想要找到用于病人管理的毫升解决方案数。如果该药物每5毫升溶液以20毫克计, 护士需要多少毫升的药物来管理, 这样病人就能得到70毫克的药物 ?

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

Please see the Appendix.
::请参看附录。

PLIX
::PLIX

Try these interactives that reinforce the concepts explored in this section:
::尝试这些强化本节所探讨概念的交互作用 :

Linear Equations	Nonlinear Equations
$\begin{align}2x+3=4\end{align}$	$\begin{align}x^2-2x=0\end{align}$
$\begin{align}-5(3x+2) -4x=15 \longrightarrow -19x - 10 = 15\end{align}$	$\begin{align}\sqrt{b}-5=0\end{align}$
$\begin{align}0.5n+2=3n-7 \longrightarrow -2.5x + 2 = -7\end{align}$	$\begin{align}\frac{2}{x} -4 = 8\end{align}$

章节大纲

Solving One-Step Equations ::解决单步方位

Linear Equations ::线性等式

Properties of Equality ::平等财产

Solving Equations by Addition or Subtraction ::以增加或减法解决的等号

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

WARNING ::警告

Solving Equations by Multiplication or Division ::按乘数或除法解决等号

Example 4 ::例4

Example 5 ::例5

Example 6 ::例6

WARNING ::警告

Example 7 ::例7

How to Solve Linear Equations With Desmos ::如何用脱mos 来解析线性等式

How to Solve Linear Equation With a TI-83/84 ::如何用 TI-83/84 解决线性赤道

Feature: Where the Cassowary Roams ::Cassowary Roams 的地方

Summary ::摘要

Review ::回顾

Explore More ::探索更多

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

PLIX ::PLIX

Solving One-Step Equations
::解决单步方位

Linear Equations

::线性等式

Properties of Equality
::平等财产

Solving Equations by Addition or Subtraction
::以增加或减法解决的等号

Example 1
::例1

Example 2
::例2

Example 3
::例3

WARNING
::警告

Solving Equations by Multiplication or Division
::按乘数或除法解决等号

Example 4
::例4

Example 5
::例5

Example 6
::例6

WARNING
::警告

Example 7
::例7

How to Solve Linear Equations With Desmos
::如何用脱mos 来解析线性等式

How to Solve Linear Equation With a TI-83/84
::如何用 TI-83/84 解决线性赤道

Feature: Where the Cassowary Roams
::Cassowary Roams 的地方

Summary
::摘要

Review
::回顾

Explore More
::探索更多

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

PLIX
::PLIX