Section: 解决单步和两步不平等 | 2.6 解决一和二分不平等

Section outline

On average, infants gain one pound per month between the time that they are 6 months old until they are 2 years old. Assuming that an infant is 6 months old and weighs 16 pounds, how many months will pass where they weigh less than 25 pounds? ^{1, 2}
::平均而言,婴儿从6个月大到2岁,每个月增加1英镑。假设婴儿6个月大,体重16磅,那么他们体重不足25磅的月数会过去多少个月? 1, 2。

Notice that there is a range of numbers that the infant can weigh, for example, 18 pounds, 21 pounds, 24 pounds, etc. All of these weights are less than 25 pounds. In this example, we are not dealing with a linear equation with one solution. Instead, we are dealing with a linear inequality with many solutions.
::请注意, 婴儿有一定的体重, 比如18磅, 21磅, 24磅等等。所有这些体重都不到25磅。在这个例子中, 我们不是用一个解决方案处理线性方程式。相反, 我们处理线性不平等, 有很多解决方案。

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Solving One- and Two-Step Inequalities
::解决单步和两步不平等

An inequality is a statement in which the expression on one side of the inequality sign is greater than or less than the expression on the other side. Recall from Chapter 1, the symbols we have to express inequalities.
::不平等是一种声明,其中不平等标志一方的表达大于或小于或小于另一侧的表达。回顾第一章,我们必须从第一章中回顾表示不平等的象征。

$\begin{aligned} < \end{aligned}$ Less than
::<小于

Example: $\begin{aligned} 4 < 6 \end{aligned}$
::示例:4 < 6

$\begin{aligned} > \end{aligned}$ Greater than
::> 大于

Example: $\begin{aligned} 6 > 4 \end{aligned}$
::示例:6>4

$\begin{aligned} \leq \end{aligned}$ Less than or equal to
::小于或等于

Example: $\begin{aligned} 4 \leq 6 OR 4 \leq 4 \end{aligned}$
::示例:4_____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

$\begin{aligned} \geq \end{aligned}$ Greater than or equal to
::大于或等于

Example: $\begin{aligned} 6 \geq 4 OR 6 \geq 6 \end{aligned}$
::示例: 6 4 OR 6 6

The inequality $\begin{aligned} x > - 1 \end{aligned}$ would be read, “ $\begin{aligned} x \end{aligned}$ is greater than negative one.” Notice there are many values that would make this statement true. For example, $\begin{aligned} x = 2 \end{aligned}$ or $\begin{aligned} x = 5 \end{aligned}$ or $\begin{aligned} x = 10. \end{aligned}$ In fact, any number to the right of -1 on the number line would be a solution to this inequality. We can describe the solution set as an interval of real numbers. In this case, the interval $\begin{aligned} (- 1, \infty) \end{aligned}$ .
::不平等 x% 1 将读为 , “ x 大于负 ” 。注意有许多值可以使此语句成为事实。例如, x=2 或 x=5 或 x=10。事实上, 数字行右侧的 -1 的任何数字都会解决这种不平等问题。我们可以将解决方案描述为实际数字的间隔。在这种情况下, 间隔( - 1 ) 。

We can graph this interval on a number line. The graph of $\begin{aligned} x > - 1 \end{aligned}$ looks like this:
::我们可以用数字线绘制此间距。 x% 1 的图形看起来是这样 :

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On the leftmost side of the interval, there is an open circle at -1 to indicate that -1 is not included in the solution. If the endpoint was included in the interval, then the circle would be closed, or filled in. We draw a line to the right or shade to show that any number greater than -1 is included in this interval. Note, the shading replaces drawing an infinite number of points since there are an infinite number of solutions.
::在间隔的最左侧, -1 处有一个开放的圆圈, 以表示 -1 不包含在解决方案中。如果端点包含在间隔中, 则圆将被关闭或填满。我们向右或阴影绘制一条线以显示此间隔中包含大于 - 1 的任何数字。注意, 阴影将取代绘制无限多点的点, 因为有无限数的解决方案。

Remember from Chapter 1 the possibilities for intervals.
::从第一章中记起间隔的可能性。

Bounded Intervals

Unbounded Intervals

Inequality: $\begin{aligned} a \leq x \leq b \end{aligned}$
::不平等:axxb

Interval: $\begin{aligned} [a, b] \end{aligned}$
::间间:[a,b]

Set-Builder: $\begin{aligned} {x \in R | a \leq x \leq b} \end{aligned}$
::设置为 {x{R} {x} {a} x} b} 。

Inequality: $\begin{aligned} a < x \end{aligned}$
::不平等:a<x

Interval: $\begin{aligned} (a, \infty) \end{aligned}$
::间隔a,)

Set-Builder: $\begin{aligned} {x \in R | a < x} \end{aligned}$
::设置构建器 : {x{R} a<x}

Inequality: $\begin{aligned} a < x < b \end{aligned}$
::不平等:a<x<b>

Interval: $\begin{aligned} (a, b) \end{aligned}$
::间隔a,b)

Set-Builder: $\begin{aligned} {x \in R | a < x < b} \end{aligned}$
::设置构建器 : {x{R} a <x<b}

Inequality: $\begin{aligned} a \leq x \end{aligned}$
::不平等:xx

Interval: $\begin{aligned} [a, \infty) \end{aligned}$
::间歇性:[a,]

Set-Builder: $\begin{aligned} {x \in R | a \leq x} \end{aligned}$
::设置- 构建器 : {x} R { ax}

Inequality: $\begin{aligned} a < x \leq b \end{aligned}$
::不平等: a<xb

Interval: $\begin{aligned} (a, b] \end{aligned}$
::间隔a,b)

Set-Builder: $\begin{aligned} {x \in R | a < x \leq b} \end{aligned}$
::设置构建器 : {x{R} a<x}

Inequality: $\begin{aligned} a > x \end{aligned}$
::不平等:a>x

Interval: $\begin{aligned} (- \infty, a) \end{aligned}$
::间隔,a)

Set-Builder: $\begin{aligned} {x \in R | a > x} \end{aligned}$
::设置- 构建器 : {x{R} {a>x}

Inequality: $\begin{aligned} a \leq x < b \end{aligned}$
::不平等:ax<b

Interval: $\begin{aligned} [a, b) \end{aligned}$
::间间:[a,b]

Set-Builder: $\begin{aligned} {x \in R | a \leq x < b} \end{aligned}$
::设置构建器 : {xR } ax<b}

Inequality: $\begin{aligned} a \geq x \end{aligned}$
::不平等:xx

Interval: $\begin{aligned} (- \infty, a] \end{aligned}$
::间隔,a)

Set-Builder: $\begin{aligned} {x \in R | a \geq x} \end{aligned}$
::设置- 构建器 : {x} R { ax}

Solving linear inequalities is similar to solving linear equations. In fact, all the properties that we used to solve linear equations also hold for linear inequalities.
::解决线性不平等与解决线性方程式相似。事实上,我们用来解决线性方程式的所有属性也都存在线性不平等。

The Properties of Inequality
(Note: These properties also hold for $\begin{aligned} \geq and \leq . \end{aligned}$ )

Addition Property of Inequality
::增加不平等财产

If $\begin{aligned} a < b \end{aligned}$ , then $\begin{aligned} a + c < b + c \end{aligned}$ for real numbers $\begin{aligned} a, b, c . \end{aligned}$
::如果a<b,则a+c>b+c等于实际数字a,b,c。

Multiplication Property of Inequality
::不平等的乘法特性

If $\begin{aligned} a < b \end{aligned}$ and $\begin{aligned} c \geq 0 \end{aligned}$ , then $\begin{aligned} a \times c < b \times c \end{aligned}$ for real numbers $\begin{aligned} a, b, c . \end{aligned}$
::a. b. c. c. a. a. b. c. a. a. a. b. c. a. a. b. c. a. a. b. c. a. a. b. c. a. a. b. c. c. a. a. a. a. a. b. a. b. c. c. c. a. a. a. a. a. a. a. b. b. c. c. a. a. a. a. a. a. a. a. a. a. b. c. c. a. a. a. a. a. a. c. a. a. a. a. a. a. a. a. a. a. a. a. a. a. a. a. a. b. b. b. c. c. c. c. c. a. a. a. a. a. a. a. a. a. a. a. a. a. a. a. a. a. a. a. a. a. c. c. c. c. c. c. c. c.c.c.c.c.c.c.c.c.c.c.c.c.c.c.c.c.c.c.c.c.c.c.c.a.c.c.c.c.c.c.c.c.c.c.c.c.c.c.c.c.c.c.c.c.c.c.c.c.c.a.c.c.c.c.c.c.c.c.c.c.a.a.a.a.a.a.c.c.a.a.c.a.c.c.a.a.a.a.a.a.c.a.a.a.a.a.a.a.a.a.a.a.c.a.a.a.a.a.a.a.a.a.a.a.a.a.a.a.a.a.a.a.a.a.a.a.a.a.a.a.a.a.a.a.a.a.a.a.a.

However, if $\begin{aligned} c < 0 \end{aligned}$ , then $\begin{aligned} a \times c > b \times c \end{aligned}$ .
::但是,如果c<0,那么axc>bxc。

Subtraction follows from addition :
::减法如下:

If $\begin{aligned} a < b \end{aligned}$ , then $\begin{aligned} a + (- c) < b + (- c) \end{aligned}$ or $\begin{aligned} a - c < b - c . \end{aligned}$
::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 $\begin{aligned} a < b \end{aligned}$ and $\begin{aligned} c \geq 0 \end{aligned}$ , then $\begin{aligned} a \times \frac{1}{c} < b \times \frac{1}{c} \end{aligned}$ or $\begin{aligned} \frac{a}{c} < \frac{b}{c} \end{aligned}$ for $\begin{aligned} c \neq 0. \end{aligned}$
::如果 a<b 和 c0 ,则 a×1c < b×1c > b×1c 或 ac < bc for c0 。

However, if $\begin{aligned} c < 0 \end{aligned}$ , then $\begin{aligned} \frac{a}{c} > \frac{b}{c} \end{aligned}$ .
::然而,如果c<0,则ac>bc。

Notice these properties are the same as the properties of equality with one exception: If you multiply or divide both sides of an inequality by a negative number , then you need to switch the direction of the inequality. Why is this true? Let's consider what happens with numbers. We can agree $\begin{aligned} 3 < 5. \end{aligned}$ However, if we multiply both sides of the inequality by -1, we get
::注意这些属性与平等属性相同,只有一个例外:如果将不平等的两侧乘以负数或将不平等的两侧除以负数,那么你就需要改变不平等的方向。为什么这是真实的?让我们想想数字会发生什么。我们可以同意 3 < 5。但是,如果我们将不平等的两侧乘以-1,我们得到的是

\begin{aligned} - 1 \times 3 & < - 1 \times 5 \\ - 3 & ≮ - 5 \end{aligned}

The same inequality would result if we divided both sides by -1. So, when we multiply or divide both sides of an inequality by a negative number, we need to switch the direction of the inequality.
::因此,当我们把不平等的两边扩大或除以负数时,我们需要改变不平等的方向。

Solving linear inequalities works the exact same way as solving linear equations with this one exception.
::解决线性不平等的方法与解决线性方程的方法完全相同,只有一个例外。

Solving One-Step Inequalities
::解决单步不平等

Example 1
::例1

On average, infants gain one pound per month between the time that they are 6 months old until they are 2 years old. Assuming that an infant is 6 months old and weighs 16 pounds, how many months will pass where they weigh less than 25 pounds?
::平均而言,婴儿从6个月大到2岁之间每月获得1英镑。假设婴儿6个月大,体重16磅,那么他们体重不到25磅的月数会过去多少个月?

Solution: Let's begin by letting m represent the age of the infant in months. We set up an inequality to solve. We have the 16 pounds that the average baby is at 6 months and then we have the 1 pound gained per month: $\begin{aligned} 1 \cdot m = m \end{aligned}$ .
::解决方案: 让我们首先让 m 代表婴儿数月中的年龄。我们设置了一个不平等的解决方案。我们有16磅的婴儿平均为6个月, 然后我们每月增加1磅:1磅。

\begin{aligned} 16 + m < 25 \\ \underline{- 16 - 16} & addition property of inequality \\ m < 9 \end{aligned}

::16+m<25-16_16_不平等的附加属性m9

After 6 months, it will take less than 9 months for the infant to weigh 25 pounds. Or, if we want to start from when the infant is born, we can add the 6 months from when the infant was born, and we have $\begin{aligned} m < 15 \end{aligned}$ . So, in the first 15 months of life, the average infant will weigh less than 25 pounds. Realistically, solutions less than 0 do not make sense in this context, so the solution is $\begin{aligned} [0, 15) \end{aligned}$ .
::6个月后,婴儿体重为25磅需要不到9个月的时间。或者,如果我们想从婴儿出生时开始,我们可以加上婴儿出生后6个月的时间,我们只有m < 15。因此,在婴儿出生后的头15个月中,平均婴儿体重将低于25磅。现实地说,0磅以下的解决方案在这方面没有意义,所以解决办法是[0,15]。

We can also graph our solution. Since 15 is not included in the solution, we put an open circle at 15 and shade all of the numbers to the left until the closed circle at 0.
::我们还可以绘制我们的解决方案。由于解决方案中没有15个选项,我们设置了15个选项的开放圆圈,左侧所有数字的阴影值为15个,直到关闭的圆圈为0个。

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Example 2
::例2

Solve and graph $\begin{aligned} \frac{2}{3} t \geq 8. \end{aligned}$
::解决和图23t8。

Solution: Just as in an equation , we divide by $\begin{aligned} \frac{2}{3} \end{aligned}$ or multiply by the reciprocal $\begin{aligned} \frac{3}{2} \end{aligned}$ .
::解决方案:就像方程式一样,我们除以23,或乘以对等32。

\begin{aligned} \frac{3}{2} \cdot \frac{2}{3} t & \geq \frac{3}{2_{1}} \cdot 8^{4} & multiplication property of inequality \\ t & \geq 12 \end{aligned}

::3223t32184不平等性财产的倍增

The solution is $\begin{aligned} [12, \infty) \end{aligned}$ . To graph this, we draw a solid circle at 12 and shade all of the numbers to the right of 12 on the number line.
::答案是 [12,\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\D\\\\\\\\\\\\\D8\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\"\\\\\\\\\\\\\"\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\"\\\\\\"\\\\\\\\\\\\\\\\\\\\\\\\\"\\\\\"\"\"\"\"\"\"\"\"\"\"\"\"\"\\\\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\\\\\\\\\\\\"\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\"\"\"\"\\\\\\\\\\\\\\\\"\\\\\\"\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Example 3
::例3

Solve and graph $\begin{aligned} - \frac{d}{7} \geq 2.4. \end{aligned}$
::解决和图表-d72.4。

Solution: Here, we have to be careful. To undo the operations around $\begin{aligned} d \end{aligned}$ , we multiply by -7. Now we switch the direction of the inequality.
::解决方案: 在这里, 我们必须小心。要取消 d 周围的操作, 我们乘以 - 7 。现在我们改变不平等的方向。

\begin{aligned} - 7 \times - \frac{d}{7} & \geq - 7 \times 2.4 \\ d & \leq - 16.8 \end{aligned}

::- 7777×2.416.8

The solution is $\begin{aligned} (- \infty, - 16.8] . \end{aligned}$ The graph is shading to the left of -16.8 and a closed circle at -16.8.
::答案是 (, -16.8) 。图表向 -16.8 左侧投下阴影, 在 -16.8 时有一个封闭的圆圈。

by CK-12 demonstrates the multiplication property of inequality.
::CK-12表明不平等的多重属性。

by CK-12 demonstrates the division property of inequality .
::CK-12显示了不平等的分割属性。

WARNING
::警告

Switch the direction of the inequality when you multiply or divide by a negative number. Do not switch the direction of the inequality if you multiply or divide by a positive number. The following example is incorrect:
::当您乘以负数或除以负数时,切换不平等的方向。如果乘以正数或除以正数,则不切换不平等的方向。以下示例不正确:

\begin{aligned} \frac{n}{4} & \leq - 3 \\ 4 \times \frac{n}{4} & \leq 4 \times - 3 \\ n & \geq - 12 \end{aligned}

::n434n44312

Solving Two-Step Inequalities
::解决双步不平等

Example 4
::例4

Solve and graph the solution to $\begin{aligned} 2 x - 5 \leq 17 \end{aligned}$ .
::解析并绘制2x-517的解决方案图。

Solution: Just as we did for equations, for , we perform inverse operations to the variable in the reverse order .
::解答:就像我们对于方程式所做的一样, 因为,我们执行逆向操作, 逆向顺序对变量进行反向操作。

\begin{aligned} 2 x - 5 \leq 17 \\ \underline{+ 5 + 5} & perform addition or subtraction \\ \frac{2 x}{2} \leq \frac{22}{2} & perform multiplication or division \\ x \leq 2 \end{aligned}

::2-5-5-17+5+5+5_性能增加或减号 2x2__2222-222 性能乘法或分形x%2

Graphing the solution, $\begin{aligned} (- \infty, 2] \end{aligned}$ , we get:
::分析解决方案, (,2), 我们得到:

Example 5
::例5

Solve and graph $\begin{aligned} - 6 n + 7 \leq - 29 \end{aligned}$ .
::解决和图表 - 6n+7\\\\\29。

Solution: Again, here we have to be careful with dividing by negative numbers.
::解决方案:再次,我们必须小心以负数除以负数。

\begin{aligned} - 6 n + 7 \leq - 29 \\ \underline{- 7 - 7} & perform addition or subtraction \\ \frac{- 6 n}{- 6} \leq \frac{- 36}{- 6} & perform multiplication or division \\ n \geq 6 & switch the direction of the inequality, divided by -6 \end{aligned}

::-6n+7+729 - 7 - 7 - 7_ 性能增加或减- 6n- 636-6 性能乘法或分除 n6 切除不平等的方向,除以 - 6

The graph of $\begin{aligned} [6, \infty) \end{aligned}$ is:
::[6,]的图表如下:

Checking Solutions to Inequalities
::检查不平等的解决方案

Up until now, we have not demonstrated a check for any of the solutions we have found. Unlike linear equations, there are usually an infinite number of solutions to check. Naturally, we do not want to check all of the solutions and we cannot check all of the solutions.
::到目前为止,我们还没有展示出我们找到的任何解决方案。与线性方程式不同,通常有无限数量的解决方案需要检查。当然,我们不想检查所有解决方案,也不能检查所有解决方案。

However, we can check the boundary point or points. These occur when the expressions on both sides of the inequality are equal to each other. If the inequality is less than or equal to or greater than or equal to, the boundary points are included in the solution. If the inequality is less than or greater than, then these values are not included in the solution.
::然而,我们可以检查边界点或点, 它们是当不平等的两侧的表达形式彼此平等时发生的。如果不平等程度小于或等于或大于或等于或等于或等于边界点, 解决方案中包括了边界点。如果不平等程度小于或大于, 那么解决方案中不包括这些价值观。

We also need to check the values in between the boundary points. We do this by choosing a number to test, called a test point, which is a point in the solution set. Then, we substitute the test point in the inequality to see if the inequality is a true statement. If it is, we assume our interval is correct. Let's see how this works in an example.
::我们还需要检查边界点之间的值。我们这样做的方式是选择一个数字来测试, 称为测试点, 这是解决方案设定的一个点。然后, 我们用不平等的测试点来替代不平等的测试点, 看看不平等是否是一个真实的声明。如果是这样的话, 我们假设我们的间隔是正确的。让我们以一个例子来看看这是怎么回事。

Example 6
::例6

Is $\begin{aligned} x < - 6 \end{aligned}$ a solution to $\begin{aligned} \frac{1}{2} x + 6 > 3 \end{aligned}$ ?
::x% 6 是 12x+6> 3 的解决方案吗 ?

Solution: Let's start by setting up an equation for the boundary point, -6. We set both expressions in the inequality equal to each other and substitute -6.
::解决方案:让我们首先为边界点设置一个方程式, -6. 我们设置了不平等的两种表达方式, 彼此平等, 替代 -6.

\begin{aligned} \frac{1}{2} x + 6 & = 3 \\ \frac{1}{2} (- 6) + 6 & = 3 \\ - 3 + 6 & = 3 \\ 3 & = 3 \end{aligned}

We can also check a number in the interval, a number that is less than -6. Let's try

\begin{aligned} x = - 8 \end{aligned}

as our test point .
::12x+6=312(-6)+6=3-3+6=33=3 我们还可以在间隔内检查一个数字,一个小于 -6. 的号码。让我们尝试 x8 作为我们的测试点。

\begin{aligned} \frac{1}{2} (- 8) + 6 & > 3 \\ - 4 + 6 & > 3 \\ 2 & > 3 \end{aligned}

Two is not greater than 3. Therefore , -8 cannot be in the solution interval, so $\begin{aligned} x < - 6 \end{aligned}$ is not a valid solution set for this inequality.
::两个不大于3, 因此, - 8 不能在解决方案间隔内, 所以 x6 并不是为这种不平等设定的有效解决方案。

Example 7
::例7

Is $\begin{aligned} x < - \frac{5}{3} \end{aligned}$ a solution to $\begin{aligned} - 3 x + 7 > 12 \end{aligned}$ ?
::x53 是- 3x+7> 12 的解决方案吗 ?

Solution: We set up an equation to check the boundary point, $\begin{aligned} x = \frac{- 5}{3} \end{aligned}$ .
::解答:我们设置了一个方程式来检查边界点 x53。

\begin{aligned} - 3 x + 7 & = 12 \\ - 3 (- \frac{5}{3}) + 7 & = 12 \\ 5 + 7 & = 12 \\ 12 & = 12 \end{aligned}

T he boundary point checks. Now we can check a test point, say

\begin{aligned} x = - 5 \end{aligned}

. Substituting -5 into the inequality we have:
::- 3x+7=12-3(- 53)+7=125+7=1212=12边界点检查。现在我们可以检查一个测试点,比如 x5。将 - 5 替换为不平等:

\begin{aligned} - 3 (- 5) + 7 & > 12 \\ 15 + 7 & > 12 \\ 22 & > 12 \end{aligned}

This is true , so $\begin{aligned} x < - \frac{5}{3} \end{aligned}$ is the solution set.
::这是真的,所以x53是设定的解决方案。

How to Check Solutions to Linear Inequalities With Desmos
::如何检查解决脱mos 线性不平等的解决方案

1. Enter each side of the inequality into Desmos as a function in the form y=____.
::1. 将不平等的方方面面输入Desmos, 作为一种以y为形态的函数。

2. Then find the value of x where the two lines intersect.
::2. 然后在两条线交汇处找到 x 的值。

3. The line that is above the other line represents the expression that is greater.
::3. 高于另一行的直线代表更大的表达式。

Given this graph, we can solve two different inequalities.
::根据这个图表,我们可以解决两种不同的不平等。

For $\begin{aligned} x - 3 > 2 \end{aligned}$ the values of x that make this inequality true are $\begin{aligned} (5, \infty) \end{aligned}$ .
::对于 x-3>2, 使这种不平等真实化的 x 值为 (5, ) 。

For $\begin{aligned} x - 3 < 2 \end{aligned}$ , the values of x that make this inequality true are $\begin{aligned} (- \infty, 5) \end{aligned}$ .
::对于 x-3 < 2, 使这种不平等真实化的 x 的值是 (, 5)。

How to Check Solutions to Linear Inequalities With a TI-83/84
::如何用TI-83/84检查解决线性不平等的解决办法

1. Press \fbox{Y=}. Enter each side of the inequality as functions in the form \text{Y}_n=______________. Use the \fbox{X} button to represent the variable regardless of the letter used for the variable in the inequality.
::1. 按\fbox{Y}} 键。在窗体\ text{Yn} 中输入不平等的每侧函数。使用\fbox{X} 按钮代表变量, 而不考虑不平等变量所使用的字母。

2. Follow the steps in used in Section 2.2 to find the point of intersection of the two lines.
::2. 遵循第2.2节所用的步骤,找到两条线的交叉点。

3. The line that is above the other line represents the expression that is greater.
::3. 高于另一行的直线代表更大的表达式。

For $\begin{aligned} x - 3 > 2 \end{aligned}$ , the values of x that make this inequality true are $\begin{aligned} (5, \infty) \end{aligned}$ .
::对于 x-3>2, 使这种不平等真实化的 x 值为 (5, ) 。

Summary
::摘要

Solve linear inequalities just as you would linear equations with one exception.
::解决线性不平等,就像解决线性等式一样,只有一个例外。

Exception: If you multiply or divide by a negative number, you need to switch the direction of the inequality.
::例外:如果乘以负数或除以负数,您需要改变不平等的方向。

Review
::回顾

Solve and graph each inequality.
::解决每个不平等问题并绘制图表。

$\begin{aligned} x + 5 > - 6 \end{aligned}$
::x+56

$\begin{aligned} 9 > b - 2 \end{aligned}$
::9>b-2

$\begin{aligned} 2 m \geq 14 \end{aligned}$
::2时14分

$\begin{aligned} 4 < - d \end{aligned}$
::4d

$\begin{aligned} 3 f - 4 \leq 8 \end{aligned}$
::3f-48

$\begin{aligned} 21 - 8 v < 45 \end{aligned}$
::21-85 < 45

$\begin{aligned} \frac{1}{2} q + 5 \geq 12 \end{aligned}$
::12q+512

$\begin{aligned} 10 - \frac{3}{4} u < - 8 \end{aligned}$
::10-34u8

$\begin{aligned} 1.5 > - 2.7 - 0.3 s \end{aligned}$
::1.5.2.7-0.3s

$\begin{aligned} \frac{1}{2} - \frac{3}{4} x \leq - \frac{5}{8} \end{aligned}$
::12-34x58

Determine if the interval is a solution set to the inequality by checking the boundary point(s) and a test point .
::通过检查边界点和测试点,确定间隔是否是设定的不平等解决办法。

11. Is $\begin{aligned} x > 56 \end{aligned}$ a solution to $\begin{aligned} \frac{3}{8} x + 5 < 26 ? \end{aligned}$
::11. x>56 是38x+5<26的解决方案吗?

12. Is $\begin{aligned} 7 > x \end{aligned}$ a solution to $\begin{aligned} 11 < 4 - x ? \end{aligned}$
::12. 7>x是11 < 4-x的解决方案吗?

Explore More
::探索更多

1. You have a 20 gallon aquarium. The biomass limit for an aquarium this size is 325 grams. ³ The goldfish you want weigh 23 grams each. How many goldfish can you add to the aquarium without exceeding the 325 gram limit?
::1. 你有20加仑水族馆,这种规模的水族馆的生物量限为325克。3 你要的金鱼每只重23克。在不超过325克限度的情况下,可以增加多少金鱼?

2. You are buying flowers and soil for the garden at the senior center. The flowers cost $4 each, and the soil costs $50. You have $125 to purchase the soil and flowers. How many flowers can you purchase?
::2. 您在高级中心为花园购买花卉和土壤,花卉每花4美元,土壤50美元。您有125美元购买土壤和花卉。您可以购买多少花卉?

3. You drive 15,000 miles a year and the cost of gas is $3.50 a gallon. You are trying to decide whether to buy a standard or a hybrid car. The hybrid gives you 30 miles per gallon (mpg) and the standard gives you 20 mpg. The hybrid car costs you $5,000 more than the standard. How long would it take for your savings on fuel to surpass the additional cost of the hybrid vehicle?
::3. 你每年驾驶15 000英里,汽油成本为每加仑3.50美元。你试图决定是购买一辆标准车还是一辆混合车。混合型汽车每加仑30英里,标准型汽车每加仑20美元。混合型汽车的费用比标准多5 000美元。你节省燃料需要多长时间才能超过混合型汽车的额外费用?

4. You have $437.29 in your checking account. You need to write a check for $531.15. Write an inequality showing the least amount of money that you have to deposit in the checking account so the check can clear your account.
::4. 您的支票账户有437.29美元。您需要写一张531.15美元的支票。写一张不平等的支票, 显示您必须在支票账户中存款最少的金额, 这样支票就可以结清您的账户。

5. A food manufacturer can claim on a label that a food is "high" in a nutrient if it has at least 20% of your daily value of that nutrient. ⁴ Write an inequality describing how much iron must a serving of food have to claim on the label that it is "high in iron" if the daily value for iron is 18 milligrams.

::5. 食品制造商在标签上可以声称,如果食物的营养值至少达到其日价值的20%,那么它就具有营养 " 高 " 的营养。 4 写一个不平等的字,说明在标签上,如果铁的日价值为18毫克,那么食品的供货必须多少铁才能声称它是 " 高的铁 " 。

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 this interactive that reinforces the concepts explored in this section:
::尝试这一互动,强化本节所探讨的概念:

References
::参考参考资料

“Your Child’s Size and Growth Timeline,” last updated October, 2016, .
::2016年10月更新。

“Infant Growth: What’s Normal? - Mayo Clinic,” posted October 10, 2014, .
::2014年10月10日张贴的「儿童成长:什么是正常?

“Question About Stocking Limits - The Planted Tank Forum,” original post April 17, 2012, .
::2012年4月17日“关于库存限额的问题――原油轮论坛”,原文为2012年4月17日。

“Labeling & Nutrition> Guidance for Industry: A Food Labeling Guide (10. Appendix B: Additional Requirements for Nutrient Content Claims),” posted January 2013, .
::2013年1月张贴的《工业标签和营养指导:食品标签指南》(10. 附录B:对营养内容索赔的附加要求),2013年1月公布。

Section outline

Solving One- and Two-Step Inequalities ::解决单步和两步不平等

Solving One-Step Inequalities ::解决单步不平等

Example 2 ::例2

WARNING ::警告

Solving Two-Step Inequalities ::解决双步不平等

Example 4 ::例4

Example 5 ::例5

Checking Solutions to Inequalities ::检查不平等的解决方案

Example 6 ::例6

Example 7 ::例7

How to Check Solutions to Linear Inequalities With Desmos ::如何检查解决脱mos 线性不平等的解决方案

How to Check Solutions to Linear Inequalities With a TI-83/84 ::如何用TI-83/84检查解决线性不平等的解决办法

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

References ::参考参考资料

Solving One- and Two-Step Inequalities
::解决单步和两步不平等

Solving One-Step Inequalities
::解决单步不平等

Example 2
::例2

WARNING
::警告

Solving Two-Step Inequalities
::解决双步不平等

Example 4
::例4

Example 5
::例5

Checking Solutions to Inequalities
::检查不平等的解决方案

Example 6
::例6

Example 7
::例7

How to Check Solutions to Linear Inequalities With Desmos
::如何检查解决脱mos 线性不平等的解决方案

How to Check Solutions to Linear Inequalities With a TI-83/84
::如何用TI-83/84检查解决线性不平等的解决办法

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

References
::参考参考资料