Section: 两个变量的线性不平等 | 6.12 两个变量的线性不平等

Section outline

Linear Inequalities in Two Variables
::两个变量的线性不平等

The general procedure for graphing inequalities in two variables is as follows:
::将不平等情况分为两个变数的一般程序如下:

Re-write the inequality in slope-intercept form : $\begin{aligned} y = m x + b \end{aligned}$ . Writing the inequality in this form lets you know the direction of the inequality.
::以斜坡界面形式重写不平等:y=mx+b。以这种形式写出不平等, 使您了解不平等的方向。

Graph the line of the equation $\begin{aligned} y = m x + b \end{aligned}$ using your favorite method (plotting two points, using and $\begin{aligned} y - \end{aligned}$ intercept , using $\begin{aligned} y - \end{aligned}$ intercept and another point, or whatever is easiest). Draw the line as a dashed line if the equals sign is not included and a solid line if the equals sign is included.
::使用您最喜欢的方法( 绘制两个点, 使用和 y- intercept, 使用 y- intercept 和另一个点, 或其它最简单的方法) 绘制 y = mx+b 等式的线条。如果等号不包含, 则将线条画成虚线, 如果等号包含, 则绘制一条固线。

Shade the half plane above the line if the inequality is “ greater than .” Shade the half plane under the line if the inequality is “ less than .”
::如果不平等性“大于”的话,将线上的半平面遮盖起来。如果不平等性“低于”的话,则将线下的半平面遮盖起来。

Graphing Inequalities
::不平等图表

1. Graph the inequality $\begin{aligned} y \geq 2 x - 3 \end{aligned}$ .
::1. 图1. 不平等 y2x-3。

The inequality is already written in slope-intercept form, so it’s easy to graph. First we graph the line $\begin{aligned} y = 2 x - 3 \end{aligned}$ ; then we shade the half-plane above the line. The line is solid because the inequality includes the equals sign.
::不平等已经以斜坡界面形式写成,因此很容易图形化。首先,我们绘制了 y=2x-3 的线条; 然后,我们在线上方的半平面上阴影。这条线是牢固的, 因为不平等包含等号。

2. Graph the inequality $\begin{aligned} 5 x - 2 y > 4 \end{aligned}$ .
::2. 图5x-2y>4的不平等情况。

First we need to rewrite the inequality in slope-intercept form:
::首先,我们需要重写斜坡界面的不平等:

\begin{aligned} - 2 y & > - 5 x + 4 \\ y & < \frac{5}{2} x - 2 \end{aligned}

::-2y=5x+4y<52x-2

Notice that the inequality sign changed direction because we divided by a negative number.
::请注意,不平等标志改变了方向,因为我们除以负数。

To graph the equation, we can make a table of values:
::要绘制方程图,我们可以绘制一个数值表:

$\begin{aligned} x \end{aligned}$	$\begin{aligned} y \end{aligned}$
-2	$\begin{aligned} \frac{5}{2} (- 2) - 2 = - 7 \end{aligned}$
0	$\begin{aligned} \frac{5}{2} (0) - 2 = - 2 \end{aligned}$
2	$\begin{aligned} \frac{5}{2} (2) - 2 = 3 \end{aligned}$

After graphing the line, we shade the plane below the line because the inequality in slope-intercept form is less than . The line is dashed because the inequality does not include an equals sign.
::在绘制线条图后,我们将线线下方的平面蒙上阴影,因为斜坡截面的不平等性小于线条。这条线被冲破,因为不平等性不包括等号。

Solve Real-World Problems Using Linear Inequalities
::利用线性不平等解决现实世界问题

In this section, we see how linear inequalities can be used to solve real-world applications.
::在本节中,我们看到如何利用线性不平等解决现实世界的应用问题。

Real-World Application: Coffee Beans
::真实世界应用程序:咖啡豆

A retailer sells two types of coffee beans. One type costs $9 per pound and the other type costs $7 per pound. Find all the possible amounts of the two different coffee beans that can be mixed together to get a quantity of coffee beans costing $8.50 or less.
::零售商出售两种咖啡豆,一种每磅9美元,另一种每磅7美元。找出两种不同的咖啡豆的所有可能数量,这两种咖啡豆可以混合在一起,以获得一定数量的咖啡豆,价格为8.50美元或更少。

Let $\begin{aligned} x = \end{aligned}$ weight of $9 per pound coffee beans in pounds.
::重量为每磅咖啡豆9美元(磅)。

Let $\begin{aligned} y = \end{aligned}$ weight of $7 per pound coffee beans in pounds.
::每磅咖啡豆7美元的重量以磅为单位

The cost of a pound of coffee blend is given by $\begin{aligned} 9 x + 7 y \end{aligned}$ .
::1磅混合咖啡的费用由9x+7y支付。

We are looking for the mixtures that cost $8.50 or less. We write the inequality $\begin{aligned} 9 x + 7 y \leq 8.50 \end{aligned}$ .
::我们正在寻找花费8.50美元或更少的混合物。我们写下不平等 9x+7y8.50。

Since this inequality is in standard form , it’s easiest to graph it by finding the $\begin{aligned} x - \end{aligned}$ and $\begin{aligned} y - \end{aligned}$ intercepts . When $\begin{aligned} x = 0 \end{aligned}$ , we have $\begin{aligned} 7 y = 8.50 \end{aligned}$ or $\begin{aligned} y = \frac{8.50}{7} \approx 1.21 \end{aligned}$ . When $\begin{aligned} y = 0 \end{aligned}$ , we have $\begin{aligned} 9 x = 8.50 \end{aligned}$ or $\begin{aligned} x = \frac{8.50}{9} \approx 0.94 \end{aligned}$ . We can then graph the line that includes those two points.
::由于这种不平等是以标准形式呈现的,所以最容易通过查找 x- 和 y- intercuts 来图解它。当 x=0 时, 我们有 7y= 8. 50 或 y= 850\ 1. 221 。当 y= 0 时, 我们有 9x= 8. 50 或 x= 8. 50 或 x= 8. 50 n. 0. 94 。然后我们可以绘制包含这两个点的直线。

Now we have to figure out which side of the line to shade. In $\begin{aligned} y - \end{aligned}$ intercept form , we shade the area below the line when the inequality is “less than.” But in standard form that’s not always true. We could convert the inequality to $\begin{aligned} y - \end{aligned}$ intercept form to find out which side to shade, but there is another way that can be easier.
::现在我们必须找出线的哪一边是阴暗线。以y-inter-interform 的形式,当不平等“比不上 ” 时,我们将线下区域遮蔽。但标准形式并不总是真实的。我们可以将不平等转换成y-inter-inter-form 来找出阴暗的哪一边,但是还有另一种方法可以更容易一些。

The other method, which works for any linear inequality in any form, is to plug a random point into the inequality and see if it makes the inequality true. Any point that’s not on the line will do; the point (0, 0) is usually the most convenient.
::另一种方法适用于任何形式的线性不平等,就是将一个随机点插进不平等中,看看它是否使不平等成为真实。任何不在线的点都会这样做;点(0,0)通常是最方便的。

In this case, plugging in 0 for $\begin{aligned} x \end{aligned}$ and $\begin{aligned} y \end{aligned}$ would give us $\begin{aligned} 9 (0) + 7 (0) \leq 8.50 \end{aligned}$ , which is true. That means we should shade the half of the plane that includes (0, 0). If plugging in (0, 0) gave us a false inequality, that would mean that the solution set is the part of the plane that does not contain (0, 0).
::在此情况下, 插入 0 为 x 和 y 插入 0 将会给我们 9( 0)+7( 0)\\\ 850 , 这是正确的。这意味着我们应该遮盖包括 (0, 0) 的半平面。如果插入 (0, 0) 给我们一个假的不平等, 这意味着设定的解决方案是平面中不包含 (0, 0) 的部分。

Notice also that in this graph we show only the first quadrant of the coordinate plane . That’s because weight values in the real world are always nonnegative, so points outside the first quadrant don’t represent real-world solutions to this problem.
::请注意,在本图中,我们只展示了坐标平面的第一个象限。这是因为现实世界的重量值总是非负值,因此第一个象限以外的点并不代表这个问题的真实世界解决办法。

Example
::示例示例示例示例

Example 1
::例1

Julius has a job as an appliance salesman. He earns a commission of $60 for each washing machine he sells and $130 for each refrigerator he sells. How many washing machines and refrigerators must Julius sell in order to make $1000 or more in commissions?
::朱利叶斯拥有一个电器销售员的工作。他每卖一台洗衣机挣60美元,每卖一台冰箱赚130美元。朱利叶斯必须卖多少洗衣机和冰箱才能赚1000美元或以上?

Let $\begin{aligned} x = \end{aligned}$ number of washing machines Julius sells.
::朱利叶斯卖的洗衣机数量

Let $\begin{aligned} y = \end{aligned}$ number of refrigerators Julius sells.
::朱利叶斯卖的冰箱数量

The total commission is $\begin{aligned} 60 x + 130 y \end{aligned}$ .
::总委员会为60x+130y。

We’re looking for a total commission of $1000 or more, so we write the inequality $\begin{aligned} 60 x + 130 y \geq 1000 \end{aligned}$ .
::我们想要的佣金总额超过1000美元, 所以我们写下不平等60x+130y1000。

Once again, we can do this most easily by finding the $\begin{aligned} x - \end{aligned}$ and $\begin{aligned} y - \end{aligned}$ intercepts. When $\begin{aligned} x = 0 \end{aligned}$ , we have $\begin{aligned} 130 y = 1000 \end{aligned}$ , or $\begin{aligned} y = \frac{1000}{30} \approx 7.69 \end{aligned}$ . When $\begin{aligned} y = 0 \end{aligned}$ , we have $\begin{aligned} 60 x = 1000 \end{aligned}$ , or $\begin{aligned} x = \frac{1000}{60} \approx 16.67 \end{aligned}$ .
::我们再次通过查找 x - 和 y - intercuts 来更容易地做到这一点。当 x=0, 我们有130y=1000, 或者y=10030\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

We draw a solid line connecting those points, and shade above the line because the inequality is “greater than.” We can check this by plugging in the point (0, 0): selling 0 washing machines and 0 refrigerators would give Julius a commission of $0, which is not greater than or equal to $1000, so the point (0, 0) is not part of the solution; instead, we want to shade the side of the line that does not include it.
::我们绘制一条连接这些点的坚实的线条,并在线上阴影,因为不平等是“更大的 ” 。我们可以通过插入点(0,0 ) 来检查这一点:出售0台洗衣机和0台冰箱可以让朱利叶斯佣金为0美元,数额不大于或等于1000美元,因此,点(0,0)不是解决方案的一部分;相反,我们希望将不包括它的线的侧面遮盖起来。

Notice also that we show only the first quadrant of the coordinate plane, because Julius’s commission should be non-negative.
::因为朱利叶斯的委员会应该是非负面的。

Review
::回顾

Graph the following inequalities on the coordinate plane.
::图1 坐标平面上以下的不平等。

$\begin{aligned} y \leq 4 x + 3 \end{aligned}$
::y=4x+3 y=4x+3

$\begin{aligned} y > - \frac{x}{2} - 6 \end{aligned}$
::yx2-6

$\begin{aligned} 3 x - 4 y \geq 12 \end{aligned}$
::3x-44y12

$\begin{aligned} x + 7 y < 5 \end{aligned}$
::x+7y <5

$\begin{aligned} 6 x + 5 y > 1 \end{aligned}$
::6x+5y>1

$\begin{aligned} y + 5 \leq - 4 x + 10 \end{aligned}$
::y+5+4x+10

$\begin{aligned} x - \frac{1}{2} y \geq 5 \end{aligned}$
::x- 12y5

$\begin{aligned} 6 x + y < 20 \end{aligned}$
::6x+y <20

$\begin{aligned} 30 x + 5 y < 100 \end{aligned}$
::30x+5y < 100

Remember what you learned in the last chapter about families of lines.
1. What do the graphs of $\begin{aligned} y > x + 2 \end{aligned}$ and $\begin{aligned} y < x + 5 \end{aligned}$ have in common?
  ::y> x+2 和 y < x+5 的图形有什么共同点 ?
2. What do you think the graph of $\begin{aligned} x + 2 < y < x + 5 \end{aligned}$ would look like?
  ::您认为 x+2 <y < x+5 的图形看起来像什么 ?
::记住您在最后一章中学到的关于线条家族的内容。 y> x+2 和 y < x+5 的图形有什么共同之处? 您认为 x+2 < y < x+5 的图形会是什么样子 ?

How would the answer to problem 6 change if you subtracted 2 from the right-hand side of the inequality?
::如果你从不平等的右侧减去2个,那么问题6的答案会如何改变呢?

How would the answer to problem 7 change if you added 12 to the right-hand side?
::如果你在右手边加上12个,问题7的答案会如何改变?

How would the answer to problem 8 change if you flipped the inequality sign?
::如果你翻转不平等标志问题8的答案会如何改变呢?

A phone company charges 50 cents per minute during the daytime and 10 cents per minute at night. Sketch a graph showing how many daytime minutes and nighttime minutes could you use in one week if you wanted to pay less than $20.
::电话公司在白天每分钟收费50美分,在晚上每分钟收费10美分。绘制一个图表,显示如果你想支付不到20美元, 一周内可以使用多少日间分钟和夜间分钟。

Suppose you are graphing the inequality $\begin{aligned} y > 5 x \end{aligned}$ .
1. Why can’t you plug in the point (0, 0) to tell you which side of the line to shade?
  ::为何不能插上点(0,0),
2. What happens if you do plug it in?
  ::如果你插进去会怎么样?
3. Try plugging in the point (0, 1) instead. Now which side of the line should you shade?
  ::尝试插入点( 0, 1) 。现在, 您应该对线的哪一侧进行阴影 ?
::假设您正在绘制不平等 y> 5x 的图。为什么您不能插插插点( 0, 0) 来告诉你阴影线的哪一边? 如果您插插进去会怎么样? 请尝试插插点( 0, 1) 。现在您应该插插到哪一边 ?

A theater wants to take in at least $2000 for a certain matinee. Children’s tickets cost $5 each and adult tickets cost $10 each.
1. If $\begin{aligned} x \end{aligned}$ represents the number of adult tickets sold and $\begin{aligned} y \end{aligned}$ represents the number of children’s tickets, write an inequality describing the number of tickets that will allow the theater to meet their minimum take.
  ::如果x代表售出的成人票数,y代表儿童票数,写一个不平等的字,说明允许剧院达到最低票价的票数。
2. If 100 children’s tickets and 100 adult tickets have already been sold, what inequality describes how many more tickets of both types the theater needs to sell?
  ::如果100张儿童票和100张成人票已经售出,那么什么不平等可以说明剧院需要售出多少两种类型的票?
3. If the theater has only 300 seats (so only 100 are still available), what inequality describes the maximum number of additional tickets of both types the theater can sell?
  ::如果剧院只有300张座位(所以现在只有100张),那么,什么不平等可以说明剧院可以出售的两种类型的额外票最多数量?
::一个剧院至少要为某一场婚礼购买2000美元。儿童票每张要5美元,成人票每张要10美元。如果x代表售出的成人票数,y代表儿童票数,那么写一个不平等的描述让剧院达到最低票数的票数。如果100张儿童票和100张成人票已经售出,那么什么不平等描述剧院需要出售的两种类型的票数?如果剧院只有300张座位(因此只有100张),那么什么不平等描述剧院可以出售的两种类型的新票最多数量?

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。

Texas Instruments Resources
::得克萨斯州工具资源

In the CK-12 Texas Instruments Algebra I FlexBook® resource, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See .
::在CK-12得克萨斯州仪器代数I FlexBook资源中,有图表计算活动,旨在补充本章某些经验教训的目标。

Section outline

Linear Inequalities in Two Variables ::两个变量的线性不平等

Graphing Inequalities ::不平等图表

Solve Real-World Problems Using Linear Inequalities ::利用线性不平等解决现实世界问题

Real-World Application: Coffee Beans ::真实世界应用程序:咖啡豆

Example ::示例示例示例示例

Example 1 ::例1

Review ::回顾

Review (Answers) ::回顾(答复)

Texas Instruments Resources ::得克萨斯州工具资源