节: 通过乘数和加法解决等号的解决系统 | 4.6 通过乘数和加法解决等号的解决系统

章节大纲

Mattie wants to plant some flowers in her yard. She has space for 15 plants. She buys pansies and daisies at her local garden center. The pansies are each $2.75 and the daisies are each $2.00. How many of each does she buy if she spends a total of $35.25?
::Mattie想在院子里种花。她有15种植物的空间。她在当地的园艺中心购买葡萄园和小菊。葡萄园各为2.75美元,小菊每为2.00美元。如果她总共花35.25美元,她要买多少?

We can answer questions like this by solving a system of equations . In this section, we expand the previous approach of elimination by addition by creating terms that are additive inverses.
::我们可以通过解决一个方程式系统来回答这样的问题。在本节中,我们扩大了先前的消除方法,又增加了一些词,这些词是复数。

Creating Additive Inverses
::正在创建补充反省

It is not always the case that the coefficients of one variable in a system of linear equations are additive inverses. When this is not the case, we may be able to multiply one of the equations by a constant so that we have additive inverses of one variable and then can add the equations to eliminate that variable. To do this, we need the least common multiple (LCM).
::在线性方程系统中,一个变量的系数并非总能成为复数。如果情况并非如此,我们也许能够将一个方程乘以一个常数,这样我们就可以将一个变量的复数乘以一个变量的复数,然后添加方程来消除该变量。为此,我们需要最小常见的倍数(LCM ) 。

How to Find the Least Common Multiple
::如何找到最小常见多重

To find the least common multiple:
::要找到最不常见的多个 :

1. Find the prime factorization of each of the numbers.
::1. 找出每个数字的基数乘数。

2. List each prime factor in both numbers with the highest exponent that they have in all of the factored numbers.
::2. 列出每个质因数,以两个数列出,在所有因数中列出最高指数。

3. Multiply the prime numbers to get the least common multiple.
::3. 乘以质数以获得最不常见的倍数。

Example 1
::例1

Find the least common multiple of 14, 20, and 35.
::找到最不常见的乘数 14 20 和35。

Solution: First, we find the prime factorization of each of the numbers.
::解决方案:首先,我们发现每个数字的质因子化。

$\begin{aligned} 14 20 35 \\ ↙↘ ↙↘ ↙↘ \\ 2 7 2 10 5 7 \\ ↙↘ \\ 2 5 \\ 14 = 2 \cdot 7 \\ 20 = 2^{2} \cdot 5 \\ 35 = 5 \cdot 7 \end{aligned}$

The LCM is the product of each factor raised to the highest power in the factors . $\begin{aligned} LCM = 2^{2} \cdot 5 \cdot 7 = 140 \end{aligned}$ .
::业连管是每个因素的产物,在各种因素中,每个因素的功率最高。 LCM=2257=140。

by Jason Bradley demonstrates how to find the LCM using prime factorization. This is the best method when finding the LCM of three or more numbers.
::杰森·布拉德利(Jason Bradley)展示了如何使用质因数来找到 LCM。这是找到三个或三个以上数字的 LCM的最佳方法。

Solving Systems of Equations by Multiplication and Addition
::通过乘数和加法解决等号的解决系统

If we do not have additive inverse coefficients, we can create them using the least common multiple.
::如果我们没有添加反系数,我们可以使用最不常见的倍数来创造这些系数。

To Solve Systems of Equations by Multiplication and Addition
::通过乘法和加法解决等式系统

1. Choose a variable to eliminate.
::1. 选择要消除的变量。

2. Find the LCM of the coefficients of that variable.
::2. 找到该变量系数的 LCM。

3. Multiply both sides of the equations to make the coefficient of the variable you want to eliminate the LCM. Make sure these terms have opposite signs, so they are additive inverses of each other.
::3. 将方程的两边乘以使变量的系数成为您想要消除 LCM 的系数,确保这些术语有相反的符号,因此它们是相互的添加反相。

4. Add the equations and solve for the remaining variable if necessary.
::4. 必要时添加余下变量的方程和解答。

5. Once you have found one value, substitute into either equation to find the value of the variable you choose to eliminate.
::5. 一旦找到一个值,就替换成任一方程,以找到您选择删除的变量的值。

Example 2
::例2

Solve the system of equations by elimination by multiplication and addition.
::通过乘法和加法消除等式系统。

$\begin{aligned} 4 x + y & = 0 \\ x - 3 y & = 26 \end{aligned}$

::4x+y=0x-3y=26

Solution: One of the properties of additive inverses to look for is opposite signs. The coefficients of $\begin{aligned} y \end{aligned}$ have opposite signs—one is positive and one is negative. We can create additive inverses if we multiply the first equation by 3. Be careful, make sure to multiply the entire equation by 3:
::解决方案: 需要寻找的添加反函数的一个属性是相反的符号。 y 的系数有相反的符号—— 一个是正数, 一个是负数。如果我们将第一个方程乘以3, 我们可以创建添加反函数。注意, 确保将整个方程乘以3 :

$\begin{aligned} 3 (4 x + y & = 0) \\ 12 x + 3 y & = 0 \end{aligned}$

::3(4x+y=0)12x+3y=0

Now we can use this new equation in our system to eliminate $\begin{aligned} y \end{aligned}$ and solve for $\begin{aligned} x \end{aligned}$ :
::现在我们可以使用我们系统中的这个新方程式来消除y 并解决 x:

$\begin{aligned} 12 x + 3 y = 0 \\ + \underline{x - 3 y = 26} \\ 13 x = 26 \\ x = 2 \end{aligned}$

::12x+3y=0+ x-3y=26_ 13x=26 x=2

Now, find $\begin{aligned} y \end{aligned}$ :
::现在,找到y:

$\begin{aligned} 4 (2) + y & = 0 \\ 8 + y & = 0 \\ y & = - 8 \end{aligned}$

::4(2)+y=08+y=0y8

The solution is (2, -8).
::解决办法是2,8。

To check whether this is the solution, we need to make sure it is a solution for both equations.
::为了检查这是否是解决办法,我们需要确保它是两种方程式的解决办法。

$\begin{aligned} 4 (2) + (- 8) & = 8 - 8 = 0 \\ (2) - 3 (- 8) & = 2 + 24 = 26 \end{aligned}$

Note that we could have used the other equation to find $\begin{aligned} y \end{aligned}$ in the final step. From the beginning, we could have multiplied the second equation by -4 instead to cancel out the $\begin{aligned} x \end{aligned}$ variable.
::请注意, 我们本可以用其它方程式在最后一步找到 y。从一开始, 我们本可以乘以 - 4 来取消 x 变量。

by CK-12 demonstrates how to solve a linear system by elimination and multiplication.
::通过 CK-12 演示如何通过除法和乘法解决线性系统。

Example 3
::例3

Solve the system using elimination by multiplication and addition.
::利用乘法和加法来消除系统。

$\begin{aligned} 2 x - 5 y & = 15 \\ 3 x + 7 y & = 8 \end{aligned}$

::2x-5y=153x+7y=8

Solution: None of the pairs of coefficients in this problem are additive inverses, so we have to create additive inverses. We choose to eliminate x. (You could also choose to eliminate y ). T he coefficients of $\begin{aligned} x \end{aligned}$ are 2 and 3, so the least common multiple of these numbers is 6. We want the coefficients of $\begin{aligned} x \end{aligned}$ to be 6 and -6 so that they are additive inverses and will cancel when we add the two equations together. In order to get coefficients of 6 and -6, we multiply the first equation by 3 and the second equation by -2. (We could also multiply the first equation by -3 and the second equation by 2.)
::解答:这个问题的系数对中,没有一个是复数反数,所以我们必须创造复数反数。我们选择去掉x.(你也可以选择去掉y)。x的系数是2和3,因此这些数字中最不常见的倍数是6。我们希望x的系数是6和-6,这样它们就是复数反数,当我们把两个方程加在一起时,它们就会取消。为了将系数乘以6和-6,我们把第一个方程乘以3,第二个方程乘以-2。 (我们也可以把第一个方程乘以-3,第二个方程乘以2)

$\begin{aligned} 3 (2 x - 5 y = 15) \Rightarrow 6 x - 15 y = 45 \\ - 2 (3 x + 7 y = 8) + \underline{- 6 x - 14 y = - 16} \\ - 29 y = 29 \\ y = - 1 \end{aligned}$

::3(2x-5y=15) 6x-15y=45-2(3x+7y=8)+-6x-14y}1629y=29y=1

Now we substitute this value into the original first equation to find $\begin{aligned} x \end{aligned}$ :
::现在,我们把这个值替换为最初的第一个方程式, 以找到 x:

$\begin{aligned} 2 x - 5 (- 1) & = 15 \\ 2 x + 5 & = 15 \\ 2 x & = 10 \\ x & = 5 \end{aligned}$

::2x-5( - 1) = 152x+5= 152x= 1525x= 10x=5

The solution is (5, -1).
::解决办法是(5,-1) 。

Checking the solution, we have
::检查解决方案,我们有

$\begin{aligned} 2 (5) - 5 (- 1) & = 10 + 5 = 15 \\ 3 (5) + 7 (- 1) & = 15 - 7 = 8 \end{aligned}$

This problem could also be solved by eliminating $\begin{aligned} y \end{aligned}$ . To do this, find the least common multiple of the coefficients of $\begin{aligned} y \end{aligned}$ , 5 and 7. The least common multiple is 35, so we would multiply the first equation by 7 and the second equation by 5. Since one of them is already negative, we do not have to multiply by a negative number.
::这个问题也可以通过消除y.来解决。要做到这一点,找到y、5和7等系数中最不常见的倍数,最不常见的倍数是35,因此我们将第一个方程乘以7,第二个方程乘以5。由于其中之一已经是负数,我们不必乘以负数。

Example 4
::例4

Solve the system using elimination by multiplication and addition.
::利用乘法和加法来消除系统。

$\begin{aligned} 14 x - 6 y & = - 3 \\ 16 x - 9 y & = - 7 \end{aligned}$

::14 - 6y316x-9y7

Solution: This time, we choose to eliminate $\begin{aligned} y \end{aligned}$ . We need to find the LCM of 6 and 9.
::解决方案:这次,我们选择消除y。我们需要找到6和9的LCM。

$\begin{aligned} 6 & = 2 \cdot 3 \\ 9 & = 3 \cdot 3 = 3^{2} \\ LCM & = 2 \cdot 3^{2} = 18 \end{aligned}$

::6=239=33=32LCM=232=18

The LCM is 18, so we will multiply the first equation by 3 and the second equation by -2. Again, it does not matter which equation we multiply by a negative value.
::LCM是18, 所以我们将把第一个方程乘以3, 第二个方程乘以-2。再说一遍, 我们乘以负值的哪个方程并不重要。

$\begin{aligned} 3 (14 x - 6 y = - 3) \Rightarrow + 42 x - 18 y = - 9 \\ - 2 (16 x - 9 y = - 7) \underline{- 32 x + 18 y = 14} \\ 10 x = 5 \\ x = \frac{1}{2} \end{aligned}$

::3(14x-6y3)+42x-18y9-2(16x-9y7)-32x+18y=14_10x=5x=12

Now find $\begin{aligned} y \end{aligned}$ by substituting into either equation. We choose the first equation.
::现在通过替换两个方程中的任何一个方程来找到y。我们选择第一个方程。

$\begin{aligned} 14 (\frac{1}{2}) - 6 y & = - 3 \\ 7 - 6 y & = - 3 \\ - 6 y & = - 10 \\ y & = \frac{10}{6} = \frac{5}{3} \end{aligned}$

::14(12)-6y37-6y3-6y10y=106=53

The solution is $\begin{aligned} (\frac{1}{2}, \frac{5}{3}) \end{aligned}$
::解决办法是(12,53)

Checking the solution , we have
::检查解决方案,我们有

$\begin{aligned} 14 (\frac{1}{2}) - 6 (\frac{5}{3}) & = 7 - 10 = - 3 \\ 16 (\frac{1}{2}) - 9 (\frac{5}{3}) & = 8 - 15 = - 7 \end{aligned}$

by Bigbend Emporium demonstrates how to solve a system of equation by elimination and multiplication.
::Bigbend Emporium 演示如何通过消除和乘法解决等式系统。

Example 5
::例5

Mattie wants to plant some flowers in her yard. She has space for 15 plants. She buys pansies and daisies at her local garden center. The pansies are each $2.75 and the daisies are each $2.00. How many of each does she buy if she spends a total of $35.25?
::Mattie想在院子里种花。她有15种植物的空间。她在当地的园艺中心购买葡萄园和小菊。葡萄园各为2.75美元,小菊每为2.00美元。如果她总共花35.25美元,她要买多少?

Solution: The system of linear equations represented by this situation is below, where p is the number of pansies and d is the number of daisies.
::解决办法:这种情况所代表的线性方程式系统如下,p为板块数,d为小菊数。

$\begin{aligned} p + d & = 15 \\ 2.75 p + 2 d & = 35.25 \end{aligned}$

::p+d=152.75p+2d=35.25

If we multiply the first of these two equations by $\begin{aligned} - 2 \end{aligned}$ , we get a new first equation and our equations are set up to eliminate d .
::如果我们将这两个方程式中的第一个方程式乘以-2, 我们就会得到一个新的第一个方程式, 我们的方程式是用来消除 d 的。

$\begin{aligned} - 2 p - 2 d & = - 30 \\ 2.75 p + 2 d & = 35.25 \end{aligned}$

::-2p-2d302.75p+2d=35.25

Now we can add these two equations to cancel out the d -terms . We get
::现在我们可以加上这两个方程式来取消 d 条件。

$\begin{aligned} 0.75 p = 5.25 \end{aligned}$ or $\begin{aligned} p = 7 \end{aligned}$
::0.75p=5.25或p=7

Finally, we can substitute $\begin{aligned} p = 7 \end{aligned}$ into either of our original equations to get the value of d .
::最后,我们可以用p=7 来替代我们原来的方程中的任何一个方程来获得d的值。

$\begin{aligned} 7 + d = 15 \end{aligned}$ or $\begin{aligned} d = 8 \end{aligned}$
::7+d=15或d=8

Therefore Mattie buys 7 pansies and 8 daisies.
::所以Mattie买了7个罐子和8个菊花

WARNING
::警告

Make sure to multiply both sides of the equation by a number. If you multiply only one side, the equation will not be balanced and will represent a different equation.

We can recognize our special cases similar to how we recognize them in substitution.
::我们可以承认我们的特殊情况,类似于我们如何以替代方式承认它们。

Example 6
::例6

Solve the system using elimination by multiplication and addition.
::利用乘法和加法来消除系统。

$\begin{aligned} 4 x - 6 y & = - 12 \\ y & = \frac{2}{3} x + 2 \end{aligned}$

::4-6y12y=23x+2

Solution: Rearrange the second equation in this system to get it in general form. We can do this by subtracting $\begin{aligned} \frac{2}{3} x \end{aligned}$ on both sides to get the following system:
::解答: 重新排列此系统中的第二个方程, 以获得一般形式。我们可以通过在两侧减去 23x 来做到这一点, 以获得以下系统 :

$\begin{aligned} 4 x - 6 y & = - 12 \\ - \frac{2}{3} x + y & = 2 \end{aligned}$

::4 - 6y 12 - 23x+y=2

Let's choose to eliminate y . Multiply the second equation by 6 to eliminate y :
::将第二个方程式乘以 6 来删除 y:

$\begin{aligned} 6 (- \frac{2}{3} x + y = 2) \\ - 4 x + 6 y = 12 \end{aligned}$

::6(- 23x+y=2) - 4x+6y=12

Replace the second equation in the problem and add it to the first equation.
::将问题中的第二个方程式替换为第二个方程式,并将其添加到第一个方程式中。

$\begin{aligned} 4 x - 6 y = - 12 \\ + \underline{- 4 x + 6 y = 12} \\ 0 x = 0 \\ 0 = 0 \end{aligned}$

::4 - 6y 12 + - 4x+6y=12_ 0x=0=0

Here, both variables were eliminated and we wound up with 0 = 0. Recall that this is a true statement and thus this system has infinite solutions.
::这里,两个变量都被删除了, 我们最后以 0 = 0 结束。提醒大家注意这是一个真实的语句, 因此这个系统有无限的解决方案。

Example 7
::例7

Solve the systems using elimination by multiplication and addition.
::利用乘法和加法来消除系统。

$\begin{aligned} x - 3 y & = 5 \\ y & = \frac{1}{3} x + 8 \end{aligned}$

::x-3y=5y=13x+8

Solution: We rewrite the second equation in general form.
::解决方案:我们以一般形式重写第二个方程。

$\begin{aligned} x - 3 y & = 5 \\ - \frac{1}{3} x + y & = 8 \end{aligned}$

::x-3y=5-13x+y=8

Now we can multiply the second equation by 3 to get coefficients of $\begin{aligned} x \end{aligned}$ that are additive inverses.
::现在我们可以把第二个方程乘以3 来获得 x 的系数,而x 是添加反函数。

$\begin{aligned} 3 (- \frac{1}{3} x + y = 8) \Rightarrow - x + 3 y = 24 \end{aligned}$

::3(- 13x+y=8) x+3y=24

Now our system is
::现在我们的系统是

$\begin{aligned} x - 3 y = 5 \\ + \underline{- x + 3 y = 24} \\ 0 = 29 \end{aligned}$

::x-3y=5+-x+3y=24_0=29

When we add these equations together, both variables are eliminated and the result 0 = 29 is a false statement. Therefore, this system has no solution.
::当我们把这些方程加在一起时,两个变量都被删除,结果0=29是一个虚假的报表。因此,这个系统没有解决办法。

Feature: Quantum Computing
::特点:量子计算

by Tariq Ali
::塔里克·阿里(签名)

You have learned a few different methods for solving a system of linear equations with just two or three variables and equations. If you want, you can apply the same procedures to solve a larger number of equations. But how large of a system can you handle? Five equations? Ten? Twenty? How about a thousand, or a million? Well, time to bring your laptop out!
::您已经学会了几种不同的方法来解决一个只有两三个变量和方程式的线性方程式系统。如果您想要的话, 您可以应用同样的程序来解决更多的方程式。但是您能处理多大的系统? 5个方程式? 10个方程式? 20个方程式? 大约一千个或者100万个方程式? 那么, 该是时候拿出你的笔记本电脑了!

Solving linear systems of equations is a common problem that often arises with a very large number of equations. While systems of three or four equations can be readily solved by hand, computers are usually used to help solve larger systems. Computational algorithms for finding such solutions are an important part of numerical linear algebra.
::解决等式线性系统是一个常见问题,经常在大量方程式中出现。虽然三、四个方程式的系统可以通过人工手解答,但计算机通常用来帮助解决更大的系统。找到这些解决方案的计算算法是数字线性代数的一个重要部分。

What you have done by hand so far (eliminating, substituting, etc.) is what the computer does too. The difference is that computers store the numbers in the problem in array forms, known as matrices, and work based on these. The elimination and substitution methods, when applied on matrix representations, is known as the Gaussian elimination method . However, this is not the only method computers use; there are also methods based on matrix inversions or Cramer’s rule. Different methods have different speeds and different degrees of precision. (If you would like to learn more about matrices, see the related reads on the CK-12 site.)
::到目前为止,你亲手做的(除去、替换等)是计算机也做的事情。区别在于计算机以阵列形式存储问题中的数字, 称为矩阵, 并以此为基础工作。当在矩阵表示中应用时, 消除和替代方法被称为Gaussian消除方法。然而, 这不是唯一的计算机方法; 也有一些基于矩阵反转或Cramer规则的方法。不同的方法速度不同,精确度也不同。 (如果你想更多地了解矩阵,请见CK-12站点的相关读物。 )

The best-known classical computer algorithms require times that are proportional to the number of variables in a linear set of equations. But with rapidly growing data sets, keeping up with the enormous numerical burdens is becoming difficult. A recently proposed quantum algorithm shows that quantum supercomputers could solve linear systems at a speed exponentially better than traditional computers.
::最著名的经典计算机算法需要与一系列线性方程式中的变量数量成正比的时间。但是,随着数据集的迅速增长,要跟上巨大的数字负担正变得越来越困难。最近提出的量子算法表明,量子超级计算机能够以比传统计算机更快的速度解决线性系统。

by Australian National University (ANU) demonstrates how physicists at ANU have taken a step closer to quantum computing by stopping light in a new experiment.
::澳大利亚国立大学(ANU)的研究表明,ANU的物理学家如何通过在新的实验中停止光线,更加接近量子计算。

Summary
::摘要

To solve a system of equations by elimination by multiplication and addition, choose a variable to eliminate and find the LCM of the coefficients of that variable.
::通过乘法和加法消除等式系统,选择一个变量来消除和找到该变量系数的 LCM。

Then, multiply both sides of the equations to make the coefficient of the variable you want to eliminate the LCM. Make sure these terms have opposite signs, so they are additive inverses of each other.
::然后将方程的两边乘以使变量的系数成为您想要消除 LCM 的系数。确保这些词有相反的符号, 所以它们是相互的添加反函数。

Add the equations and solve for the variable if necessary.
::必要时添加变量的方程式和解析符。

Once you have found one value, substitute into either equation to find the value of the variable you choose to eliminate.
::一旦找到一个值, 将替换为任意一个方程, 以找到您选择删除的变量的值。

Review
::回顾

Solve the following systems using elimination by multiplication and addition.
::利用乘法和加法来消除下列系统。

1.

$\begin{aligned} \begin{aligned} x - 7 y & = 27 \\ 2 x + y & = 9 \end{aligned} \end{aligned}$

2.

$\begin{aligned} \begin{aligned} 10 x + y & = - 6 \\ - 7 x - 5 y & = - 13 \end{aligned} \end{aligned}$

3.

$\begin{aligned} \begin{aligned} 2 x + 6 y & = 8 \\ 3 x + 2 y & = - 23 \end{aligned} \end{aligned}$

4.

$\begin{aligned} \begin{aligned} 2 x + 4 y & = 24 \\ - 3 x - 2 y & = - 26 \end{aligned} \end{aligned}$

5.

$\begin{aligned} 8 x + 2 y = - 4 \\ 3 y = - 16 x + 2 \end{aligned}$

6.

$\begin{aligned} \begin{aligned} \frac{1}{3} x - \frac{2}{3} y & = - 8 \\ \frac{1}{2} x - \frac{1}{3} y & = 12 \end{aligned} \end{aligned}$

7.

$\begin{aligned} \begin{aligned} 6 x + 5 y & = 3 \\ - 4 x - 2 y & = - 14 \end{aligned} \end{aligned}$

8.

$\begin{aligned} \begin{aligned} 9 x - 7 y & = - 19 \\ 5 x - 3 y & = - 15 \end{aligned} \end{aligned}$

9.

$\begin{aligned} \begin{aligned} 15 x - 21 y = - 63 \\ 7 y = 5 x + 21 \end{aligned} \end{aligned}$

10.

$\begin{aligned} \begin{aligned} 3 x + 4 y & = - 16 \\ 5 x + 5 y & = - 5 \end{aligned} \end{aligned}$

11.

$\begin{aligned} \begin{aligned} 15 x + 2 y & = 23 \\ 18 x - 9 y & = - 18 \end{aligned} \end{aligned}$

12.

$\begin{aligned} \begin{aligned} 12 x + 8 y & = 64 \\ 17 x - 12 y & = 9 \end{aligned} \end{aligned}$

13.

$\begin{aligned} \begin{aligned} 11 x - 3 y & = 12 \\ 33 x - 36 & = 9 y \end{aligned} \end{aligned}$

14.

$\begin{aligned} \begin{aligned} - 6 x + 11 y & = - 109 \\ 8 x - 15 y & = 149 \end{aligned} \end{aligned}$

15.

$\begin{aligned} \begin{aligned} 8 x = - 5 y - 1 \\ - 32 x + 20 y = 8 \end{aligned} \end{aligned}$

Explore More
::探索更多

1. Lia is making a mixture of chlorine and water in her lab. She needs to make 13 ml of a 60% chlorine solution from a solution that is 35% chlorine and a second solution which is 75% chlorine. How many milliliters of each solution does she need?
::1. Lia正在实验室制造氯和水的混合物,她需要从35%氯的溶液和75%氯的第二个溶液中,制成60%氯溶液的13毫升,每个溶液需要多少毫升?

2. A one pound mix consisting of 30% cashews and 70% pistachios sells for $6.25. A one pound mix consisting of 80% cashews and 20% pistachios sells for $7.50. How much would a mix consisting of 50% of each type of nut sell for?
::2. 一磅组合,包括30%的腰果和70%的Pistachios售价6.25美元;一磅组合,包括80%的腰果和20%的pistachios售价7.50美元;每类坚果售价50%的组合,包括多少?

3. A mix of 35% almonds and 65% peanuts sells for $5.70. A mix of 75% almonds and 25% peanuts sells for $6.50. How much would a mix of 60% almonds and 40% peanuts sell for?
::3. 35%的杏仁和65%的花生以5.70美元售出,75%的杏仁和25%的花生以6.50美元售出,60%的杏仁和40%的花生的组合将卖出多少?

4. The Robinson family pays $19.75 at the movie theater for 3 medium popcorns and 4 medium drinks. The Jamison family pays $33.50 at the same theater for 5 medium popcorns and 7 medium drinks. How much would it cost for a couple to get 2 medium drinks and 2 medium popcorns?
::4. Robinson一家在电影院支付19.75美元,购买3个中度爆米花和4个中度饮料,Jamison一家在同一剧院支付33.50美元,购买5个中度爆米花和7个中度饮料,一对买2个中度饮料和2个中度爆米花要多少钱?

5. A cell phone company charges extra when users exceed their included call time and text message limits. One user paid $3.24 extra having talked for 240 extra minutes and sending 12 additional texts. A second user talked for 120 extra minutes and sent 150 additional texts and was charged $4.50 above the regular fee. How much extra would a user be charged for talking 140 extra minutes and sending 200 additional texts?
::5. 当用户超过包括通话时间和短信限制在内的电话用户时,手机公司收取额外费用;一名用户在多讲240分钟并发送12份新文本后,额外支付3.24美元;另一名用户多讲120分钟,多发150份新文本,并收取超出正常收费4.50美元;如果多讲140分钟并发送200份新文本,需要收取多少额外费用?

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:

章节大纲

Creating Additive Inverses ::正在创建补充反省

How to Find the Least Common Multiple ::如何找到最小常见多重

Example 1 ::例1

Solving Systems of Equations by Multiplication and Addition ::通过乘数和加法解决等号的解决系统

To Solve Systems of Equations by Multiplication and Addition ::通过乘法和加法解决等式系统

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

WARNING ::警告

Example 6 ::例6

Example 7 ::例7

Feature: Quantum Computing ::特点:量子计算

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

Creating Additive Inverses
::正在创建补充反省

How to Find the Least Common Multiple
::如何找到最小常见多重

Example 1
::例1

Solving Systems of Equations by Multiplication and Addition
::通过乘数和加法解决等号的解决系统

To Solve Systems of Equations by Multiplication and Addition
::通过乘法和加法解决等式系统

Example 2
::例2

Example 3
::例3

Example 4
::例4

Example 5
::例5

WARNING
::警告

Example 6
::例6

Example 7
::例7

Feature: Quantum Computing
::特点:量子计算

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