Section: 具有乘法的线性系统 | 7.5 乘法线性系统

Section outline

Linear Systems with Multiplication
::具有乘法的线性系统

So far, we’ve seen that the elimination method works well when the coefficient of one variable happens to be the same (or opposite) in the two equations. But what if the two equations don’t have any coefficients the same?
::到目前为止,我们已经看到,当一个变量的系数在两个方程式中恰好相同(或相反)时,消除方法效果很好。但是,如果这两个方程式没有任何相同的系数呢?

It turns out that we can still use the elimination method; we just have to make one of the coefficients match. We can accomplish this by multiplying one or both of the equations by a constant .
::事实证明,我们仍然可以使用消除方法;我们只需要使一个系数匹配。我们可以通过将一个或两个方程式乘以一个常数来实现这一点。

Here’s a quick review of how to do that. Consider the following questions:
::以下是对如何做到这一点的快速回顾。考虑以下问题:

If 10 apples cost $5, how much would 30 apples cost?
::如果10个苹果要5美元 30个苹果要多少钱?

If 3 bananas plus 2 carrots cost $4, how much would 6 bananas plus 4 carrots cost?
::如果3个香蕉加2个胡萝卜要4美元 6个香蕉加4个胡萝卜要多少钱?

If you look at the first equation , it should be obvious that each apple costs $0.50. So 30 apples should cost $15.00.
::如果你看一下第一个方程,应该很明显每个苹果要花0.50美元。所以30个苹果要花15美元。

The second equation is trickier; it isn’t obvious what the individual price for either bananas or carrots is. Yet we know that the answer to question 2 is $8.00. How?
::第二个等式更狡猾;香蕉或胡萝卜的个别价格并不明显。但我们知道问题2的答案是8.00美元。如何?

If we look again at question 1, we see that we can write an equation: $\begin{aligned} 10 a = 5 \end{aligned}$ ( $\begin{aligned} a \end{aligned}$ being the cost of 1 apple). So to find the cost of 30 apples, we could solve for $\begin{aligned} a \end{aligned}$ and then multiply by 30—but we could also just multiply both sides of the equation by 3. We would get $\begin{aligned} 30 a = 15 \end{aligned}$ , and that tells us that 30 apples cost $15.
::如果我们再看一下问题1,我们可以看到我们可以写一个方程:10a=5(一个苹果的成本是1个苹果 ) 。因此,为了找到30个苹果的成本,我们可以解决一个,然后乘以30 — — 但我们也可以将方程的两边乘以3。我们会得到30a=15, 这告诉我们30个苹果的成本是15美元。

And we can do the same thing with the second question. The equation for this situation is $\begin{aligned} 3 b + 2 c = 4 \end{aligned}$ , and we can see that we need to solve for $\begin{aligned} (6 b + 4 c) \end{aligned}$ , which is simply 2 times $\begin{aligned} (3 b + 2 c) \end{aligned}$ ! So algebraically, we are simply multiplying the entire equation by 2:
::我们可以对第二个问题做同样的事情。这个情况的方程式是 3b+2c=4, 我们可以看到我们需要解决 (6b+4c), 也就是2倍(3b+2c)! 代数如此之大, 我们只是将整个方程式乘以 2:

$\begin{aligned} 2 (3 b + 2 c) & = 2 \cdot 4 & d i s t r i b u t e a n d m u l t i p l y : \\ 6 b + 4 c & = 8 \end{aligned}$

::2(3b+2c)=24分配和乘数:6b+4c=8

So when we multiply an equation, all we are doing is multiplying every term in the equation by a fixed amount.
::所以当我们乘以一个方程时, 我们所做的就是将方程中的每一个条件乘以一个固定的金额。

Solving a Linear System by Multiplying One Equation
::以乘法一等法解决线性系统

If we can multiply every term in an equation by a fixed number (a scalar ), that means we can use the addition method on a whole new set of linear systems. We can manipulate the equations in a system to ensure that the coefficients of one of the variables match.
::如果我们能在方程式中将每个术语乘以一个固定数字( scalar) , 这意味着我们可以在全套新的线性系统中使用添加法。我们可以在一个系统中操纵方程式, 以确保变量之一的系数匹配。

This is easiest to do when the coefficient as a variable in one equation is a multiple of the coefficient in the other equation.
::当一个方程式中作为变量的系数是另一个方程式中系数的倍数时,这是最容易做到的。

Solve the system:
::解决系统:

$\begin{aligned} 7 x + 4 y & = 17 \\ 5 x - 2 y & = 11 \end{aligned}$

::7x+4y=175x-2y=11

You can easily see that if we multiply the second equation by 2, the coefficients of $\begin{aligned} y \end{aligned}$ will be +4 and -4, allowing us to solve the system by addition:
::如果将第二个方程乘以2, y的系数将为+4和4, 这样我们就可以通过加法解决这个系统:

2 times equation 2:
::2 次方程 2 :

$\begin{aligned} 10 x - 4 y = 22 & n o w a d d t o e q u a t i o n o n e : \\ \underline{+ (7 x + 4 y) = 17} \\ 17 x = 34 \\ d i v i d e b y 17 t o g e t : x = 2 \end{aligned}$

::10 - 4y=22 现在添加到方程一 : + (7x+4y) = 17_ 17_ 17x= 34divide 乘以 17 , 获得: x=2

Now simply substitute this value for $\begin{aligned} x \end{aligned}$ back into equation 1:
::现在简单地用此值替换 x 返回方程 1 :

$\begin{aligned} 7 \cdot 2 + 4 y & = 17 & s i n c e 7 \times 2 = 14, s u b t r a c t 14 f r o m b o t h s i d e s : \\ 4 y & = 3 & d i v i d e b y 4 : \\ y & = 0.75 \end{aligned}$

::72+4y=17自 7x2=14 起,减去 14, 双方: 4y= 3divide 4:y= 0.75

Converting a Word Problem into a System of Equations
::将单词问题转换成等式系统

Anne is rowing her boat along a river. Rowing downstream, it takes her 2 minutes to cover 400 yards. Rowing upstream, it takes her 8 minutes to travel the same 400 yards. If she was rowing equally hard in both directions, calculate, in yards per minute, the speed of the river and the speed Anne would travel in calm water.
::游到下游,她需要2分钟才能覆盖400码。在上游,她需要8分钟才能在400码之间行驶。如果她在双向行驶时同样坚硬,则计算每分钟的河速和安妮在平静的水中行走的速度。

Step one: first we convert our problem into equations. We know that distance traveled is equal to speed $\begin{aligned} \times \end{aligned}$ time . We have two unknowns, so we’ll call the speed of the river $\begin{aligned} x \end{aligned}$ , and the speed that Anne rows at $\begin{aligned} y \end{aligned}$ . When traveling downstream, her total speed is her rowing speed plus the speed of the river, or $\begin{aligned} (x + y) \end{aligned}$ . Going upstream, her speed is hindered by the speed of the river, so her speed upstream is $\begin{aligned} (x - y) \end{aligned}$ .
::第一步:我们首先将问题转换成方程式。我们知道所走的距离与速度x时间相等。我们有两个未知点, 因此我们称其为河速x, 以及Anne排在 y 的速度。当她下游旅行时, 她的总速度是她的划行速度加上河速或(x+y ) 。上上游时,她的速度受到河速的阻碍, 因此她的上游速度是(x-y ) 。

Downstream Equation: $\begin{aligned} 2 (x + y) = 400 \end{aligned}$
::下游等量 : 2( x+y) = 400

Upstream Equation: $\begin{aligned} 8 (x - y) = 400 \end{aligned}$
::上流等量 : 8(x-y) = 400

Distributing gives us the following system:
::发行给我们的系统如下:

$\begin{aligned} 2 x + 2 y & = 400 \\ 8 x - 8 y & = 400 \end{aligned}$

::2x+2y=4008x-8y=400

Right now, we can’t use the method of elimination because none of the coefficients match. But if we multiplied the top equation by 4, the coefficients of $\begin{aligned} y \end{aligned}$ would be +8 and -8. Let’s do that:
::现在,我们无法使用消除方法,因为任何系数都没有匹配。但如果我们乘以顶等方乘以4, y的系数就是+8和8。让我们这样做:

$\begin{aligned} 8 x + 8 y = 1, 600 \\ \underline{+ (8 x - 8 y) = 400} \\ 16 x = 2, 000 \end{aligned}$

::8x+8y=1 600+(8x-8y)=400_16x=2 000

Now we divide by 16 to obtain $\begin{aligned} x = 125 \end{aligned}$ .
::现在,我们除以16 获得 x=125。

Substitute this value back into the first equation:
::将此值替换回第一个方程 :

$\begin{aligned} 2 (125 + y) & = 400 & d i v i d e b o t h s i d e s b y 2 : \\ 125 + y & = 200 & s u b t r a c t 125 f r o m b o t h s i d e s : \\ y & = 75 \end{aligned}$

::2(125+y) = 400divivide 双方减少 2 125+y= 200 减号125 双方减少:y= 75

Anne rows at 125 yards per minute, and the river flows at 75 yards per minute.
::安妮以每分钟125码的速度排行,河水以每分钟75码的速度流行。

Solving a Linear System by Multiplying Both Equations
::以乘法两个等数解决线性系统

So what do we do if none of the coefficients match and none of them are simple multiples of each other? We do the same thing we do when we’re adding fractions whose denominators aren’t simple multiples of each other. Remember that when we add fractions, we have to find a lowest common denominator —that is, the lowest common multiple of the two denominators—and sometimes we have to rewrite not just one, but both fractions to get them to have a common denominator. Similarly, sometimes we have to multiply both equations by different constants in order to get one of the coefficients to match.
::那么,如果我们没有一种系数匹配,而其中任何一个系数都不是简单的相互多重,我们该怎么办呢?当我们增加分数时,当分母不是简单的相互多重时,我们也这样做。记住,当我们增加分数时,我们必须找到一个最低的共同分母 — — 即两个分母中最小的共同倍数 — — 有时我们不仅要重写一个,而且要重写两个分数才能让它们有一个共同分母。同样,我们有时不得不用不同的常数乘两个等式,才能让一个系数相匹配。

Andrew and Anne both use the I-Haul truck rental company to move their belongings from home to the dorm rooms on the University of Chicago campus. I-Haul has a charge per day and an additional charge per mile. Andrew travels from San Diego, California, a distance of 2060 miles in five days. Anne travels 880 miles from Norfolk, Virginia, and it takes her three days. If Anne pays $840 and Andrew pays $1845, what does I-Haul charge
::Andrew和Anne都使用I-Haul卡车租赁公司把他们的物品从家里搬到芝加哥大学校园的宿舍。I-Haul每天收费,每英里额外收费。Andrew从加利福尼亚州圣地亚哥旅行,五天内距离2060英里。Anne从弗吉尼亚州诺福克旅行880英里,需要三天时间。如果Anne支付840美元,Andrew支付1845美元,那么I-Haul收取什么费用?

a) per day?
:a) 每天?

b) per mile traveled?
:b) 每行驶的里程?

Solution
::解决方案

First, we’ll set up our equations. Again we have 2 unknowns: the daily rate (we’ll call this $\begin{aligned} x \end{aligned}$ ), and the per-mile rate (we’ll call this $\begin{aligned} y \end{aligned}$ ).
::首先,我们将设置我们的方程式。再说一遍,我们有两个未知数:每日费率(我们称之为x)和每英里费率(我们称之为y ) 。

Anne’s equation: $\begin{aligned} 3 x + 880 y = 840 \end{aligned}$
::安妮的方程式: 3x+880y=840

Andrew’s Equation: $\begin{aligned} 5 x + 2060 y = 1845 \end{aligned}$
::安德鲁的方程式: 5x+2060y=1845

We can’t just multiply a single equation by an integer number in order to arrive at matching coefficients. But if we look at the coefficients of $\begin{aligned} x \end{aligned}$ (as they are easier to deal with than the coefficients of $\begin{aligned} y \end{aligned}$ ), we see that they both have a common multiple of 15 (in fact 15 is the lowest common multiple ). So we can multiply both equations.
::我们不能仅仅将一个单方程式乘以一个整数来达到匹配系数。但如果我们看一下 x 的系数(因为它们比y 的系数更容易处理 ) , 我们就会发现它们都有一个共同的乘数为 15 ( 事实上, 15 是最小的公数倍数 ) 。因此我们可以将两个方程式乘以两个方程式。

Multiply the top equation by 5:
::乘以顶等式 5 :

$\begin{aligned} 15 x + 4400 y = 4200 \end{aligned}$

::15x+4400y=4200

Multiply the lower equation by 3:
::乘以下方方程乘以 3 :

$\begin{aligned} 15 x + 6180 y = 5535 \end{aligned}$

::15x+6180y=5535

Subtract:
::减:

$\begin{aligned} 15 x + 4400 y = 4200 \\ \underline{- (15 x + 6180 y) = 5535} \\ - 1780 y = - 1335 \\ Divide by - 1780 : y = 0.75 \end{aligned}$

::15x+4400y=4200 - (15x+6180y)=5535_ - 1780y1335Divide by - 1780: y=0.75

Substitute this back into the top equation:
::把这个换回顶级方程式

$\begin{aligned} 3 x + 880 (0.75) & = 840 \\ 3 x & = 180 \\ x & = 60 \end{aligned}$

::3x+880(0.75)=8403x=180x=60

I-Haul charges $60 per day plus $0.75 per mile.
::I -Haul每天收费60美元加上每英里0.75美元。

Example
::示例示例示例示例

Example 1
::例1

Solve the system $\begin{aligned} {\begin{cases} 4 x + 7 y = 6 \\ 6 x + 5 y = 20 \end{cases} \end{aligned}$ .
::解决系统 {4x+7y=66x+5y=20。

Neither $\begin{aligned} x \end{aligned}$ nor $\begin{aligned} y \end{aligned}$ have additive inverse coefficients, but the $\begin{aligned} x \end{aligned}$ -variables do share a common factor of 2. Thus we can most easily eliminate $\begin{aligned} x \end{aligned}$ .
::x和y都没有添加的反系数,但X变量的确有一个共同的2系数。因此,我们最容易消除x。

In order to make $\begin{aligned} x \end{aligned}$ have the same coefficient in each equation, we must multiply one equation by the factor not shared in the coefficient of $\begin{aligned} x \end{aligned}$ in the other equation. We need to multiply the first equation by 3 and the second equation by 2, making one of them negative as well:
::为了让 x 在每个方程中具有相同的系数, 我们必须将一个方程乘以在另一个方程中 x 系数中未分享的系数。我们需要将第一个方程乘以3, 第二个方程乘以2, 使其中之一为负数 :

$\begin{aligned} {\begin{cases} 3 (4 x + 7 y = 6) \\ - 2 (6 x + 5 y = 20) \end{cases} & \to {\begin{cases} 12 x + 21 y = 18 \\ - 12 x - 10 y = - 40 \end{cases} \end{aligned}$

::{3( 4x+7y=6)- 2( 6x+5y=20)\\\ 12x+21y=18- 12x- 10y=40

$\begin{aligned} Add the two equations. & 11 y & = - 22 \\ Divide by 11. & y & = - 2 \end{aligned}$

::添加两个方程式 11 y 22 Divide 乘 11. y 2

To find the $\begin{aligned} x - \end{aligned}$ value, use the Substitution Property in either equation.
::为了找到 x - 值,在任一方程中使用替代财产。

$\begin{aligned} 4 x + 7 (2) & = 6 \\ 4 x + 14 & = 6 \\ 4 x & = - 8 \\ x & = - 2 \end{aligned}$

::4x+7(2)=64x+14=64x=64x=8x=2

The solution to this system is $\begin{aligned} (- 2, - 2) \end{aligned}$ .
::这一系统的解决办法是(-2,-2002)。

Explore More
::探索更多

Solve the following systems using multiplication.
::使用乘法解决下列系统。

$\begin{aligned} 5 x - 10 y = 15 \\ 3 x - 2 y = 3 \end{aligned}$

::5x-10y=153x-2yy=3

$\begin{aligned} 5 x - y = 10 \\ 3 x - 2 y = - 1 \end{aligned}$

::5-y=103x-2y1

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

::5x+7y=157x-3y=5

$\begin{aligned} 9 x + 5 y = 9 \\ 12 x + 8 y = 12.8 \end{aligned}$

::9x+5y=912x+8y=12.8

$\begin{aligned} 4 x - 3 y = 1 \\ 3 x - 4 y = 4 \end{aligned}$

::4x-3y=13x-4y=4

$\begin{aligned} 7 x - 3 y = - 3 \\ 6 x + 4 y = 3 \end{aligned}$

::7-3y36x+4y=3

$\begin{aligned} x = 3 y \\ x - 2 y = - 3 \end{aligned}$

::x=3yx-2y3

$\begin{aligned} y = 3 x + 2 \\ y = - 2 x + 7 \end{aligned}$

::y=3x+2y=2x+7 y=3x+2y=2x+7

$\begin{aligned} 5 x - 5 y = 5 \\ 5 x + 5 y = 35 \end{aligned}$

::5x-5y=55x+5y=35

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

::y3x- 33x- 2y+12=0

$\begin{aligned} 3 x - 4 y = 3 \\ 4 y + 5 x = 10 \end{aligned}$

::3x-4y=34y+5x=10

$\begin{aligned} 9 x - 2 y = - 4 \\ 2 x - 6 y = 1 \end{aligned}$

::9 - 2y% 42x-6y=1

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。

Section outline

Linear Systems with Multiplication ::具有乘法的线性系统

Converting a Word Problem into a System of Equations ::将单词问题转换成等式系统

Example ::示例示例示例示例

Example 1 ::例1

Explore More ::探索更多

Review (Answers) ::回顾(答复)

Linear Systems with Multiplication
::具有乘法的线性系统

Converting a Word Problem into a System of Equations
::将单词问题转换成等式系统

Example
::示例示例示例示例

Example 1
::例1

Explore More
::探索更多

Review (Answers)
::回顾(答复)