4.11 线性系统矩阵公式的编写和解决

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  线性系统矩阵公式的书写和解析

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  On Friday, an ice cream shop sold 15 small chocolate cones and 25 X-large chocolate cones. It also sold 20 small vanilla cones and 50 X-large vanilla cones. The shop's chocolate sales for the day totaled $220 and its vanilla sales totaled $410. How much did the shop charge for a small cone versus an X-large cone?
  ::星期五,一家冰淇淋店出售了15个小巧克力锥和25个X大巧克力锥,还出售了20个小香草锥和50个X大香草锥,该店当天的巧克力销售总额为220美元,香草销售总额为410美元。

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  Matrix Equations
  ::矩阵数

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  Look at the system:
  ::看看这个系统:

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  $$\begin{array}{rl}3x-5y& =11\\ 7x-y& =15.\end{array}$$

  
  ::3x-5y=117x-y=15。

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  Working backwards , this system can be written as the matrix equation :
  ::向后工作,这个系统可以写成矩阵等式:

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  $$\begin{array}{r}\left[\begin{array}{cc}3& -5\\ 7& -1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}11\\ 15\end{array}\right]\end{array}$$

  where the variables, $\begin{array}{r}x\end{array}$ and $\begin{array}{r}y\end{array}$ are the components of a variable matrix for which we can solve using an inverse as shown:
  ::[3-57-1][xy]=[1115],其中变量、 x 和 y 是变量矩阵的组成部分,我们可以用所示的反向解决:

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  $$\begin{array}{rl}\frac{1}{32}\left[\begin{array}{cc}-1& 5\\ -7& 3\end{array}\right]\left[\begin{array}{cc}3& -5\\ 7& -1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]& =\frac{1}{32}\left[\begin{array}{cc}-1& 5\\ -7& 3\end{array}\right]\left[\begin{array}{c}11\\ 15\end{array}\right]\\ \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]& =\frac{1}{32}\left[\begin{array}{c}64\\ -32\end{array}\right]\\ \left[\begin{array}{c}x\\ y\end{array}\right]& =\left[\begin{array}{c}2\\ -1\end{array}\right]\end{array}$$

  
  ::132[-15-73][3-57-1][xy]=132[-15-73][1115][1001][xy]=132[64-32][xy]=[2-1]

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  In general, any system of linear equations in two variables can be written as a matrix equation that can then be solved using inverses. The coefficients of $\begin{array}{r}x\end{array}$ and $\begin{array}{r}y\end{array}$ form a $\begin{array}{r}2\times 2\end{array}$ matrix and the constants form a $\begin{array}{r}2\times 1\end{array}$ matrix as shown below.
  ::一般而言,两个变量中的任何线性方程系统都可以写成矩阵方程,然后通过反向解答。如下文所示, x 和 y 的系数形成 2x2 矩阵,而常数形成 2x1 矩阵。

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  $$\begin{array}{rl}ax+by& =e\Rightarrow \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}e\\ f\end{array}\right]\\ cx+dy& =f\end{array}$$

  
  ::ax+by=e[abcd][xy]=[ef]cx+dy=f [abcd][xy]=[ef]cx+dy=f

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  If we let

  $$\begin{array}{r}A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right],X=\left[\begin{array}{c}x\\ y\end{array}\right]\end{array}$$

  and

  $$\begin{array}{r}B=\left[\begin{array}{c}e\\ f\end{array}\right],\end{array}$$

  then we could write the equation as $\begin{array}{r}AX=B\end{array}$ . Now we can see what happens when we multiply both sides by $\begin{array}{r}{A}^{-1}\end{array}$ :
  ::如果我们让 A = [abcd] 、 X = [xy] 和 B = [ef] , 我们就可以将方程式写成 AX= B 。 现在我们可以看到当我们把两边乘以 A- 1 时会发生什么 :

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  $$\begin{array}{rl}{A}^{-1}AX& =A^{-1}B\\ IX& =A^{-1}B\\ X& =A^{-1}B\end{array}$$

  
  ::A-1AX=A-1BIX=A-1BX=A-1BX=A-1BB

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  So, we can find the matrix, $\begin{array}{r}X\end{array}$ , by “left multiplying” the matrix, $\begin{array}{r}B\end{array}$ , by $\begin{array}{r}{A}^{-1}\end{array}$ .
  ::因此,我们能找到矩阵X,通过“左乘”矩阵B,A-1。

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  Solve the following systems using matrices.
  ::使用矩阵解决以下系统。

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  1.       
     $$\begin{array}{rl}3x+4y& =-14\\ -2x+y& =-9\end{array}$$
     
     ::3x+4y14-2x+y9

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  First, let’s translate the system into a matrix equation:
  ::首先,让我们把系统变成一个矩阵等式:

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  $$\begin{array}{r}\left[\begin{array}{cc}3& 4\\ -2& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-14\\ -9\end{array}\right]\end{array}$$

  
  ::[34-21][xy]=[-14-9]

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  Now we can “left multiply” both sides by the inverse of the coefficient matrix and solve:
  ::现在我们可以用系数矩阵的反面“左倍”解决:

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  $$\begin{array}{rl}\left[\begin{array}{c}x\\ y\end{array}\right]& =\frac{1}{11}\left[\begin{array}{cc}1& -4\\ 2& 3\end{array}\right]\left[\begin{array}{c}-14\\ -9\end{array}\right]\\ \left[\begin{array}{c}x\\ y\end{array}\right]& =\frac{1}{11}\left[\begin{array}{c}22\\ -55\end{array}\right]\\ \left[\begin{array}{c}x\\ y\end{array}\right]& =\left[\begin{array}{c}2\\ -5\end{array}\right]\end{array}$$

  
  ::[xy]=111[1-423][-14-9][xy]=111[22-55][xy]=[2-5]

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  Therefore , the solution is (2, -5).
  ::因此,解决办法是(2,5)。

  2.         

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  $$\begin{array}{rl}-5x-2y& =1\\ 3x+y& =-2\end{array}$$

  
  ::-5x-2y=13x+y2

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  First, let’s translate the system into a matrix equation:
  ::首先,让我们把系统变成一个矩阵等式:

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  $$\begin{array}{r}\left[\begin{array}{cc}-5& -2\\ 3& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}1\\ -2\end{array}\right]\end{array}$$

  
  ::[-5-231][xy]=[1-2]

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  Now we can “left multiply” both sides by the inverse of the coefficient matrix and solve:
  ::现在我们可以用系数矩阵的反面“左倍”解决:

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  $$\begin{array}{rl}\left[\begin{array}{c}x\\ y\end{array}\right]& =\frac{1}{1}\left[\begin{array}{cc}1& 2\\ -3& -5\end{array}\right]\left[\begin{array}{c}1\\ -2\end{array}\right]\\ \left[\begin{array}{c}x\\ y\end{array}\right]& =1\left[\begin{array}{c}-3\\ 7\end{array}\right]\\ \left[\begin{array}{c}x\\ y\end{array}\right]& =\left[\begin{array}{c}-3\\ 7\end{array}\right]\end{array}$$

  
  ::[xy]=11[12-3-3-5][1-2][xy]=1[-37][xy]=[-37]

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  Therefore, the solution is (-3, 7).
  ::因此,解决办法是(-3,7)。

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  3.      
     
     $$\begin{array}{rl}3x-2y& =3\\ -6x+4y& =5\end{array}$$
     
     ::3x-2y=3-6x+4y=5

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  First, let’s translate the system into a matrix equation:
  ::首先,让我们把系统变成一个矩阵等式:

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  $$\begin{array}{r}\left[\begin{array}{cc}3& -2\\ -6& 4\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}3\\ 5\end{array}\right]\end{array}$$

  
  ::[3-2--64][xy]=[35]

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  What happens this time when we try to find the inverse of the coefficient matrix?
  ::当我们试图找出系数矩阵的反面时,这一次会发生什么情况?

  $$\begin{array}{r}\frac{1}{0}\left[\begin{array}{cc}4& 2\\ 6& 3\end{array}\right]\end{array}$$

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  The inverse does not exist so there is no unique solution to the system. Now we must use an alternative method to determine whether there are infinite solutions or no solution. Let’s use linear combination:
  ::反之亦然,因此这个系统没有独特的解决方案。 现在我们必须使用替代方法来确定是否有无限的解决方案,还是没有解决方案。 让我们使用线性组合:

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  $$\begin{array}{rl}& 2(3x-2y=3)\Rightarrow \cancel{6x}-4y=6\\ & -6x+4y=5\underset{\_}{-\cancel{6x}+4y=5}\\ & 0=11\end{array}$$

  
  ::2(3x-2y=3) 6x-4y=6-6x+4y=5-6x+4y=5-6x+4y=5_0=11

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  Therefore, there is no solution.
  ::因此,没有解决办法。

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked to find the cost of a small cone and an X-large cone.
  ::早些时候,你被要求 寻找一个小锥体和一个X大锥体的成本。

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  The system of equations represented by this situation is
  ::这种状况所代表的方程式制度是:

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  $$\begin{array}{rl}15x+25y& =220\\ 20x+50y& =410\end{array}$$

  
  ::15x+25y=22020x+50y=410

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  First, let’s translate the system into a matrix equation:
  ::首先,让我们把系统变成一个矩阵等式:

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  $$\begin{array}{r}\left[\begin{array}{cc}15& 25\\ 20& 50\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}220\\ 410\end{array}\right]\end{array}$$

  
  ::[15252050][xy]=[220410]

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  Now we can “left multiply” both sides by the inverse of the coefficient matrix and solve:
  ::现在我们可以用系数矩阵的反面“左倍”解决:

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  $$\begin{array}{rl}\left[\begin{array}{c}x\\ y\end{array}\right]& =\frac{1}{250}\left[\begin{array}{cc}50& -25\\ -20& 15\end{array}\right]\left[\begin{array}{c}220\\ 410\end{array}\right]\\ \left[\begin{array}{c}x\\ y\end{array}\right]& =\frac{1}{250}\left[\begin{array}{c}750\\ 750\end{array}\right]\\ \left[\begin{array}{c}x\\ y\end{array}\right]& =\left[\begin{array}{c}3\\ 7\end{array}\right]\end{array}$$

  
  ::[xy]=1250[50-25-2015][220410][xy]=1250[750750][xy]=[37]

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  Therefore, the shop charges $3 for a small cone and $7 for an X-large cone.
  ::因此,商店为一个小锥体收取3美元,一个大锥体收取7美元。

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  Example 2
  ::例2

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  Solve the following system using matrices.
  ::使用矩阵表解决以下系统 。

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  $$\begin{array}{rl}2x+y& =-5\\ -3x-2y& =2\end{array}$$

  
  ::2x+y=2x5-3x-2y=2

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  Translate to a matrix equation and solve.
  ::转换为矩阵方程式和解析 。

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  $$\begin{array}{rl}\left[\begin{array}{cc}2& 1\\ -3& -2\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]& =\left[\begin{array}{c}-5\\ 2\end{array}\right]\\ \left[\begin{array}{c}x\\ y\end{array}\right]& =\frac{1}{-1}\left[\begin{array}{cc}-2& -1\\ 3& 2\end{array}\right]\left[\begin{array}{c}-5\\ 2\end{array}\right]\\ \left[\begin{array}{c}x\\ y\end{array}\right]& =-1\left[\begin{array}{c}8\\ -11\end{array}\right]\\ \left[\begin{array}{c}x\\ y\end{array}\right]& =\left[\begin{array}{c}-8\\ 11\end{array}\right]\end{array}$$

  
  ::[21-3-2][xy]=[-52][xy]=1-1[-2-2-132][-52][xy]_1[8-11][xy]=[-8-811][xy]=[-811]

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  Therefore, the solution is (-8, 11).
  ::因此,解决办法是(8,11)。

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  Example 3
  ::例3

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  Solve the following system using matrices.
  

  $$\begin{array}{rl}7x-y& =8\\ 3y& =21x-24\end{array}$$

  
  ::使用 7x-y=83y=21x-24 矩阵解决以下系统

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  First, rewrite the second equation as $\begin{array}{r}-21x+3y=-24\end{array}$ so the coefficient matrix can be identified.
  ::首先,将第二个方程重写为-21x+3y24,以便确定系数矩阵。

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  $$\begin{array}{r}\left[\begin{array}{cc}7& -1\\ -21& 3\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}8\\ -24\end{array}\right],\end{array}$$

  
  ::[7-1-213][xy]=[8-24],

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  However

  $$\begin{array}{r}{\left[\begin{array}{cc}7& -1\\ -21& 3\end{array}\right]}^{-1}\end{array}$$

  does not exist. Using linear combinations:
  ::然而,[7-1-213]-1并不存在。

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  $$\begin{array}{rl}& 3(7x-y=8)\Rightarrow \cancel{21x}-3y=24\\ & -21x+3y=-24\underset{\_}{-\cancel{21x}+3y=-24}\\ & 0=0\end{array}$$

  
  ::3(7x-y=8) 21x-3y=24-21x+3y}24-21x+3y24-21x+3y}24_0=0

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  Therefore, there are infinite solutions.
  ::因此,有无限的解决办法。

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  Example 4
  ::例4

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  Solve the following system using matrices.
  ::使用矩阵表解决以下系统 。

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  $$\begin{array}{rl}4x+5y& =5\\ -8x+15y& =5\end{array}$$

  
  ::4x+5y=5-8x+15y=5

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   Translate to a matrix equation and solve.
  ::转换为矩阵方程式和解析 。

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  $$\begin{array}{rl}\left[\begin{array}{cc}4& 5\\ -8& 15\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]& =\left[\begin{array}{c}5\\ 5\end{array}\right]\\ \left[\begin{array}{c}x\\ y\end{array}\right]& =\frac{1}{100}\left[\begin{array}{cc}15& -5\\ 8& 4\end{array}\right]\left[\begin{array}{c}5\\ 5\end{array}\right]\\ \left[\begin{array}{c}x\\ y\end{array}\right]& =\frac{1}{100}\left[\begin{array}{c}50\\ 60\end{array}\right]\\ \left[\begin{array}{c}x\\ y\end{array}\right]& =\left[\begin{array}{c}\frac{1}{2}\\ \frac{3}{5}\end{array}\right]\end{array}$$

  
  ::[45-815][xy]=[55][xy]=1100[15-584][55][xy]=1100[5060][xy]=[1235]

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  Therefore, the solution is $\begin{array}{r}\left(\frac{1}{2},\frac{3}{5}\right)\end{array}$ .
  ::因此,解决办法是(12 35)。

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  Review
  ::回顾

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  Solve the systems of linear equations using matrices.
  ::使用矩阵解决线性方程式系统。

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  1. $$\begin{array}{rl}2x+13y& =-29\\ 4x-3y& =29\end{array}$$
     
     ::2x+13y294x-3y=29
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  2. $$\begin{array}{rl}3x-7y& =98\\ -2x+5y& =-69\end{array}$$
     
     ::3x-7y=98-2x+5y69
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  3. $$\begin{array}{rl}7x-9y& =31\\ 10x+5y& =-45\end{array}$$
     
     ::7-9y=3110x+5y45
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  4. $$\begin{array}{rl}3x-2y& =1\\ 7x-5y& =-2\end{array}$$
     
     ::3x-2y=17x-5y2
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  5. $$\begin{array}{rl}3x-5y& =2\\ -9x+15y& =4\end{array}$$
     
     ::3x-5y=2-9x+15y=4
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  6. $$\begin{array}{rl}4x-3y& =-5\\ 12x+9y& =-3\end{array}$$
     
     ::4 - 3y512x+9y3
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  7. $$\begin{array}{rl}6x+9y& =-3\\ -5x-7y& =-1\end{array}$$
     
     ::6x+9y3-5x-7y1
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  8. $$\begin{array}{rl}x+4y& =7\\ -2x& =8y-14\end{array}$$
     
     ::x+4y=7-2x=8y-14
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  9. $$\begin{array}{rl}x+y& =45\\ -3x+2y& =-10\end{array}$$
     
     ::x+y=45-3x+2y10
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  10. $$\begin{array}{rl}4x+6y& =8\\ -9y& =2x-2\end{array}$$
      
      ::4x+6y=8-9y=2x-2
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  11. $$\begin{array}{rl}2x-y& =-30\\ x+2y& =10\end{array}$$
      
      ::2 - y% 30x+2y=10
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  12. $$\begin{array}{rl}14x-7y& =-1\\ 7x+21y& =10\end{array}$$
      
      ::14 - 7y% 17x+21y=10
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  13. $$\begin{array}{rl}-3x+4y& =5\\ 2x-y& =-10\end{array}$$
      
      ::-3x+4y=52x-y10
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  14. $$\begin{array}{rl}6x-12y& =-18\\ 5x-10y& =-15\end{array}$$
      
      ::6-12y185x-10y15
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  15. Tommy and Max start a lawn care business. They both charge by the hour for mowing and raking leaves. One week Tommy mowed lawns for 10 hours and raked leaves for 6 hours and earned a total of $114. In the same week, Max mowed lawns for 8 hours and raked leaves for 9 hours and earned a total of $118.50. Assuming they both charge the same rates for each of their services, how much do they charge per hour for mowing? For raking leaves?
      ::Tommy 和 Max 开始草坪护理业务。 两人按小时收取修剪和开垦叶的收费。 一个星期, Tommy 修剪草坪的收费为10小时, 6小时的修剪草坪, 总共挣到114美元。 同一周, Max 修剪草坪的收费为8小时, 9小时的修剪草坪, 总共赚到118.5美元。 假设他们每件服务都收取同样的收费, 那么他们每小时的修剪草坪的收费是多少? 开垦叶的收费是多少?

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  Review (Answers)
  ::回顾(答复)

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  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 。