3.12 三个变量中的溶解线性线性系统

Section outline

• Select section General

  Collapse Expand

  General

  Collapse all Expand all

  • Select activity 新闻通告

    

    新闻通告 Forum
• Select section 三个变量中的解决线性系统

  Collapse Expand

  三个变量中的解决线性系统

  播放段落

  You want to make a fruit salad for a summer picnic. Three pounds of strawberries plus five pounds of grapes plus one pound of melon cost $20. Three pounds of strawberries plus two pounds of grapes plus two pounds of melon cost $21. Four pounds of strawberries plus three pounds of grapes plus three pounds of melon cost $30. How much does each fruit cost?
  ::3磅草莓加5磅葡萄加1磅瓜 3磅草莓加2磅葡萄加2磅瓜 21美元 4磅草莓加3磅葡萄加3磅瓜 30美元

  播放段落

  Linear Systems in Three Variables
  ::三个变量中的线性系统

  播放段落

  An equation in three variables, such as $\begin{array}{r}2x-3y+4z=10\end{array}$ , is an equation of a plane in three dimensions . In other words, this equation expresses the relationship between the three coordinates of each point on a plane. The solution to a system of three equations in three variables is a point in space which satisfies all three equations. When we add a third dimension we use the variable , $\begin{array}{r}z\end{array}$ , for the third coordinate. For example, the point (3, -2, 5) would be $\begin{array}{r}x=3,y=-2\end{array}$ and $\begin{array}{r}z=5\end{array}$ . A solution can be verified by substituting the $\begin{array}{r}x,y,\end{array}$ and $\begin{array}{r}z\end{array}$ values into the equations to see if they are valid.
  ::三个变量中的方程式, 如 2x-3y+4z=10, 是平面的三个维度的方程式。 换句话说, 这个方程式表示平面上每个点的三个坐标之间的关系。 三个变量中三个方程式的解答是满足所有三个方程式的空间点。 当我们添加第三个维度时, 我们使用第三个维度, z。 例如, 点( 3, 2, 5) 将是 x=3, y2 和 z=5 。 一个解决方案可以通过在方程式中替换 x, y 和 z 值来验证它们是否有效 。

  播放段落

  A system of three equations in three variables consists of three planes in space. These planes could intersect with each other or not as shown in the diagrams below.
  ::三个变量中的三个方程式系统由三平面组成。这些方程式可以相互交叉,也可以不相互交叉,如下图所示。

  

  播放段落
  • In the first diagram, the three planes intersect at a single point and thus a unique solution exists and can be found.
    ::在第一个图表中,三平面在一个点交叉,因此存在一个独特的解决办法,可以找到。
  播放段落
  • The second diagram illustrates one way that three planes can exist and there is no solution to the system. It is also possible to have three parallel planes or two that are parallel and a third that intersects them. In any of these cases, there is no point that is in all three planes.
    ::第二个图解说明了三架飞机存在的一种方式,即三架飞机无法解决系统问题。 也可以有三架平行飞机或两架平行飞机,第三架飞机相互交错。 在任何一种情况下,这三架飞机都没有任何意义。
  播放段落
  • The third diagram shows three planes intersecting in a line. Every point on this line is a solution to the system and thus there are infinite solutions.
    ::第三张图表显示三平面在一行中交叉。这条线上的每一个点都是系统的解决办法,因此有无限的解决办法。

  播放段落

  To solve a system of three equations in three variables, we will be using the linear combination method. This time we will take two equations at a time to eliminate one variable and using the resulting equations in two variables to eliminate a second variable and solve for the third. This is just an extension of the linear combination procedure used to solve systems with two equations in two variables.
  ::要在三个变量中解决三个方程式的系统, 我们将会使用线性组合法。 这一次我们每次要用两个方程式去掉一个变量, 然后用两个变量中产生的方程式去掉第二个变量, 然后解决第三个变量。 这只是用来解决两个变量中两个方程式的系统的线性组合程序的延伸 。

  播放段落

  Let's determine whether the point, (6, -2, 5), is a solution to the system:
  ::让我们来决定点, (6, 2, 5) 是否是系统的解决办法:

  播放段落

  $$\begin{array}{rl}x-y+z& =13\\ 2x+5y-3z& =-13\\ 4x-y-6z& =-4\end{array}$$

  
  ::x- y+z=132x+5y- 3z- 134x- y-6z4

  播放段落

  In order for the point to be a solution to the system, it must satisfy each of the three equations.
  ::要使这一点成为系统的解决办法,它必须满足三个方程式中的每一个方程式。

  播放段落

  First equation: $\begin{array}{r}(6)-(-2)+(5)=6+2+5=13\end{array}$
  ::第一等式6)-(-2)+(5)=6+2+5=13

  播放段落

  Second equation: $\begin{array}{r}2(6)+5(-2)-3(5)=12-10-15=-13\end{array}$
  ::第二个方程: 2(6)+5(-2)-3(5)=12-10-1513

  播放段落

  Third equation: $\begin{array}{r}4(6)-(-2)-6(5)=24+2-30=-4\end{array}$
  ::第三等式:4(6)-(-2)-6(5)=24+2-304

  播放段落

  The point, (6, -2, 5), satisfies all three equations. Therefore , it is a solution to the system.
  ::点(6,2,5)满足了所有三个方程式,因此,这是系统的解决办法。

  播放段落

  Now, let's solve the following systems using linear combinations.
  ::现在,让我们用线性组合解决以下系统。

  1.     

  播放段落

  $$\begin{array}{rl}2x+4y-3z& =-7\\ 3x-y+z& =20\\ x+2y-z& =-2\end{array}$$

  
  ::2x+4y-3z73x-y+z=20x+2y-z%2

  播放段落

  We can start by taking two equations at a time and eliminating the same variable. We can take the first two equations and eliminate $\begin{array}{r}z\end{array}$ , then take the second and third equations and also eliminate $\begin{array}{r}z\end{array}$ .
  ::我们可以从一次两个方程式开始,同时消除同一个变量。 我们可以采取前两个方程式,删除z,然后采取第二和第三个方程式,同时消除z。

  播放段落

  $$\begin{array}{rl}& 2x+4y-3z=-7\Rightarrow 2x+4y-\cancel{3z}=-7\\ & 3(3x-y+z=20)\underset{\_}{9x-3y+\cancel{3z}=60}\\ & 11x+y=53\end{array}$$

  
  ::2x+4y-3z72x+4y-3z73(3x-y+z=20) 9x-3y+3z=60_11x+y=53

  播放段落

  Result from equations 1 and 2: $\begin{array}{r}11x+y=53\end{array}$
  ::方程1和2: 11x+y=53

  播放段落

  $$\begin{array}{rl}& 3x-y+\cancel{z}=20\\ & \underset{\_}{x+2y-\cancel{z}=-2}\\ & 4x+y=18\end{array}$$

  
  ::3 - y+z=20x+2y - z2 - 4x+y=18 3 - y+z=20x+2y - z2 - 4x+y=18

  播放段落

  Result from equations 2 and 3: $\begin{array}{r}4x+y=18\end{array}$
  ::方程2和3:4x+y=18的结果

  播放段落

  Now we have reduced our system to two equations in two variables. We can eliminate $\begin{array}{r}y\end{array}$ most easily next and solve for $\begin{array}{r}x\end{array}$ .
  ::现在我们把我们的系统简化为两个变数中的两个方程式。 我们可以最容易地删除下一个 y , 然后解决 x 。

  播放段落

  $$\begin{array}{rl}& 11x+y=53\Rightarrow 11x+\cancel{y}=53\\ & -1(4x+y=18)\underset{\_}{-4x-\cancel{y}=-18}\\ & 7x=35\\ & x=5\end{array}$$

  
  ::11x+y=5311x+y=53-1(4x+y=18)-4x-y18_7x=35 x=5

  播放段落

  Now use this value to find $\begin{array}{r}y\end{array}$ :
  ::现在使用此值查找 y:

  播放段落

  $$\begin{array}{rl}4(5)+y& =18\\ 20+y& =18\\ y& =-2\end{array}$$

  
  ::4(5)+y=1820+y=18y @%2

  播放段落

  Finally, we can go back to one of the original three equations and use our $\begin{array}{r}x\end{array}$ and $\begin{array}{r}y\end{array}$ values to find $\begin{array}{r}z\end{array}$ .
  ::最后,我们可以回到原来的三个方程中的一个, 用我们的x值和y值来找到z。

  播放段落

  $$\begin{array}{rl}2(5)+4(-2)-3z& =-7\\ 10-8-3z& =-7\\ 2-3z& =-7\\ -3z& =-9\\ z& =3\end{array}$$

  
  ::2(5)+4(-2)-3z710-8-3z72-3z7-3z7-3z9z=3

  播放段落

  Therefore the solution is (5, -2, 3).
  ::因此,解决办法是5,2,3。

  播放段落

  Don’t forget to check your answer by substituting the point into each equation.
  ::不要忘了检查你的答案, 将点换成每个方程式。

  播放段落

  Equation 1: $\begin{array}{r}2(5)+4(-2)-3(3)=10-8-9=-7\end{array}$
  ::等式1: 2(5)+4(-2-2)-3(3)=10-8-97

  播放段落

  Equation 2: $\begin{array}{r}3(5)-(-2)+(3)=15+2+3=20\end{array}$
  ::等式2: 3(5)-(-2)+(3)=15+2+3=20

  播放段落

  Equation 3: $\begin{array}{r}(5)+2(-2)-(3)=5-4-3=-2\end{array}$
  ::等式35)+2(-2)-(3)=5-4-32

  2.     

  播放段落

  $$\begin{array}{rl}x+y+z& =5\\ 5x+5y+5z& =20\\ 2x+3y-z& =8\end{array}$$

  
  ::x+y+z=55x+5y+5z=202x+3y-z=8

  播放段落

  We can start by combining equations 1 and 2 together by multiplying the first equation by -5.
  ::我们可以将方程1和2结合起来, 将第一个方程乘以 -5。

  播放段落

  $$\begin{array}{rl}& -5(x+y+z=5)\Rightarrow -5x-5y-5z=-25\\ & 5x+5y+5z=20\underset{\_}{5x+5y+5z=20}\\ & 0=-5\end{array}$$

  
  ::5(x+y+z=5) 5(x+y+z=5) 5(5) 5(5) 5(5) 5(5) 5(5) 5(5) 5(5) 5(5) 5(5) 5(5) 5(5) 5(5) 5(5) 5(5) 5(5) 5(5) 5(5) 5(5) 5(5) 5(5) 5(5) 5(5) 5(5) 5(5) 5(5) 5(5) 20(0) 5(5)

  播放段落

  Since the result is a false equation, there is no solution to the system. If you end up with 0 = 0, then there will be infinitely many solutions.
  ::由于结果是一个虚假的等式, 系统就没有解决方案。 如果您以 0 = 0 结尾, 那么将会有很多解决方案 。

  播放段落

  Examples
  ::实例

  播放段落

  Example 1
  ::例1

  播放段落

  Earlier, you were asked to find the price of each individual fruit.
  ::早些时候,有人要求你寻找每个水果的价格。

  播放段落

  The system of linear equations represented by this situation is:
  ::以这种情况为代表的线性方程式系统是:

  播放段落

  $$\begin{array}{r}2s+5g+m=20\\ 3s+2g+2m=21\\ 4s+3g+3m=30\end{array}$$

  
  ::2s+5g+m=203s+2g+2m=214s+3g+3m=30

  播放段落

  The easiest way to solve this system is to first solve $\begin{array}{r}2s+5g+m=20\end{array}$ for m . If we do so, we get:
  ::最容易解决这个系统的方法是首先解决 m 的 2s+5g+m=20。 如果我们这样做, 我们就会得到:

  播放段落

  $\begin{array}{r}m=20-2s-5g\end{array}$
  ::m=20-2s-5g

  播放段落

  Now we can substitute this value for m into the other two equations. When we do so we get a new system of linear equations:
  ::现在我们可以用这个值代替 m, 换成另外两个方程式。 当我们这样做的时候, 我们得到一个新的线性方程式系统 :

  播放段落

  $$\begin{array}{r}3s+2g+2(20-2s-5g)=21\\ 4s+3g+3(20-2s-5g)=30\end{array}$$

  
  ::3s+2g+2(20-20-2s-5g)=214s+3g+3(20-2s-5g)=30

  播放段落

  Simplifying both equations results in:
  ::简化两个方程式的结果如下:

  播放段落

  $$\begin{array}{r}-s-8g=-19\\ -2s-12g=-30\end{array}$$

  
  ::-s -8g19 -2s -12g30

  播放段落

  If we multiply the first of these equations by –2, we get the new system of equations :
  ::如果我们把第一个方程式乘以 -2, 我们就会得到新的方程式系统:

  播放段落

  $$\begin{array}{r}2s+16g=38\\ -2s-12g=-30\end{array}$$

  
  ::2s+16g=38-2s-12g30

  播放段落

  Now we can add these two equations to eliminate the s variable. When we do so, we get g = 2.
  ::现在我们可以添加这两个方程式来消除 s 变量。 当我们这样做时, 我们得到 g = 2 。

  播放段落

  Next, we can substitute this value of g into either of these two equations to get the value of s :
  ::接下来,我们可以用这个g值来替代这两个方程式中的任何一个方程式 来获得 s值:

  播放段落

  $\begin{array}{r}-2s-12(2)=-30\end{array}$ results in s = 3.
  ::-2s-12(2)30结果=3。

  播放段落

  Finally we substitute these values for g and s into one of our original equations.
  ::最后,我们用这些数值来替代g和s, 来替代我们原先的方程。

  播放段落

  $\begin{array}{r}2(3)+5(2)+m=20\end{array}$ results in m = 4
  ::2(3)+5(2)+m=20 结果m=4

  播放段落

  Therefore, strawberries cost $3 per pound, grapes cost $2 per pound, and melon costs $4 per pound.
  ::因此草莓每磅3美元,葡萄每磅2美元,瓜每磅4美元。

  播放段落

  Example 2
  ::例2

  播放段落

  Determine if the point (-3, 2, 1) is the solution to the following system:
  ::确定点(-3,2,1)是否是以下系统的解决办法:

  播放段落

  $$\begin{array}{rl}x+y+z& =0\\ 4x+5y+z& =-1\\ 3x+2y-4z& =-8\end{array}$$

  
  ::x+y+z=04x+5y+z13x+2y_4z_8

  播放段落

  Check to see if the point satisfies all three equations.
  ::检查点是否满足所有三个方程式。

  播放段落

  Equation 1: $\begin{array}{r}(-3)+(2)+(1)=-3+2+1=0\end{array}$
  ::等式 1: (-3) +(2) +(1) + 3+2+1=0

  播放段落

  Equation 2: $\begin{array}{r}4(-3)+5(2)+(1)=-12+10+1=-1\end{array}$
  ::等式 2: 4(-3)+5(2)+(1)12+10+1+1

  播放段落

  Equation 3: $\begin{array}{r}3(-3)+2(2)-4(1)=-9+4-4=-9\ne -8\end{array}$
  ::等量 33-3)、+2(2)-4(1)、9+4-4-4(9)8

  播放段落

  Since the third equation is not satisfied by the point, the point is not a solution to the system.
  ::由于第三种方程式不能满足这一点,因此问题不是系统的解决办法。

  播放段落

  Example 3
  ::例3

  播放段落

  Solve the following system using linear combinations:
  ::使用线性组合解决以下系统:

  播放段落

  $$\begin{array}{rl}5x-3y+z& =-1\\ x+6y-4z& =-17\\ 8x-y+5z& =12\end{array}$$

  
  ::5-3y+z1x+6y-4z178x-y+5z=12

  播放段落

  Combine the first and second equations to eliminate $\begin{array}{r}z\end{array}$ . Then combine the first and third equations to eliminate $\begin{array}{r}z\end{array}$ .
  ::合并第一和第二方程式来消除z。 然后合并第一和第二方程式来消除z。 然后合并第一和第二方程式来消除z。

  播放段落

  $$\begin{array}{rl}& 4(5x-3y+z=-1)\Rightarrow 20x-12y+\cancel{4z}=-4\\ & x+6y-4z=-17\underset{\_}{x+6y-\cancel{4z}=-17}\\ & 21x-6y=-21\end{array}$$

  
  ::4( 5x-3y+z1) 20x- 12y+4z4x+6y- 4z17 x+6y- 4z17_ 21x-6y21

  播放段落

  Result from equations 1 and 2: $\begin{array}{r}21x-6y=-21\end{array}$
  ::方程1和2产生结果:21x-6y21

  播放段落

  $$\begin{array}{rl}& -5(5x-3y+z=-1)\Rightarrow -25x+15y-\cancel{5z}=5\\ & 8x-y+5z=12\underset{\_}{8x-y+\cancel{5z}=12}\\ & -17x+14y=17\end{array}$$

  
  ::-5(5x-3y+z1)-25x+15y=5z=58x-y+5z=12 8x-y+5z=12x-y+5z=12+17x+14y=17

  播放段落

  Result from equations 1 and 3: $\begin{array}{r}-17x+14y=17\end{array}$
  ::方程1和3的结果: - 17x+14y=17

  播放段落

  Now we have reduced our system to two equations in two variables. We can eliminate $\begin{array}{r}y\end{array}$ most easily next and solve for $\begin{array}{r}x\end{array}$ .
  ::现在我们把我们的系统简化为两个变数中的两个方程式。 我们可以最容易地删除下一个 y , 然后解决 x 。

  播放段落

  $$\begin{array}{rl}& 7(21x-6y=-21)\Rightarrow 147x-\cancel{42y}=-147\\ & 3(-17x+14y=17)\underset{\_}{-51x+\cancel{42y}=51}\\ & 96x=-96\\ & x=-1\end{array}$$

  
  ::7(21x-6y 21) 147x-42y 14773(- 17x+14y=17) − 51x+42y=51_ 96x_ 96x_ 96x_ 1)

  播放段落

  Now find $\begin{array}{r}y\end{array}$ :
  ::现在找到 y :

  播放段落

  $$\begin{array}{rl}21(-1)-6y& =-21\\ -21-6y& =-21\\ -6y& =0\\ y& =0\end{array}$$

  
  ::21 - 6y=0y=0 21 - 21 - 21 - 6y _ 21 - 6y=0

  播放段落

  Finally, we can go back to one of the original three equations and use our $\begin{array}{r}y\end{array}$ and $\begin{array}{r}z\end{array}$ values to find $\begin{array}{r}x\end{array}$ .
  ::最后,我们可以回到原来的三个方程中的一个, 用我们的y和z值来找到x。

  播放段落

  $$\begin{array}{rl}5(-1)-3(0)+z& =-1\\ -5+z& =-1\\ z& =4\end{array}$$

  
  ::5(-1)-3(0)+z1-5+z1z=4

  播放段落

  Therefore the solution is (-1, 0, 4).
  ::因此,解决办法是(1,0,4)。

  播放段落

  Don’t forget to check your answer by substituting the point into each equation.
  ::不要忘了检查你的答案, 将点换成每个方程式。

  播放段落

  Equation 1: $\begin{array}{r}5(-1)-3(0)+(4)=-5+4=-1\end{array}$
  ::等式 1 : 5(-1)- 3(0)+(4) 5+41

  播放段落

  Equation 2: $\begin{array}{r}(-1)+6(0)-4(4)=-1-16=-17\end{array}$
  ::等式 2: (-1) +6(0)-4(4) 1-16 17

  播放段落

  Equation 3: $\begin{array}{r}8(-1)-(0)+5(4)=-8+20=12\end{array}$
  ::等式 3: 8(-1)-(0)+5(4) 8+20=12

  播放段落

  Example 4
  ::例4

  播放段落

  Solve the following system using linear combinations:
  ::使用线性组合解决以下系统:

  播放段落

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

  
  ::2x+y-z=3x-2y+z=56x+3y-3z=6

  播放段落

  Combine equations 1 and two to eliminate $\begin{array}{r}z\end{array}$ . Then combine equations 2 and 3 to eliminate $\begin{array}{r}z\end{array}$ .
  ::组合方程式1和2来消除z。然后合并方程式2和3来消除z。

  播放段落

  $$\begin{array}{rl}& 2x+y-\cancel{z}=3\\ & \underset{\_}{x-2y+\cancel{z}=5}\\ & 3x-y=8\end{array}$$

  
  ::2x+y-z=3x-2y+z=5_3x-y=8

  播放段落

  Result from equations 1 and 2: $\begin{array}{r}3x-y=8\end{array}$
  ::公式1和公式2的结果:3x-y=8

  播放段落

  $$\begin{array}{rl}& 3(x-2y+z=5)\Rightarrow 3x-6y+\cancel{3z}=15\\ & 6x+3y-3z=6\underset{\_}{6x+3y-\cancel{3z}=6}\\ & 9x-3y=21\end{array}$$

  
  ::3(x-2y+z=5)3x-6y+3z=156x+3y-3z=66x+3y-3z=6_9x-3y=21)

  播放段落

  Result from equations 2 and 3: $\begin{array}{r}9x-3y=21\end{array}$
  ::等方2和3: 9x-3y=21

  播放段落

  Now we have reduced our system to two equations in two variables. Now we can combine these two equations and attempt to eliminate another variable.
  ::现在我们把我们的系统简化为两个变量中的两个方程式。 现在我们可以把这两个方程式结合起来, 并尝试去掉另一个变量。

  播放段落

  $$\begin{array}{rl}& -3(3x-y=8)\Rightarrow -9x+3y=-24\\ & 9x-3y=21\underset{\_}{9x-2y=21}\\ & 0=-3\end{array}$$

  
  ::- 3(3x-y=8)-9x+3y24 9x-3y=219x-2y=21_03

  播放段落

  Since the result is a false equation, there is no solution to the system.
  ::由于结果是虚假的等式,对该系统没有解决办法。

  播放段落

  Review
  ::回顾

  播放段落
  1. Determine if the point (2, -3, 5) is the solution to the system:
     ::确定点(2,3,5)是否是系统的解决办法:

  播放段落

  $$\begin{array}{rl}2x+5y-z& =\text{-}16\\ 5x-y-3z& =\text{-}2\\ 3x+2y+4z& =20\end{array}$$

  
  ::2x+5y-z=-165x-y-3z=-23x+2y+4z=20

  播放段落
  2. Determine if the point (-1, 3, 8) is the solution to the system:
     ::确定点(-1、3、8)是否是系统的解决办法:

  播放段落

  $$\begin{array}{rl}8x+10y-z& =14\\ 11x+4y-3z& =\text{-}23\\ 2x+3y+z& =10\end{array}$$

  
  ::8x+10y-z=1411x+4y-3z=232x+3y+z=10

  播放段落
  3. Determine if the point (0, 3, 5) is the solution to the system:
     ::确定点(0、3、5)是否是系统的解决办法:

  播放段落

  $$\begin{array}{rl}5x-3y+2z& =1\\ 7x+2y-z& =1\\ x+4y-3z& =\text{-}3\end{array}$$

  
  ::5x-3y+2z=17x+2y-z=1x+4y-3z=3

  播放段落
  4. Determine if the point (1, -1, 1) is the solution to the system:
     ::确定点(1,-1,1)是否是系统的解决办法:

  播放段落

  $$\begin{array}{rl}x-2y+2z& =5\\ 6x+y-4z& =1\\ 4x-3y+z& =8\end{array}$$

  
  ::x-2y+2z=56x+y-4z=14x-3y+z=8

  播放段落

  Solve the following systems in three variables using linear combinations.
  ::使用线性组合,用三个变量解决以下系统。

  5. .

  $$\begin{array}{rl}3x-2y+z& =0\\ 4x+y-3z& =\text{-}9\\ 9x-2y+2z& =20\end{array}$$

  6. .

  $$\begin{array}{rl}11x+15y+5z& =1\\ 3x+4y+z& =\text{-}2\\ 7x+13y+3z& =3\end{array}$$

  7. .

  $$\begin{array}{rl}2x+y+7z& =5\\ 3x-2y-z& =\text{-}1\\ 4x-y+3z& =5\end{array}$$

  8. .

  $$\begin{array}{rl}x+3y-4z& =\text{-}3\\ 2x+5y-3z& =3\\ \text{-}x-3y+z& =\text{-}3\end{array}$$

  9. .

  $$\begin{array}{rl}3x-2y-5z& =\text{-}8\\ 3x+2y+5z& =\text{-}8\\ 6x+4y-10z& =\text{-}16\end{array}$$

  10. .

  $$\begin{array}{rl}x+2y-z& =\text{-}1\\ 2x+4y+z& =10\\ 3x-y+8z& =6\end{array}$$

  11. .

  $$\begin{array}{rl}x+y-z& =\text{-}3\\ 2x-y-z& =6\\ 4x+y+z& =0\end{array}$$

  12. .

  $$\begin{array}{rl}4x+y+3z& =8\\ 8x+2y+6z& =15\\ 3x-3y-z& =5\end{array}$$

  13. .

  $$\begin{array}{rl}2x+3y-z& =\text{-}1\\ x+2y+3z& =\text{-}4\\ \text{-}x+y-2z& =3\end{array}$$

  14. .

  $$\begin{array}{rl}x-3y+4z& =14\\ \text{-}x+2y-5z& =\text{-}13\\ 2x+5y-3z& =\text{-}5\end{array}$$

  15. .

  播放段落

  $$\begin{array}{rl}x+y+z& =3\\ x+y-z& =3\\ 2x+2y+z& =6\end{array}$$

  
  ::x+y+z=3x+y-z=32x+2y+z=6

  播放段落

  Review (Answers)
  ::回顾(答复)

  播放段落

  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 。