7.2 替代系统

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  使用替代系统的系统

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  Systems Using Substitution 
  ::使用替代系统的系统

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  Solving Linear Systems Using Substitution of Variable Expressions
  ::使用可变表达式替代法解决线性系统

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  Consider  this  problem regarding Peter and Nadia racing:
  ::想想Peter和Nadia赛事的这个问题:

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  Peter and Nadia like to race each other. Peter can run at a speed of 5 feet per second and Nadia can run at a speed of 6 feet per second. To be a good sport, Nadia likes to give Peter a head start of 20 feet. How long does Nadia take to catch up with Peter? At what distance from the start does Nadia catch up with Peter?
  ::彼得和纳迪亚喜欢相互竞争。 彼得每秒可以跑5英尺的速度, 纳迪亚每秒可以跑6英尺的速度。 作为一项好运动, Nadia喜欢让Peter先跑20英尺。 Nadia要多久才能赶上Peter? Nadia从什么时候开始会赶上Peter?

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  This can be described with  with two equations:
  ::这可以用两个方程式来描述:

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  Nadia’s equation : $\begin{align*}d = 6t\end{align*}$
  ::Nadia的方程: d=6t

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  Peter’s equation: $\begin{align*}d = 5t + 20\end{align*}$
  ::彼得的方程: d=5t+20

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  Each equation would produce its own line on a graph, and to solve the system, you could find the point at which the lines intersected - the point where the values for $\begin{align*}d\end{align*}$ and $\begin{align*}t\end{align*}$ satisfied both relationships. When the values for $\begin{align*}d\end{align*}$ and $\begin{align*}t\end{align*}$ are equal, that means that Peter and Nadia are at the same place at the same time.
  ::每个方程式都会在图表上生成自己的线条, 并解决系统, 您可以找到线条交叉的点, 即 d 和 t 的值能满足两个关系。 当 d 和 t 的值相等时, 这意味着 Peter 和 Nadia 在同一时间点 。

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  However,  there’s a faster way than graphing to solve this system of equations . Since we want the value of $\begin{align*}d\end{align*}$ to be the same in both equations, we could just set the two right-hand sides of the equations equal to each other to solve for $\begin{align*}t\end{align*}$ . That is, if $\begin{align*}d = 6t\end{align*}$ and $\begin{align*}d = 5t + 20\end{align*}$ , and the two $\begin{align*}d\end{align*}$ ’s are equal to each other, then by the transitive property we have $\begin{align*}6t = 5t + 20\end{align*}$ . We can solve this for $\begin{align*}t\end{align*}$ :
  ::然而,解决这个方程式系统比图形化更快的方法。 既然我们希望 d 值在两个方程式中都是一样的, 我们只能为 t 设定两个对等方程式的对立面, 以便解决。 也就是说, 如果 d=6t 和 d=5t+20 , 并且两个 d 等值是等值的, 那么通过过渡性属性, 我们只有 6t=5t+20 。 我们可以解决这个问题 :

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  $$\begin{align*}6t &= 5t + 20 && \text{subtract 5}t \text{ from both sides}\\ t &= 20 && \text{substitute this value for } t \text{ into Nadia's equation}\\ d &= 6 \cdot 20 = 120\end{align*}$$
  ::6t=5t+20从两边提取5t =20 将t的这个值替换为 Nadia 的方程= 620=120

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  Even if the equations weren’t so obvious, we could use simple algebraic manipulation to find an expression for one variable in terms of the other. If we rearrange Peter’s equation to isolate $\begin{align*}t\end{align*}$ :
  ::即使方程式不那么明显,我们也可以使用简单的代数操控来找到一个变量在另一个变量中的表达方式。 如果我们重新排列彼得的方程式来孤立 t:

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  $$\begin{align*}d &= 5t + 20 && \text{subtract 20 from both sides}\\ d - 20 &= 5t && \text{divide by 5}\\ \frac{d - 20}{5} &= t\end{align*}$$
  ::d=5t+20 双向减量 20 = 5d-20= 5tdivide 5d-205= t

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  Now you know that, for the purposes of this situation,  $\begin{align*}\frac{d-20}{5}\end{align*}$  is the same as  $\begin{align*}t,\end{align*}$  so you  can  substitute this expression for $\begin{align*}t\end{align*}$ into Nadia’s equation $\begin{align*}(d = 6t)\end{align*}$ to solve:
  ::现在你知道,为了这个情况的目的, d - 205 与 t 相同, 所以您可以用这个表达方式代替 t 在 Nadia 的方程式( d=6t) 中, 来解决 :

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  $$\begin{align*}d &= 6 \left ( \frac{d - 20}{5} \right ) && \text{multiply both sides by 5}\\ 5d &= 6(d - 20) && \text{distribute the 6}\\ 5d &= 6d - 120 && \text{subtract 6}d \text{ from both sides}\\ \text{-}d &= \text{-}120 && \text{divide by -1}\\ d &= 120 && \text{substitute value for } d \text{ into the expression for } t\\ t &= \frac{120 - 20}{5} = \frac{100}{5} = 20\end{align*}$$
  ::d=6(d-205) 将两边乘以55d=6(d-20) 将65d=6d - 120从两侧- d=- 120divide的65d= 6d - 120提要6d乘以 -1d=120divide 乘以 -1d=120 120 替代值, d 乘以 tt=120- 205=1005=20

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  So we find that Nadia and Peter meet 20 seconds after they start racing, at a distance of 120 feet away.
  ::我们发现Nadia和Peter在他们开始赛车20秒后 相遇 距离120英尺远

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  The method we just used is called the Substitution Method. In this lesson you’ll learn several techniques for isolating variables in a system of equations, and for using those expressions to solve systems of equations that describe situations like this one.
  ::我们刚刚使用的方法被称为替代方法。 在这个教训中,你会学到几种技术,在方程系统中分离变量,用这些表达方式解析描述这种情形的方程系统。

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  Solving a System of Equations 
  ::解决方平系统

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  Let’s look at an example where the equations are written in standard form .
  ::让我们看看一个方程式以标准格式书写的例子。

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  1. Solve the system
  ::1. 解决系统

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  $$\begin{align*}2x + 3y & = 6\\ -4x + y & = 2\end{align*}$$
  ::2x+3y=6-4x+y=2

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  Again, we start by looking to isolate one variable in either equation. If you look at the second equation, you should see that the coefficient of $\begin{align*}y\end{align*}$ is 1. So the easiest way to start is to use this equation to solve for $\begin{align*}y\end{align*}$ .
  ::再次,我们从在任何一个方程中分离一个变量开始。如果你看第二个方程,你应该看到y的系数是1。所以最容易开始的方法就是使用这个方程来解决y。

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  Solve the second equation for $\begin{align*}y\end{align*}$ :
  ::解决y的第二个方程式:

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  $$\begin{align*}-4x + y &= 2 && \text{add 4}x \text{ to both sides}\\ y &= 2 + 4x\end{align*}$$
  ::− 4x+y=2 向两侧加4x=2+4x

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  Substitute this expression into the first equation:
  ::将这个表达式替换为第一个方程式 :

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  $$\begin{align*}2x + 3(2 + 4x) &= 6 && \text{distribute the 3}\\ 2x + 6 + 12x &= 6 && \text{collect like terms}\\ 14x + 6 &= 6 && \text{subtract 6 from both sides}\\ 14x &= 0\\ x &= 0\end{align*}$$
  ::2 x+3 (2+4x) = 6 分配 32 x+6+6+12x=6 和 type14x+6=6 6 从两侧14x=0

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  Substitute back into our expression for $\begin{align*}y\end{align*}$ :
  ::取代我们表达的y:

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  $$\begin{align*}y = 2 + 4 \cdot 0 = 2\end{align*}$$
  ::y=2+40=2

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  You  end up with the same solution,  $\begin{align*}(x = 0, y = 2)\end{align*}$ , that you  would if you  graphed both lines to see where they cross. So long as you are careful with the algebra, the substitution method can be a very efficient way to solve systems.
  ::最后,你用同样的解决方案,(x=0,y=2),即如果用图形显示两条线,看两条线的交叉位置。只要你小心代数,替代方法可以成为解决系统问题的一个非常有效的方法。

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  Next, let’s look at a more complicated example. Here, the values of $\begin{align*}x\end{align*}$ and $\begin{align*}y\end{align*}$ we end up with aren’t whole numbers, so they would be difficult to read off a graph!
  ::接下来,让我们看看一个更复杂的例子。 这里, x 和 y 的值最终不是全部数字, 因此它们很难从图表上读出来 !

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  2. Solve the system
  ::2. 解决系统

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  $$\begin{align*}2x + 3y & = 3\\ 2x - 3y & = -1\end{align*}$$
  ::2x+3y=32x-3y1

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  Again, we start by looking to isolate one variable in either equation. In this case it doesn’t matter which equation we use , since all the variables look about equally easy to solve for.
  ::再说一遍,我们首先寻求在任何一个方程式中分离一个变量。 在这种情况下,我们使用哪个方程式并不重要,因为所有变量看起来都同样容易解决。

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  S olve the first equation for $\begin{align*}x\end{align*}$ :
  ::解析 x 的第一个方程式 :

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  $$\begin{align*}2x + 3y &= 3 && \text{subtract 3}y \text{ from both sides}\\ 2x &= 3 - 3y && \text{divide both sides by 2}\\ x &= \frac{1}{2}(3 - 3y)\end{align*}$$
  ::2 x+3y = 3 3 乘以 2 x= 12 (3- 3y)

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  Substitute this expression into the second equation:
  ::将这个表达式替换为第二个方程式 :

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  $$\begin{align*}\cancel{2} \cdot \frac{1}{2}(3 - 3y)-3y &= -1 && \text{cancel the fraction and re-write terms}\\ 3 - 3y - 3y &= -1 && \text{collect like terms}\\ 3 - 6y &= -1 && \text{subtract 3 from both sides}\\ -6y &= -4 && \text{divide by -6}\\ y &= \frac{2}{3}\end{align*}$$
  ::212( 3- 3- 3y) - 3- 3- 3 1 分数和重写术语 3- 3- 3- 3- 3 1 集合类似术语 3- 6- 6 1 分数 3 双侧 - 6y 4divide by - 6y= 23

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  Substitute into the expression for $\begin{align*}x\end{align*}$ :
  ::替代 x 的表达式:

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  $$\begin{align*}x & = \frac{1}{2} \left ( 3 - \cancel{3} \left ( \frac{2}{\cancel{3}} \right ) \right )\\ x & = \frac{1}{2}\end{align*}$$
  ::x=12( 3- 3( 23) x=12)

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  The  solution is $\begin{align*}x = \frac{1}{2}, y = \frac{2}{3}.\end{align*}$  You can see how the graphical solution $\begin{align*}\left ( \frac{1}{2}, \frac{2}{3} \right )\end{align*}$ might have been difficult to read accurately off a graph!
  ::答案是 x= 12,y= 23。 您可以看到图形解决方案 (12, 23) 可能很难从图表上准确读取 !

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  Solving Real-World Problems Using Linear Systems
  ::使用线性系统解决现实世界问题

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  Simultaneous equations can help solve many real-world problems. C onsidering a purchase, for example - trying to decide whether it’s cheaper to buy an item online where you pay shipping or at the store where you do not. Or you may wish to join a CD music club, but aren’t sure if you would really save any money by buying a new CD every month in that way. Or you might be considering two different phone contracts. Let’s look at an example of that now.
  ::同时的方程式可以帮助解决许多现实世界问题。 比如,考虑购买 — — 试图决定在网上购买一个支付运费的项目或者在不支付运费的商店购买一个项目是否更便宜。 或者你可能希望加入CD音乐俱乐部,但不确定你是否真的会通过每月以这种方式购买一张新的CD来节省任何资金。 或者你可能会考虑两种不同的电话合同。 让我们看看现在的例子。

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  Anne is trying to choose between two phone plans. The first plan, with Vendafone, costs $20 per month, with calls costing an additional 25 cents per minute. The second company, Sellnet, charges $40 per month, but calls cost only 8 cents per minute. Which should she choose?
  ::Anne试图在两个电话计划之间做出选择。第一个计划是Vendafone,每月费用20美元,电话费每分钟增加25美分。第二家公司Sellnet每月收费40美元,但电话费每分钟只收费8美分。她应该选择哪家?

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  You should see that Anne’s choice will depend upon how many minutes of calls she expects to use each month. We start by writing two equations for the cost in dollars in terms of the minutes used. Since the number of minutes is the independent variable, it will be our $\begin{align*}x\end{align*}$ . Cost is dependent on minutes – the cost per month is the dependent variable and will be assigned $\begin{align*}y.\end{align*}$
  ::你应该看到,安妮的选择将取决于她每个月要使用多少分钟的电话。 我们从用分钟以美元计算成本的两种方程式开始。 由于分钟数是独立的变量,这将是我们的 x。 成本取决于分钟 — — 每月的成本是依附变量并将被分配到 y 。

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  For Vendafone: $\begin{align*}y = 0.25x + 20\end{align*}$
  ::Vendfone:y=0.25x+20

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  For Sellnet: $\begin{align*}y = 0.08x + 40\end{align*}$
  ::Sellnet:y=0.08x+40

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  By writing the equations in slope-intercept form $\begin{align*}(y = mx + b)\end{align*}$ , you can sketch a graph to visualize the situation:
  ::通过以斜坡界面形式(y=mx+b)写出方程式,您可以绘制一个图表来直观地描述局势:

  

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  The line for Vendafone has an intercept of 20 and a of 0.25. The Sellnet line has an intercept of 40 and a slope of 0.08 (which is roughly a third of the Vendafone line’s slope). In order to help Anne decide which to choose, we’ll find where the two lines cross, by solving the two equations as a system.
  ::Vendafone的线路拦截量为20和0.25。 Sellnet的线路拦截量为40和0.08(约为Vendfone线斜坡的三分之一 ) 。 为了帮助Anne决定选择哪条路,我们将通过将两个方程式作为一个系统解开来找到这两条线的交叉点。

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  E quation 1 specifies that  $\begin{align*}y\end{align*}$  is the same as  $\begin{align*}(0.25x + 20),\end{align*}$  so we can substitute this expression directly into equation 2 in place of $\begin{align*}y:\end{align*}$
  ::等式 1 指定 Y 与 (0. 25x+20) 相同, 所以我们可以直接将这个表达式替换为 y 等式 2 :

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  $$\begin{align*}0.25x + 20 &= 0.08x + 40 && \text{subtract 20 from both sides}\\ 0.25x &= 0.08x + 20 && \text{subtract 0.08} x \text{ from both sides}\\ 0.17x &= 20 && \text{divide both sides by 0.17}\\ x &= 117.65 \ \text{minutes} && \text{rounded to 2 decimal places.}\end{align*}$$
  ::0.25x20=0.08x+40次量 20 从两侧0.25x=0.08x+20次量 0.08x从两侧0.17x=20divide 双侧0.17x=117.65四舍五入到小数点2位。

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  So if Anne uses 117.65 minutes a month (although she can’t really do exactly that, because phone plans only count whole numbers of minutes), the phone plans will cost the same. Now we need to look at the graph to see which plan is better if she uses more minutes than that, and which plan is better if she uses fewer. You can see that the Vendafone plan costs more when she uses more minutes, and the Sellnet plan costs more with fewer minutes.
  ::因此,如果安妮每月使用117.65分钟(尽管她无法真正做到这一点,因为电话计划只计算整个分钟),那么电话计划的成本也是一样的。 现在我们需要看看图表,看看如果她使用分钟比这个多,哪个计划更好,如果她使用更少,哪个计划更好。 你可以看到,万达丰计划在她使用更多分钟时成本更高,而Sellnet计划花费更少分钟。

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  So, if Anne will use 117 minutes or less every month, she should choose Vendafone. If she plans on using 118 or more minutes, she should choose Sellnet.
  ::所以,如果安妮每月使用117分钟或更少, 她应该选择Vendafone。 如果她计划使用118分钟或更多分钟, 她应该选择Sellnet。

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  Example
  ::示例示例示例示例

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  Example 1
  ::例1

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  Solve the system
  ::解决系统

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  $$\begin{align*}8x + 10y & = 2\\ 4x - 15y & = -19\end{align*}$$
  ::8x+10y=24x-15y19

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  Again, start by looking to isolate one variable in either equation. In this case it doesn’t matter which equation we use - all the variables appear  equally easy to isolate .
  ::再次,首先在任何一个方程式中选择一个变量。 在这种情况下,我们使用哪个方程式并不重要,所有变量似乎都同样容易分离。

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  So let’s solve the first equation for $\begin{align*}x\end{align*}$ :
  ::因此,让我们解决x的第一个方程式:

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  $$\begin{align*}8x + 10y & = 2 && \text{subtract 10}y \text{ from both sides}\\ 8x &= 2 - 10y && \text{divide both sides by 8}\\ x &= \frac{1}{8}(2 - 10y)\end{align*}$$
  ::8x+10y=2 从两侧8x=2-10ydivide 双侧增加8x=18(2-10yyy)

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  Substitute this expression into the second equation:
  ::将这个表达式替换为第二个方程式 :

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  $$\begin{align*}4 \cdot \frac{1}{8}(2 - 10y)-15y &= \text{}-19 && \text{simplify the fraction}\\ \frac{1}{2}(2 - 10y)-15y &= -19 && \text{distribute the fraction and re-write terms}\\ 1 - 5y - 15y &= -19 && \text{collect like terms}\\ 1 - 20y &= -19 && \text{subtract 1 from both sides}\\ -20y &= -20 && \text{divide by -20}\\ y &= 1 \end{align*}$$
  ::418( 2- 10y) - 15y19 简化分数 12( 2- 10y) - 15y19 分配分数和重写术语1 和重写术语1-5y- 15y19 集合类似术语1 - 20y19 分数1 双向- 20y20divide by - 20y=1

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  Substitute into the expression for $\begin{align*}x\end{align*}$ :
  ::替代 x 的表达式:

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  $$\begin{align*}x &= \frac{1}{8}(2 - 10y) && \text{Substitute the }y\text{-value into the }x \text{ equation}\\ x &= \frac{1}{8}(2 - 10(1)) && \text{Simplify}\\ x &= \frac{1}{8}(2 - 10) \\ x &= \frac{1}{8}(-8) \\ x & = -1\end{align*}$$
  ::x=18( 2- 10y) 将 Y 值制成 x 等式= 18( 2- 10(1)) 的 Y 值 。Simplifiex= 18( 2- 10x) = 18( 8) x @ 1

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  The  solution is $\begin{align*}x = -1, y = 1.\end{align*}$
  ::答案是 x1,y=1 。

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  Review 
  ::回顾

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  1. Solve the system: $\begin{align*}\cases{x + 2y =9\\ 3x + 5y = 20}\\\end{align*}$
     ::解决系统: {x+2y=93x+5y=20
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  2. Solve the system: $\begin{align*}\ \\ \cases{x - 3y =10\\ 2x + y = 13}\\ \ \\\end{align*}$
     ::解决系统: {x--3y=102x+y=13
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  3. Solve the system: $\begin{align*}\cases{2x + 0.5y = -10\\ x - y = -10}\\\end{align*}$
     ::解决系统 : {2x+0. 5} @ 10x- y @ 10
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  4. Solve the system: $\begin{align*}\ \\ \cases{2x + 0.5y = 3\\ x + 2y = 8.5}\\ \ \\\end{align*}$
     ::解决系统: {2x+0.5y=3x+2y=8.5
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  5. Solve the system: $\begin{align*}\cases{3x + 5y = -1\\ x + 2y = -1}\\\end{align*}$
     ::解决系统: {3x+5y}% 1x+2y}% 1
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  6. Solve the system: $\begin{align*}\ \\ \cases{3x + 5y = -3\\ x + 2y = - \frac{4}{3}}\\ \ \\\end{align*}$
     ::解决系统 : {3x+5y}%3x+2y}{43
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  7. Solve the system: $\begin{align*}\cases{x - y = - \frac{12}{5}\!\\ 2x + 5y = -2}\\ \ \\\end{align*}$
     ::解析系统 : {x- y\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
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  8. Of the two non-right angles in a right angled triangle, one measures twice as many degrees as the other. What are the angles?
     ::在右角三角形的两个非右角度中,一个是位数的两倍。什么是角?
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  9. The sum of two numbers is 70. They differ by 11. What are the numbers?
     ::两个数字的总和是70,它们除以11。 数字是多少?
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  10. A number plus half of another number equals 6; twice the first number minus three times the second number equals 4. What are the numbers?
      ::数字加另一数字的半数等于6;第一个数字的两倍减去第二个数字的三倍等于4。 数字是多少?
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  11. A rectangular field is enclosed by a fence on three sides and a wall on the fourth side. The total length of the fence is 320 yards. If the field has a total perimeter of 400 yards, what are the dimensions of the field?
      ::长方形田被三面的栅栏和第四面的墙围住,围栏的总长度为320码,如果围栏的周界总长为400码,那么野外的尺寸是多少?
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  12. A ray cuts a line forming two angles. The difference between the two angles is $\begin{align*}18^\circ\end{align*}$ . What does each angle measure?
      ::一个射线切开一条由两个角度构成的线条。两个角度的区别是 18\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
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  13. Jason is five years older than Becky, and the sum of their ages is 23. What are their ages?
      ::Jason比Becky大五岁, 他们年龄的总和是23岁, 他们几岁?

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  ::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 。