课程： 5.5 利用消除先进问题-interactive

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利用消除先进问题

When to Use Elimination
::何时使用删除

There are three methods for solving systems of equations: graphing, substitution, and elimination. Take a break from solving and discuss when each method is most efficient.
::解答方程式系统有三种方法:图形化、替代和清除。在每种方法效率最高时,先从解答中抽出一个空位,然后讨论。

Graphing: The best time to use graphing is when you need to know more information about a system of equations than just the solution.
::图形化:使用图形化的最佳时机是需要了解更多关于方程式系统的信息时,而不是仅仅了解解决方案。

Substitution: The most efficient time to use substitution is when at least one variable is already isolated. When the equations are in slope-intercept form , the variable $\begin{aligned} y \end{aligned}$ will be isolated.
::替代:使用替代的最有效时间是至少一个变量已经孤立的时候。当方程式以斜度干扰形式出现时,变量 y 将被孤立。

Elimination: The most efficient time to use elimination is when the like terms , constants and equal signs are lined up and you have a clear additive inverse . The method is commonly used when the equations are in standard form .
::消除:最有效的消除时间是等用词、常数和等效符号排成一行,并且有明显的添加反差。当方程式以标准形式出现时,通常使用这种方法。

Multiplying Equations
::倍数等数

In the previous section, you used an interactive to add equations. However, what if neither of the terms adds to 0?
::在上一节中,您使用交互方式添加方程式。但是,如果这两个词都不增加 0 呢?

Example
::示例示例示例示例

Solve the system of equations:
::解决方程式系统:

\begin{aligned} 4 x - y = 6 \\ - 2 x + 3 y = 7 \end{aligned}

::4-y=6-2x+3y=7

When you add these equations you get the equation $\begin{aligned} 2 x + 2 y = 13 \end{aligned}$ . Y ou can’t solve this equation because you still have two variables. To handle situations like this, you can multiply both sides of an equation by a number so that you have an additive inverse. In this case, if you multiply both sides of the bottom equation by 2, you get the equation $\begin{aligned} - 4 x + 6 y = 14 \end{aligned}$ . The $\begin{aligned} x \end{aligned}$ terms become additive inverses and will cancel when you add this with the first equation.
::当添加这些方程式时,您可以得到 2x+2y=13 方程式。您无法解答此方程式, 因为您仍然有两个变量。要处理这样的情况, 您可以将方程式的两边乘以一个数字, 这样您就会有一个添加方程式的反向。在这种情况下, 如果您将底方程式的两边乘以 2, 您就会得到公式 - 4x+6y=14。 x 条件会变成添加方程式的反义, 当您在第一个方程式中添加此公式时将会取消。

\begin{aligned} 4 x - y = 6 & \to 4 x - y = 6 \\ 2 (- 2 x + 3 y = 7) & \to - 4 x + 6 y = 14 \end{aligned}

::4x - Y= 6 = 4x - Y= 62(-2x+3y= 7) * - 4x+6y= 14

When you add these equations together you will get the following:
::当您将这些方程式加在一起时, 将会得到以下结果:

\begin{aligned} 4 x - y & = 6 \\ + - 4 x + 6 y & = 14 \\ 5 y & = 20 \end{aligned}

::4x-y=6+-4x+6y=145y=20

From the resulting equation, you can solve $\begin{aligned} 5 y = 20 \end{aligned}$ to get $\begin{aligned} y = 4 \end{aligned}$ . Once you know one of the variables, like with substitution, you can go back and plug it into one of the original equations to solve for the other variable, $\begin{aligned} x . \end{aligned}$
::从生成的方程式中, 您可以解析 5y=20 来获取 Y= 4。一旦您知道其中的变量之一, 比如替换, 您可以回到原来的方程式中, 将其插入到另一个变量 x 的原始方程式中。

Use the interactive below to practice using this strategy to solve for unknown variables.
::使用下面的交互效果来练习使用此策略来解决未知变量。

INTERACTIVE

Systems of Equations and Elimination - Advanced

Drag the red point to see the relationship between columns and rows of numbers.
::拖曳红色点以查看列和数字行之间的关系。

Drag the green and orange sliders to multiply each number in a row by a number.
::拖动绿色和橙色滑块,将每一行的数字乘以一个数字。

Press the white button to see how the columns of numbers relate to a system of equations.
::按下白色按钮以查看数字列与方程式系统的关系。

Elimination With Multiplication
::以乘法消除

If none of the terms will cancel when you add the equations, you will need to multiply the equations by factors that will result in additive inverses. Additive inverses are two numbers that have a sum of zero. You may need to multiply one equation or both equations to produce additive inverses.
::如果在添加方程式时没有一个条件会取消, 您就需要将方程式乘以导致添加反函数的因素。添加反函数是两个数字, 其总和为零。您可能需要乘以一个方程式或两个方程式来生成添加反函数。

Example
::示例示例示例示例

Solve the system of equations:
::解决方程式系统:

\begin{aligned} 3 x + 4 y = 15 \\ 2 x + 3 y = 11 \end{aligned}

::3x+4y=152x+3y=11

The equations above have everything lined up but do not have additive inverses. You will need to make the $\begin{aligned} x \end{aligned}$ -terms or the $\begin{aligned} y \end{aligned}$ -terms additive inverses. Since there is not an easier option, make the $\begin{aligned} x \end{aligned}$ -terms additive inverses. To do this, you will need to find a multiple of their coefficients, 3 and 2. To produce additive inverses, multiply the first equation by 2 and the second equation by -3.
::上面的方程式都排整齐, 但没有添加反函数。您需要使 x 条件或 Y 条件的添加反函数。由于没有更简单的选项, 请使 x 条件的添加反函数。要做到这一点, 您需要找到多个系数, 3 和 2 。要生成添加反函数, 请将第一个方程式乘以 2, 第二个方程式乘以 - 3 。

\begin{aligned} 2 (3 x + 4 y = 15) & \to 6 x + 8 y = 30 \\ - 3 (2 x + 3 y = 11) & \to - 6 x - 9 y = - 33 \end{aligned}

::2(3x+4y=15) 6x+8y=30-3(2x+3y=11) -6x-9y=-33

When you add the equations, the $\begin{aligned} x \end{aligned}$ -terms will cancel.
::当添加方程式时, x 条件将会取消。

\begin{aligned} 6 x + 8 y & = 30 \\ + - 6 x - 9 y & = - 33 \\ - 1 y & = - 3 \\ y & = 3 \end{aligned}

::6x+8y=30+-6x-9y=-33-1y=-3y=3

To find the value of $\begin{aligned} x, \end{aligned}$ you will have to substitute 3 in place of $\begin{aligned} y \end{aligned}$ in one of the original equations. Here is how it looks when substituted into the second equation:
::要找到 x 的值, 您必须替换 3 来代替 y , 取代 y 的原始方程。下面是第二个方程替换时的外观 :

\begin{aligned} 2 x + 3 (3) & = 11 \\ 2 x + 9 & = 11 \\ 2 x & = 2 \\ x & = 1 \end{aligned}

::2x+3(3)=112x+9=112x=2x=1

The solution to the system of equations is the point (1, 3 ).
::方程系统的解决办法是点(1,3)。

Discussion Question
::讨论问题

Discuss how you might check if this solution is correct. What are the possible ways to do this? Think algebraically and visually. Try both, does this answer checkout using both methods?
::讨论您如何检查这个解决方案是否正确。可能的方法是什么 ? 考虑代数和视觉。试两种方法, 这个答案是否使用两种方法 ?

Mixing It Up
::混合起来

are chemistry-based questions commonly seen in chemistry labs. However, this type of problem occurs when developing a mixture of candy at a candy store, a mixture of metal, and more.
::化学实验室中常见的化学问题。然而,当糖果店和金属等混合物混合糖果时,就会出现这类问题。

Example
::示例示例示例示例

Adriana is a chemist who needs to make 300 mL of a 21% acid solution using Hydrogen Chloride (HCl). She has a 30% HCl solution and an 18% HCl solution. How much of each solution will she need to mix?
::Adriana是一位化学家,他需要用氯氢(HCl) 来制造300毫升的21%的酸溶液。她有30%的 HCl 溶液和18%的 HCl 溶液。她需要混合每种溶液的多少?

To solve this problem, Adriana can make a table to keep track of each solution.
::为了解决这个问题,Adriana可以安排一张桌子,跟踪每一种解决办法。

	Amount of Solution	×	Percent	=	Amount of HCl
30% Solution	$\begin{aligned} x \end{aligned}$	×	$\begin{aligned} 0.30 \end{aligned}$	=	$\begin{aligned} 0.3 x \end{aligned}$
18% Solution	$\begin{aligned} y \end{aligned}$	×	$\begin{aligned} 0.18 \end{aligned}$	=	$\begin{aligned} 0.18 y \end{aligned}$
Total	$\begin{aligned} 300 \end{aligned}$	×	$\begin{aligned} 0.21 \end{aligned}$	=	$\begin{aligned} 63 \end{aligned}$

Adriana used $\begin{aligned} x \end{aligned}$ and $\begin{aligned} y \end{aligned}$ to represent the amount of 30% and 18% solution since she does not know how much of each solution she will need. S he knows that the solution she needs must be 300 mL and be 21% HCl. The final column represents the amount of HCl in each solution. To find this, multiply the amount of the initial solution by the percent of HCl.
::Adriana使用x和y代表了30%和18%的解决方案,因为她不知道需要多少每种解决方案。她知道,她需要的解决方案必须是300毫升和21%的 HCl。最后一栏代表了每个解决方案中的 HCl 。要找到这个答案,将初始解决方案的金额乘以 HCl 的%。

To make two equations, add the amount of solution column and the amount of HCl column. The 300 mL solution is made up of the amount of 30% solution and the amount of 18% solution. The amount of solution equation will be $\begin{aligned} x + y = 300. \end{aligned}$ A 300 mL solution with 21% HCl will have 63 mL of HCl made up of the HCl from the 30% and the 18% solutions. The amount of solution equation will be $\begin{aligned} 0.3 x + 0.18 y = 63. \end{aligned}$
::要制造两个方程式,请添加溶液柱和 HCl 柱的量。300 mL 溶液由30%的溶液和18%的溶液组成。溶液方程式的量为 x+y=300。使用21% HCl 的300 mL 溶液的量为63 mL HCl,由30%的HCl和18%的溶液组成。溶液方程式的量为0.3x+0.18y=63。

To find how much HCL is in the solution, multiply the percentage by the total amount of solution.
::要想找到解决方案中氢氟烷烃含量的多少,将百分比乘以解决方案的总量。

\begin{aligned} x + y & = 300 \\ 0.3 x + 0.18 y & = 63 \end{aligned}

::x+y=3000.3x+0.18y=63

The terms are lined up so now you need to find an additive inverse which will cancel the $\begin{aligned} x \end{aligned}$ - term or the $\begin{aligned} y \end{aligned}$ -term. If you do not see it right away, clear the decimals by multiplying the bottom equation by 100.
::术语被排成一行,所以现在您需要找到一个添加反函数来取消 x-term 或 y-term 。如果您没有马上看到, 请通过将底方程乘以100来清除小数。

\begin{aligned} 100 (0.3 x + 0.18 y = 63) & \to 30 x + 18 y = 6300 \end{aligned}

::100(0.3x+0.18y=63) 30x+18y=6300

Now, multiply the top equation by -30 to get the additive inverse which will cancel the $\begin{aligned} x \end{aligned}$ -term.
::现在, 乘以 - 30 的顶部方程, 以获得取消x- 期的添加反函数。

\begin{aligned} - 30 (x + y = 300) & \to - 30 x - 30 y = - 9000 \\ 100 (0.3 x + 0.18 y = 63) & \to 30 x + 18 y = 6300 \end{aligned}

::-30(x+y=300) -30x-30y=9000100(0.3x+0.18y=63) 30x+18y=6300

A dd the equations.
::添加方程。

\begin{aligned} - 30 x - 30 y & = - 9000 \\ + 30 x + 18 y & = 6300 \\ - 12 y & = - 2700 \\ y & = 225 \end{aligned}

::-30x-30y900+30x+18y=6300-12y2700y=225

You can substitute this into the equation $\begin{aligned} x + y = 300 \end{aligned}$ to find the value of $\begin{aligned} x . \end{aligned}$
::您可以将它替换为 x+y=300 以找到 x 的值。

\begin{aligned} x + 225 & = 300 \\ x & = 75 \end{aligned}

::x+225=300x=75

The solution to the system of equations is (75, 225). This means that Adriana will need 75 mL of the 30% solution and 225 mL of the 18% solution.
::方程式的解决方案是(75,225),这意味着Adriana需要30%的解决方案75毫升和18%的解决方案225毫升。

Summary

Additive inverses are two numbers that have a sum of zero.
::相加反差是两个数字,总和为零。

To use the elimination with multiplication method multiply one or both of the equations to produce additive inverses.
::使用乘法除去法乘以一个或两个方程式,产生添加反函数。

章节大纲

常规

利用消除先进问题

When to Use Elimination ::何时使用删除

Multiplying Equations ::倍数等数

Elimination With Multiplication ::以乘法消除

Mixing It Up ::混合起来

When to Use Elimination
::何时使用删除

Multiplying Equations
::倍数等数

Elimination With Multiplication
::以乘法消除

Mixing It Up
::混合起来