线性方程组求解_如何求解线性方程组

线性方程组求解

A linear equation is an equation that graphs a line. A system of linear equations is when there are two or more linear equations grouped together.

线性方程是绘制直线的方程。 线性方程组是当两个或多个线性方程组组合在一起时。

To simplify the illustration, we will consider systems of two equations. As the name suggests, there are two unknown variables. Often they are designated by the letters x and y. If equations describe some process, the letters can be chosen by the roles they play. For example, d can stand for distance, and t for time.

为了简化说明，我们将考虑两个方程式的系统。 顾名思义，有两个未知变量。 通常，它们由字母x和y表示 。 如果方程式描述了某个过程，则可以根据字母所扮演的角色来选择字母。 例如， d可以代表距离， t可以代表时间。

In this article we will learn how to solve systems of linear equations using two fun methods. But before we start, let's see how we end up with a particular system by looking at a real life example.

在本文中，我们将学习如何使用两种有趣的方法来求解线性方程组。 但是，在开始之前，我们先来看一个真实的例子，看看如何最终使用特定的系统。

派生系统 (Deriving a system)

A boy gets on his bicycle and starts riding to school. He rides 200 yards every minute.

一个男孩骑上自行车开始上学。 他每分钟骑200码。

6 minutes later, his mother realizes her son forgot his lunch. She gets on her own bicycle and starts following the boy. She rides 500 yards every minute (She is an Olympian and a gold medalist).

6分钟后，他的母亲意识到儿子忘记了午餐。 她骑着自己的自行车，开始跟随男孩。 她每分钟骑500码(她是奥运选手和金牌得主)。

We want to figure out how long it takes the mother to catch up to the boy, and how far she needs to ride to do so.

我们想弄清楚母亲要赶上男孩多长时间，以及她需要骑多远才能赶上男孩。

Since the boy covers 200 yards every minute, in t minutes he will cover 200 times t yards, or 200t yards.

由于男孩每分钟覆盖200码，因此在t分钟内，他将覆盖200 吨t码或200t码。

His mother starts bicycling 6 minutes later, so she rides for (t - 6) minutes. Since she covers 500 yards every minute, in (t - 6) minutes she covers 500 times (t - 6) yards, or 500(t - 6) yards.

他的母亲6分钟后开始骑自行车，所以她骑了(t-6)分钟。 因为她每分钟覆盖500码，所以在(t-6)分钟内她覆盖了500倍(t-6)码，即500(t-6)码。

By the time she catches up to him, they both have covered the same distance. Let's say for now that distance is d.

当她赶上他时，他们俩已经跨越了相同的距离。 现在说距离是d 。

For the boy we have  d = 200t and for his mother we have d = 500(t - 6). We now have our system of two equations.

对于男孩，我们的d = 200t ，对于他母亲，我们的d = 500(t-6) 。 现在，我们有了两个方程式的系统。

A curly brace is often added to indicate that equations form a system.

通常添加大括号以指示方程式构成一个系统。

Now let's see how we can solve this system.

现在让我们看看如何解决这个系统。

替代解决 (Solving by substitution)

The first method we will consider uses substitution.

我们将考虑的第一种方法是使用替换 。

We have two unknowns here, d and t. The idea is to get rid of one variable by expressing it using the other variable.

这里有两个未知数d和t 。 这个想法是通过使用另一个变量来表达它来摆脱一个变量。

The top equation tells us that d = 200t, so let's plug in 200t for the d in the bottom equation. As a result, we have an equation with just the t variable.

顶部方程式告诉我们d = 200t ，因此让我们在底部方程式中为d插入200t 。 结果，我们有了一个仅包含t变量的方程。

First we expand the right side: 500(t -6) = 500t - 500*6 = 500t - 3000.

首先我们扩展右侧：500(t -6)= 500t-500 * 6 = 500t-3000 。

Then we simplify by moving the unknown members to one side and the known members to the other. The result is: 500t - 200t = 3000.

然后，我们将未知成员移到一侧，将已知成员移到另一侧，从而简化了操作。 结果是： 500t-200t = 3000 。

Solving for t gives us t = 10, or since we measure time in minutes, t = 10 minutes. In other words, the mother will catch up to her son in 10 minutes.

求解t得出t = 10 ，或者因为我们以分钟为单位测量时间，所以t = 10分钟 。 换句话说，母亲将在10分钟内赶上她的儿子。

The second part of our problem is to find out how far she had to cycle to catch up with him.

我们问题的第二部分是找出她必须骑自行车多远才能赶上他。

To answer that question, we need to find d. Substituting t = 10 in either equation will give us that answer.

要回答这个问题，我们需要找到d 。 将t = 10代入任一等式将得到答案。

To make it easier, let use the top equation, d = 200t = 200 * 10 = 2000. Since we measure distance in yards, d = 2000 yards.

为了简化起见 ，让我们使用最上面的方程， d = 200t = 200 * 10 = 2000 。 由于我们以码为单位测量距离，因此d = 2000码 。

Let's test your understanding so far – try to solve the next system on your own:

到目前为止，让我们测试您的理解–尝试自己解决下一个系统：

In the system above, the unknown variables are x and y.

在上面的系统中，未知变量是x和y 。

From the top equation we know that y = 2x. Substituting that to the bottom equation gives us 2(2x) = 3(x + 1).

从最上面的方程式我们知道y = 2x 。 将其代入底部方程式，我们得到2(2x)= 3(x + 1) 。

Once we expand and simplify, we get 4x = 3x + 3. Or x = 3. Therefore, y = 2 * 3 = 6.

扩展和简化后，我们得到4x = 3x + 3 。 或x = 3 。 因此， y = 2 * 3 = 6 。

通过图形求解 (Solving by graphing)

The second method we will consider uses graphing, where we find the solution to a system of equations by graphing them out.

我们将考虑的第二种方法是使用绘图 ， 通过绘制方程式找到方程组的解。

For example, take this system: y = 2x + 3 and y = 9 - x.

例如，使用以下系统： y = 2x + 3和y = 9-x 。

A graph of each equation will be a line. The first one for y = 2x + 3 looks like this:  

每个方程的图形将是一条线。 y = 2x + 3的第一个看起来像这样：

Next, we can graph a line for y = 9 - x:  

接下来，我们可以绘制一条y = 9-x的线 ：

These two lines intersect at exactly one point. This point is the only solution to both equations:

这两条线恰好在一个点处相交 。 这是两个方程式的唯一解决方案：

The ordered pair (2, 7) gives us the coordinates of our point of intersection. This pair is the solution to the system. Substituting x = 2 and y = 7 will let us verify this.

有序对(2，7)为我们提供了相交点的坐标。 这对是系统的解决方案。 代入x = 2和y = 7，我们将对此进行验证。

What if the graphs are parallel and do not intersect at all? For example:

如果图形是平行的并且根本不相交怎么办？ 例如：

When graphs of the equations do not intersect, that means our system has no solution. Trying to solve by substitution will prove that.

当方程的图不相交时，这意味着我们的系统没有解。 试图通过替代解决将证明这一点。

The result of x - 1 = x - 3 will be 0 = -2, which is always false.

x-1 = x-3的结果将为0 = -2 ，这始终为 false 。

But what if two graphs are the same and are directly on top of each other?

但是，如果两个图相同并且彼此直接位于顶部，该怎么办？

In such cases there are an infinite number of points of intersection. That means our system has an infinite number of solutions. Using the substitution method will prove that.

在这种情况下，会有无限多个相交点。 这意味着我们的系统拥有无限数量的解决方案。 使用替代方法将证明这一点。

The result of x - 2 = x - 2 is 0 = 0, which is always true.

x-2 = x-2的结果是0 = 0 ，这始终是正确的 。

多练 (More practice)

Try using both the substitution and graphing methods to solve the following systems. These methods complement each other and will help you solidify your knowledge.

尝试同时使用替代方法和图形方法来解决以下系统。 这些方法相互补充，将帮助您巩固知识。

Choosing a particular variable to use in substitution should make finding a solution easier.

选择用于替换的特定变量将使查找解决方案变得容易。

Try expressing x with two other members in the top equation, then substitute the result into the bottom equation. That way you'll avoid dealing with fractions.

尝试用顶部方程式中的其他两个成员表示x ，然后将结果代入底部方程式。 这样，您将避免处理分数。

Let's do one more challenge:

让我们再做一个挑战：

Now that you know enough about substitution and graphing, get out there and solve more linear equations.

既然您对替代和图形学已经足够了解，那么请走出去并解决更多的线性方程式。

翻译自: https://www.freecodecamp.org/news/how-to-solve-a-system-of-linear-equations/

线性方程组求解