📚 Solving Two-Step Equations | 解两步方程
In KS3 algebra, solving equations is one of the most useful skills you will develop. A two-step equation is a simple but powerful type of equation that prepares you for more advanced topics such as inequalities, linear graphs and simultaneous equations.
在 KS3 代数中，解方程是你将掌握的最实用的技能之一。两步方程是一种简单但重要的方程类型，能为你学习不等式、线性图像和联立方程等更高级的内容打下基础。
This article explains what two-step equations are, how to solve them systematically and how to check your answers with confidence.
本文将解释什么是两步方程，如何系统地求解，以及如何自信地检验答案。
1. What Is a Two-Step Equation? | 什么是两步方程？
t often appears in the form ax + b = c, where a, b and c are numbers and x is the unknown.
两步方程是需要恰好进行两次相反运算才能隔离未知数的方程。它通常以 ax + b = c 的形式出现，其中 a、b、c 是已知数，x 是未知数。
0 is a two-step equation because you must first subtract 5 and then divide by 3 to find x.
例如，3x + 5 = 20 就是一个两步方程，因为你必须先减去 5，再除以 3，才能求出 x。
The word 'two-step' tells you that the solution process has two clear stages, not that the equation has two terms.
“两步”这个词告诉你，求解过程有两个清晰的阶段，而不是说方程有两项。
2. Inverse Operations: The Key Idea | 反运算：核心思想
Solving an equation means finding the value of the unknown that makes the equation true. To do this, we use inverse operations, which are operations that undo each other.
解方程就是找出使方程成立的未知数的值。为此，我们使用逆运算，也就是能够互相撤销的运算。
Addition and subtraction are inverse operations. Multiplication and division are inverse operations.
加法和减法互为逆运算。乘法和除法互为逆运算。
The golden rule of equation solving is: whatever you do to one side of the equation, you must also do to the other side. This keeps the equation balanced.
解方程的黄金法则是：你对等式的一边做了什么运算，对另一边也必须做同样的运算。这样才能保持等式平衡。
When solving a two-step equation, undo the operation that is furthest from the variable first, usually the constant term being added or subtracted. Then undo the operation closest to the variable, usually multiplication or division.
在解两步方程时，先撤销离未知数最远的运算，通常是加上或减去的常数项；然后再撤销离未知数最近的运算，通常是乘法或除法。
3. Worked Example 1: Adding First | 例题1：先处理加法
Solve the equation:
解方程：
3x + 5 = 20
Step 1: Undo the addition by subtracting 5 from both sides.
第一步：撤销加法，两边同时减去 5。
3x + 5 - 5 = 20 - 5
3x = 15
Step 2: Undo the multiplication by dividing both sides by 3.
第二步：撤销乘法，两边同时除以 3。
3x ÷ 3 = 15 ÷ 3
x = 5
So the solution is x = 5.
所以方程的解是 x = 5。
4. Worked Example 2: Subtracting First | 例题2：先处理减法
Solve the equation:
解方程：
x/4 - 2 = 3
Step 1: Undo the subtraction by adding 2 to both sides.
第一步：撤销减法，两边同时加上 2。
x/4 - 2 + 2 = 3 + 2
x/4 = 5
Step 2: Undo the division by multiplying both sides by 4.
第二步：撤销除法，两边同时乘以 4。
x/4 × 4 = 5 × 4
x = 20
The solution is x = 20.
方程的解是 x = 20。
Notice that we always perform the same operation on both sides, which preserves the equality.
注意，我们总是在两边进行相同的运算，以保持等式成立。
5. Worked Example 3: Handling a Negative Coefficient | 例题3：处理负系数
Sometimes the variable has a negative coefficient. Solve:
有时未知数前面是负系数。解下面这个方程：
7 - 2x = 1
Step 1: Undo the addition of 7 by subtracting 7 from both sides.
第一步：撤销加 7，两边同时减去 7。
7 - 2x - 7 = 1 - 7
-2x = -6
Step 2: Undo the multiplication by -2 by dividing both sides by -2.
第二步：撤销乘以 -2，两边同时除以 -2。
-2x ÷ (-2) = -6 ÷ (-2)
x = 3
Remember that a negative divided by a negative gives a positive result.
请记住，负数除以负数得到正数。
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