Solving Simple Equations | 解简单方程

📚 Solving Simple Equations | 解简单方程

In KS3 mathematics, solving equations is a foundational skill that opens the door to algebra. An equation is like a balance scale – both sides must be equal. This guide will walk you through solving simple linear equations step by step, covering one‑step, two‑step, and equations with brackets. By mastering these techniques, you will build confidence for more advanced topics later.

在KS3数学中，解方程是开启代数大门的基础技能。方程就像一个天平——两边必须相等。本指南将逐步带你解简单线性方程，涵盖一步方程、两步方程和带括号的方程。掌握这些技巧后，你将更有信心应对后续更深入的内容。

1. Understanding Equations | 理解方程

An equation is a mathematical statement that shows two expressions are equal, using the equals sign (=). For example, x + 4 = 10 is an equation. The letter x is called a variable, representing an unknown number. The goal of solving an equation is to find the value of the variable that makes the statement true.

方程是一个数学陈述，用等号（=）表示两个表达式相等。例如，x + 4 = 10 是一个方程。字母 x 称为变量，代表一个未知数。解方程的目标就是找出使这个陈述成立的变量的值。

Think of the equation as a balanced seesaw. Whatever you do to one side, you must do to the other to keep it balanced. If you add 3 to the left, you must add 3 to the right. This idea underpins every step of equation solving.

把方程想象成一个平衡的跷跷板。无论你对一边做什么操作，都必须对另一边做同样操作以保持平衡。如果你在左边加3，就必须在右边也加3。这个思想支撑着解方程的每一步。

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2. The Balancing Method | 天平法

The balancing method is the most visual way to solve equations. If you add, subtract, multiply, or divide a number on one side, you must do exactly the same on the other side. This keeps the equation balanced and helps isolate the variable.

天平法是最直观的解方程方法。如果你在一边加上、减去、乘以或除以一个数，就必须在另一边做完全相同的操作。这样能保持方程平衡，并帮助分离出变量。

For example, to solve x – 3 = 5, you add 3 to both sides: x – 3 + 3 = 5 + 3, which simplifies to x = 8. Always remember to perform the operation on the entire side, not just part of it.

例如，解 x – 3 = 5，两边都加3：x – 3 + 3 = 5 + 3，化简得 x = 8。始终记得要对整个边进行运算，而不只是其中的一部分。

x – 3 = 5 → x = 8

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3. Inverse Operations | 逆运算

To isolate the variable, we use inverse (opposite) operations. Addition and subtraction are inverses; multiplication and division are inverses. For instance, if the equation has ‘+ 7’, we use ‘- 7’ to undo it. Understanding these pairs is crucial for selecting the correct step.

为了分离出变量，我们使用逆运算（相反运算）。加法和减法互为逆运算；乘法和除法互为逆运算。例如，如果方程中有“+ 7”，我们就用“- 7”来抵消它。理解这些运算对是选择正确步骤的关键。

Operation	Inverse
+ a	– a
– a	+ a
x a	÷ a
÷ a	x a

The table above summarises common operations and their inverses. When you see a number added to the variable, subtract that number from both sides. When the variable is multiplied by a number, divide both sides by that number.

上表总结了常见运算及其逆运算。当你看到变量加上一个数，就从两边减去这个数。当变量乘以一个数，就将两边除以这个数。

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4. Solving One-Step Equations | 解一步方程

One-step equations require only one inverse operation to solve. The key is to identify the operation affecting the variable and apply its inverse. Let’s look at several examples.

一步方程只需一次逆运算即可求解。关键在于识别影响变量的运算，并应用其逆运算。我们来看几个例子。

Addition example: x + 9 = 15. Subtract 9 from both sides: x + 9 – 9 = 15 – 9, so x = 6.

加法例子：x + 9 = 15。两边减9：x + 9 – 9 = 15 – 9，得 x = 6。

Subtraction example: y – 4 = 10. Add 4 to both sides: y – 4 + 4 = 10 + 4, giving y = 14.

减法例子：y – 4 = 10。两边加4：y – 4 + 4 = 10 + 4，得 y = 14。

Multiplication example: 3z = 18. Divide both sides by 3: 3z ÷ 3 = 18 ÷ 3, so z = 6.

乘法例子：3z = 18。两边除以3：3z ÷ 3 = 18 ÷ 3，得 z = 6。

Division example: a ÷ 5 = 7. Multiply both sides by 5: a ÷ 5 x 5 = 7 x 5, so a = 35.

除法例子：a ÷ 5 = 7。两边乘以5：a ÷ 5 x 5 = 7 x 5，得 a = 35。

When dealing with negative numbers, be careful with signs. Example: -2k = 14. Divide both sides by -2: -2k ÷ -2 = 14 ÷ -2, yielding k = -7. Another example: x + (-5) = -3, which is the same as x – 5 = -3. Add 5 to both sides: x = 2.

处理负数时，要小心符号。例子：-2k = 14，两边除以-2，得 k = -7。另一个例子：x + (-5) = -3 等同于 x – 5 = -3，两边加5得 x = 2。

Equations with fractions can be solved in one step. Example: x/8 = 1.5. Multiply both sides by 8: x = 12. Similarly, (2/3)y = 10 can be solved by multiplying by the reciprocal: y = 10 x (3/2) = 15.

含有分数的方程也可以一步求解。例如：x/8 = 1.5，两边乘以8得 x = 12。类似地，(2/3)y = 10 可通过乘以倒数求解：y = 10 x (3/2) = 15。

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5. Solving Two-Step Equations | 解两步方程

Two-step equations involve two operations. The general strategy is to first undo the addition or subtraction (the constant term), and then undo the multiplication or division (the coefficient). This isolates the variable efficiently.

两步方程包含两个运算。一般策略是首先抵消加减法（常数项），然后抵消乘除法（系数）。这样可以高效地分离出变量。

Example: 2x + 3 = 11. Step 1: subtract 3 from both sides → 2x = 8. Step 2: divide both sides by 2 → x = 4.

例子：2x + 3 = 11。第一步：两边减3 → 2x = 8。第二步：两边除以2 → x = 4。

2x + 3 = 11 → 2x = 8 → x = 4

When the variable term is written as a fraction, still follow the same order. Example: (y/4) – 6 = 2. Add 6 to both sides: y/4 = 8. Multiply by 4: y = 32.

当变量项写成分数形式时，仍按相同顺序操作。例子：(y/4) – 6 = 2。两边加6得 y/4 = 8，乘以4得 y = 32。

If the equation has a negative coefficient, handle it carefully. Example: 7 – 3x = 19. Here, subtract 7 from both sides first: -3x = 12. Then divide by -3: x = -4. Alternatively, add 3x to both sides: 7 = 19 + 3x, then subtract 19: -12 = 3x, so x = -4. Both routes give the same result.

如果方程有负系数，小心处理。例子：7 – 3x = 19。首先两边减7：-3x = 12，然后除以-3得 x = -4。或者两边加上3x：7 = 19 + 3x，再减19得 -12 = 3x，所以 x = -4。两种路径结果一致。

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6. Equations with Brackets | 带括号的方程

When an equation contains brackets, such as 3(x + 2) = 15, you have two main approaches: expand the brackets first, or divide both sides by the coefficient preceding the brackets if it is a factor. Both methods are valid.

当方程含有括号，如 3(x + 2) = 15，你主要有两种方法：先展开括号，或者如果括号前有系数，先两边除以这个系数。两种方法都有效。

Method 1 – Expand: 3(x + 2) = 15 becomes 3x + 6 = 15. Subtract 6: 3x = 9. Divide by 3: x = 3.

方法一——展开：3(x + 2) = 15 变成 3x + 6 = 15。减6得 3x = 9，除以3得 x = 3。

Method 2 – Divide first: divide both sides by 3 to get x + 2 = 5. Subtract 2: x = 3. This method is often quicker when the coefficient divides the right‑hand side evenly.

方法二——先除以系数：两边除以3得 x + 2 = 5。减2得 x = 3。当系数能整除右边时，这种方法通常更快。

Be cautious with negative signs outside brackets. Example: -2(x – 4) = 6. Expand carefully: -2x + 8 = 6. Subtract 8: -2x = -2. Divide by -2: x = 1. If dividing first, remember to divide the entire left side by -2, leading to x – 4 = -3, then x = 1.

注意括号外的负号。例子：-2(x – 4) = 6。仔细展开：-2x + 8 = 6。减8得 -2x = -2，除以-2得 x = 1。如果先除以-2，记得将整个左边除以-2，得到 x – 4 = -3，然后 x = 1。

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7. Checking Your Solution | 检验你的解

Always substitute your solution back into the original equation to verify it satisfies the equality. A correct answer makes the left‑hand side equal the right‑hand side. For x = 4 in 2x + 3 = 11: 2(4) + 3 = 8 + 3 = 11, which matches the right side, confirming the solution.

始终将解代回原方程，验证它满足等式。正确的答案使左边等于右边。对于 2x + 3 = 11 中的 x = 4：2(4) + 3 = 8 + 3 = 11，与右边一致，确认解正确。

Checking is especially important when you have dealt with multiple steps, negative numbers, or brackets. It helps catch errors such as sign mistakes or performing operations in the wrong order. Even a quick mental check saves mark losses in exams.

当你处理了多个步骤、负数或括号时，检验尤为重要。它有助于发现诸如符号错误或运算顺序错误等问题。即使一个快速的心算检验也能避免考试丢分。

For an equation with brackets, e.g., 3(2x – 1) = 9, if you found x = 2, check: 3(4 – 1) = 3 x 3 = 9, correct. Get into the habit of checking every solution.

对于带括号的方程，例如 3(2x – 1) = 9，若你求得 x = 2，检验：3(4 – 1) = 3 x 3 = 9，正确。养成检验每个解的习惯。

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8. Common Mistakes to Avoid | 常见错误避免

Many errors in solving equations come from rushing or forgetting the balancing rule. Here are the most frequent pitfalls and how to avoid them.

解方程中的许多错误源于匆忙或忘记平衡规则。以下是常见陷阱及如何避免。

• Not doing the same to both sides. Always apply the operation to the entire expression on each side.
• Using the wrong inverse operation. Double‑check whether you are adding, subtracting, multiplying, or dividing correctly.
• Losing track of negative signs. Treat negative numbers with extra care, especially when expanding brackets.
• Expanding brackets incorrectly. Multiply each term inside the bracket by the factor outside, including the sign.
• Forgetting to check the solution. A quick substitution can reveal mistakes instantly.

• 没有对两边做相同操作。始终对每一边的整个表达式进行运算。
• 使用了错误的逆运算。仔细检查你是正确地在加、减、乘还是除。
• 忽略了负号。处理负数时要格外小心，特别是展开括号时。
• 括号展开不正确。将外面的因数乘以括号内的每一项，包括符号。
• 忘记检验解。快速代入能立即发现错误。

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9. Practice Problem Walkthrough | 练习题讲解

Let’s work through a few examples step by step to reinforce the techniques.

我们逐步解答几个例子，巩固技巧。

Problem 1: Solve 5(2y – 1) = 15. Step 1: Divide both sides by 5 → 2y – 1 = 3. Step 2: Add 1 to both sides → 2y = 4. Step 3: Divide by 2 → y = 2. Check: 5(2×2 – 1) = 5(4 – 1) = 5 x 3 = 15. Correct.

问题1: 解 5(2y – 1) = 15。第一步：两边除以5 → 2y – 1 = 3。第二步：两边加1 → 2y = 4。第三步：除以2 → y = 2。检验：5(2×2 – 1) = 5(4 – 1) = 5 x 3 = 15。正确。

Problem 2: Solve (x/3) + 4 = 9. Subtract 4 from both sides: x/3 = 5. Multiply by 3: x = 15. Check: 15/3 + 4 = 5 + 4 = 9, correct.

问题2: 解 (x/3) + 4 = 9。两边减4得 x/3 = 5，乘以3得 x = 15。检验正确。

Problem 3: Solve 4 – x = 10. This can be tricky. Add x to both sides: 4 = 10 + x. Subtract 10: -6 = x, so x = -6. Check: 4 – (-6) = 4 + 6 = 10.

问题3: 解 4

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