Solving Linear Equations | 解一元一次方程

📚 Solving Linear Equations | 解一元一次方程

Solving linear equations is one of the most important algebraic skills you will develop in KS3 mathematics. It involves finding the unknown value that makes an equation true, using inverse operations and the balance method. Mastering this topic will give you a secure foundation for more advanced work, such as working with inequalities, simultaneous equations, and quadratic functions.

解一元一次方程是你在 KS3 数学中将培养的最重要的代数技能之一。它涉及使用逆运算和平衡法找出使方程成立的未知值。掌握这一主题将为你更深入的学习奠定坚实基础，例如处理不等式、联立方程和二次函数。

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1. Understanding Equations | 理解方程

An equation is a mathematical sentence that states two expressions are equal, using an equals sign (=). The expression on the left has the same value as the expression on the right. In 2x + 5 = 13, the letter x is the unknown, and our task is to discover its value.

方程是一个用等号 (=) 表示两个表达式相等的数学语句。左边的表达式与右边的表达式具有相同的值。在 2x + 5 = 13 中，字母 x 是未知数，我们的任务就是求出它的值。

Equations can be simple or more complex, but all linear equations in one variable can be solved using a set of logical steps. The key idea is that any operation we perform must keep the equation balanced.

方程可以简单或复杂，但所有一元线性方程都可以通过一套逻辑步骤求解。关键理念在于，我们进行的任何运算都必须保持方程平衡。

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2. The Balance Method | 平衡法

Imagine an equation as an old-fashioned pair of scales in perfect balance. If we add or remove weight from one side only, the scales tip. To maintain balance, whatever we do to one side of the equation, we must do exactly the same to the other side.

将方程想象成一架完全平衡的老式天平。如果只在一侧增加或移去重量，天平就会倾斜。为了保持平衡，对方程一边进行的任何操作，必须同时对另一边进行完全相同的操作。

This is sometimes called the ‘golden rule’ of algebra. If our equation is x + 4 = 9, we can subtract 4 from both sides to isolate x, giving x = 5. The balance is preserved because we have subtracted 4 from both sides.

这有时被称为代数的“黄金法则”。如果方程是 x + 4 = 9，我们可以从两边同时减去 4 来分离 x，得出 x = 5。因为两边都减去了 4，平衡得以维持。

x + 4 – 4 = 9 – 4 → x = 5

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3. Solving One-Step Equations | 解一步方程

One-step equations require a single operation to reveal the value of the unknown. There are four basic types: addition, subtraction, multiplication and division equations.

一步方程只需一步运算即可显露出未知数的值。共有四种基本类型：加法、减法、乘法和除法方程。

For an addition equation like x + 7 = 15, subtract 7 from both sides to obtain x = 8. For a subtraction equation such as x – 3 = 9, add 3 to both sides, giving x = 12.

对于像 x + 7 = 15 这样的加法方程，两边同时减去 7，得到 x = 8。对于像 x – 3 = 9 这样的减法方程，两边同时加上 3，得到 x = 12。

When the unknown is multiplied by a number, we divide. For 5x = 35, divide both sides by 5: x = 7. When the unknown is divided by a number, we multiply. For x/4 = 6, multiply both sides by 4: x = 24.

当未知数乘以一个数时，我们用除法。对于 5x = 35，两边同时除以 5：x = 7。当未知数除以一个数时，我们用乘法。对于 x/4 = 6，两边同时乘以 4：x = 24。

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4. Two-Step Equations | 两步方程

Two-step equations involve two inverse operations. The general approach is to undo any addition or subtraction first, then undo multiplication or division. This reverses the order of operations (BIDMAS/BODMAS) in which the expression was built.

两步方程涉及两步逆运算。一般的做法是首先消去任何加法或减法，然后再消去乘法或除法。这是对构建表达式时所使用的运算顺序（BIDMAS/BODMAS）的逆序。

Consider 3x + 4 = 19. First, subtract 4 from both sides to get 3x = 15. Next, divide both sides by 3 to find x = 5.

考虑 3x + 4 = 19。首先，两边同时减去 4，得到 3x = 15。接着，两边同时除以 3，求出 x = 5。

3x + 4 – 4 = 19 – 4 → 3x = 15 → x = 15 ÷ 3 = 5

Another example: (x/2) – 1 = 7. Add 1 to both sides: x/2 = 8. Multiply by 2: x = 16. Always present your steps clearly to avoid errors.

另一个例子：(x/2) – 1 = 7。两边同时加 1：x/2 = 8。再乘以 2：x = 16。始终清晰地书写步骤以避免错误。

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5. Equations Containing Brackets | 含括号的方程

When an equation has brackets, you should usually expand them first using the distributive law. Multiply the term outside the bracket by each term inside, paying careful attention to signs.

当方程含有括号时，通常应先用分配律将其展开。用括号外的项乘以括号内的每一项，并仔细注意符号。

For 4(x + 3) = 28, expand to 4x + 12 = 28. Then subtract 12: 4x = 16. Finally divide by 4: x = 4.

对于 4(x + 3) = 28，展开得到 4x + 12 = 28。接着减去 12：4x = 16。最后除以 4：x = 4。

Sometimes you may need to deal with a negative multiplier. For -2(3x – 5) = 14, expand to -6x + 10 = 14. Subtract 10: -6x = 4. Divide by -6: x = -2/3. This example also shows that solutions can be fractions.

有时你需要处理负数乘子。对于 -2(3x – 5) = 14，展开为 -6x + 10 = 14。减去 10：-6x = 4。除以 -6：x = -2/3。这个例子也表明解可以是分数。

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6. Variables on Both Sides | 变量在方程两边

When the unknown appears on both sides of the equals sign, you need to collect all the variable terms on one side and the constant terms on the other. Choose the side that makes the variable term positive, if possible.

当未知数出现在等号两边时，需要将所有含变量的项集中到一边，常数项集中到另一边。如果可能，选择能使变量项为正的一边。

Take 7x + 2 = 3x + 18. Subtract 3x from both sides: 4x + 2 = 18. Subtract 2: 4x = 16. Divide by 4: x = 4.

以 7x + 2 = 3x + 18 为例。两边同时减去 3x：4x + 2 = 18。减去 2：4x = 16。除以 4：x = 4。

A more challenging example: 5 – 2x = x + 11. Add 2x to both sides: 5 = 3x + 11. Subtract 11: -6 = 3x. Divide by 3: x = -2.

一个更具挑战性的例子：5 – 2x = x + 11。两边同时加 2x：5 = 3x + 11。减去 11：-6 = 3x。除以 3：x = -2。

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7. Equations Involving Fractions | 含分数的方程

Fractions can make equations look more intimidating, but the strategy is straightforward: multiply every term by the lowest common denominator (LCD) of all the fractions. This clears the fractions and turns the equation into an integer form.

分数会使方程看起来更复杂，但策略很简单：将所有项乘以所有分数的最小公分母 (LCD)。这样就去除了分数，将方程转化为整数形式。

For (2x)/3 + 1 = 5/3, the LCD is 3. Multiply each term: 3×(2x/3) + 3×1 = 3×(5/3) → 2x + 3 = 5. Solve: 2x = 2, x = 1.

对于 (2x)/3 + 1 = 5/3，LCD 为 3。每项乘以 3：3×(2x/3) + 3×1 = 3×(5/3) → 2x + 3 = 5。解得 2x = 2，x = 1。

When fractions have different denominators, find the LCD carefully. To solve (x+1)/4 = (x-2)/3, cross-multiply or multiply by 12. Cross-multiplication gives 3(x+1) = 4(x-2) → 3x + 3 = 4x – 8 → 11 = x, so x = 11.

当分数有不同的分母时，要仔细找出 LCD。解 (x+1)/4 = (x-2)/3 时，可以交叉相乘或乘以 12。交叉相乘得 3(x+1) = 4(x-2) → 3x + 3 = 4x – 8 → 11 = x，因此 x = 11。

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8. Negative Coefficients and Solutions | 负系数与解

When the variable ends up with a negative coefficient, you can multiply the entire equation by -1 to make the coefficient positive. This often simplifies the final step and reduces mistakes.

当变量最终带有负系数时，你可以将整个方程乘以 -1，使系数变为正数。这通常能简化最后一步并减少错误。

Example: 15 – 2x = 3x + 30. Subtract 15 from both sides: -2x = 3x + 15. Subtract 3x: -5x = 15. Multiply by -1: 5x = -15 → x = -3.

例如：15 – 2x = 3x + 30。两边同时减去 15：-2x = 3x + 15。减去 3x：-5x = 15。乘以 -1：5x = -15 → x = -3。

Remember that a negative solution is perfectly valid; it simply tells us the value is less than zero. Interpret it according to the real-world context if the equation models a situation.

记住，负解是完全有效的；它只是表明该值小于零。如果方程模拟的是一个实际情境，要根据该情境加以解释。

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9. Checking Your Answer | 检验答案

Substitution is a simple yet powerful way to confirm that your solution is correct. Replace the variable in the original equation with the value you have found and simplify both sides. If they are equal, your solution works.

代入法是确认解答正确的一种简单而有效的方法。用你求得的值替换原方程中的变量，然后化简两边。如果两边相等，你的解就是正确的。

For 5x – 7 = 3x + 9, we got x = 8. Check: left side = 5(8) – 7 = 40 – 7 = 33; right side = 3(8) + 9 = 24 + 9 = 33. Both sides match.

对于 5x – 7 = 3x + 9，我们得到 x = 8。检验：左边 = 5(8) – 7 = 40 – 7 = 33；右边 = 3(8) + 9 = 24 + 9 = 33。两边相等。

Developing the habit of checking takes only a few seconds and can prevent losing marks in tests. It also builds confidence in your algebraic manipulation.

养成检验的习惯只需几秒钟，却能避免在考试中失分。它还能增强你对自己代数操作能力的信心。

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10. Common Pitfalls | 常见错误

Students often forget to apply the same operation to both sides, especially when the operation is subtraction or division by a negative number. Another frequent error is mishandling signs when expanding brackets like -2(x – 3).

学生们常常忘记对两边施加相同的运算，特别是减法或除以负数的时候。另一个常见错误是在展开括号如 -2(x – 3) 时，符号处理不当。

Take care when moving terms: a term that moves from one side to the other changes its sign. For example, 3x + 5 = x – 7, subtracting x gives 2x + 5 = -7, not ‘2x + 5 = 7’.

移项时要小心：从一边移到另一边的项要改变符号。例如，3x + 5 = x – 7，减去 x 得到 2x + 5 = -7，而不是“2x + 5 = 7”。

Always write each step on a new line and avoid doing too much mentally. This clarity helps you spot and correct mistakes quickly.

始终将每一步写在新的一行，避免过多心算。这样清晰的书写有助于快速发现和纠正错误。

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11. Real-World Contexts | 实际背景

Linear equations are not just abstract; they model real-life situations. For instance, if a taxi charges a £3 flag-down fee plus £2 per kilometre, the cost C for d kilometres is C = 2d + 3. We can solve 2d + 3 = 15 to find the distance travelled for £15.

线性方程并非只是抽象的，它们能模拟现实生活中的情况。例如，出租车起步价为 3 英镑，每公里加收 2 英镑，则行驶 d 公里的费用 C = 2d + 3。我们可以解方程 2d + 3 = 15 来求 15 英镑能行驶的距离。

Solving gives 2d = 12 → d = 6 km. This demonstrates the power of algebraic problem-solving in everyday decisions.

解方程得 2d = 12 → d = 6 公里。这展示了代数解题在日常生活中决策的力量。

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12. Building Fluency and Next Steps | 培养熟练度与后续学习

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