📚 Page 286: Solving Linear Equations | 第286页：求解线性方程
Welcome to Page 286 of your Cambridge Lower Secondary Mathematics textbook. This page focuses on solving linear equations, a fundamental skill that underpins much of algebra. You will learn to isolate the unknown variable using inverse operations, maintain the balance of an equation, and apply these methods to a variety of equation types including one-step, two-step, equations with brackets, and those with unknowns on both sides. Mastering this page prepares you for more advanced topics such as simultaneous equations, inequalities, and algebraic problem-solving across the Key Stage 3 curriculum.
欢迎来到剑桥初中数学教材第286页。本页重点讲解线性方程的求解，这是代数中的基础技能。你将学习如何使用逆运算分离未知数，保持等式的平衡，并将这些方法应用于各种方程类型，包括一步方程、两步方程、带括号的方程以及未知数在等号两边的方程。掌握本页内容将为你学习联立方程、不等式以及 KS3 课程中的代数解题打下坚实的基础。
1. What is an Equation? | 什么是方程？
An equation is a statement that two mathematical expressions have the same value. It includes an equals sign (=), a left-hand side (LHS) and a right-hand side (RHS). For example, x + 4 = 10 is an equation. The letter x represents an unknown number that we need to find.
方程是说明两个数学表达式具有相同值的陈述。它包括等号（=）、左侧表达式（LHS）和右侧表达式（RHS）。例如，x + 4 = 10 就是一个方程。字母 x 代表我们需要求出的未知数。
The key principle of solving equations is to keep the scale balanced—whatever you do to one side, you must also do to the other. Imagine a balance scale: if we add or remove weights equally from both sides, the balance remains level. Similarly, any operation applied to the equation must be performed on both sides.
解方程的核心原则是保持等号两边的平衡——你对一边做什么，就必须对另一边做同样的操作。想象一架天平：如果从两边同时添加或移除相同的重量，天平仍然保持水平。同样，对方程施加的任何运算都必须同时在等号两边进行。
2. One-Step Equations – Addition and Subtraction | 一步方程：加法和减法
To solve an equation like x + 7 = 15, we need to undo the addition of 7. The inverse operation of addition is subtraction, so we subtract 7 from both sides: x + 7 – 7 = 15 – 7, giving x = 8.
要求解如 x + 7 = 15 的方程，我们需要抵消加上 7 的操作。加法的逆运算是减法，因此我们在等号两边同时减去 7：x + 7 – 7 = 15 – 7，得到 x = 8。
Similarly, for x – 3 = 9, we undo the subtraction by adding 3 to both sides: x – 3 + 3 = 9 + 3, so x = 12. Always remember that the equality must be preserved.
同样，对于 x – 3 = 9，我们通过在两边同时加上 3 来抵消减法：x – 3 + 3 = 9 + 3，因此 x = 12。请始终记住必须保持等号两边的相等关系。
Example: x – 8 = 5 → x = 13
3. One-Step Equations – Multiplication and Division | 一步方程：乘法和除法
If the equation involves multiplication, such as 4x = 20, use division to isolate x. Divide both sides by the coefficient of x: 4x ÷ 4 = 20 ÷ 4, so x = 5.
如果方程涉及乘法，例如 4x = 20，则用除法将 x 分离。在等号两边同时除以 x 的系数：4x ÷ 4 = 20 ÷ 4，因此 x = 5。
For division equations like x/3 = 6, the inverse operation is multiplication. Multiply both sides by 3: (x/3) × 3 = 6 × 3, resulting in x = 18. In general, to remove a denominator, multiply both sides by that number.
对于如 x/3 = 6 的除法方程，逆运算是乘法。将两边同时乘以 3：(x/3) × 3 = 6 × 3，得到 x = 18。通常，为了去掉分母，可以将两边同时乘以该数。
Example: x/5 = 7 → x = 35
4. Two-Step Equations – Combining Operations | 两步方程：复合运算
A two-step equation contains two different operations affecting the unknown, e.g., 2x + 3 = 11. The order of unwrapping is crucial: first undo the addition or subtraction, then undo the multiplication or division. Think of it as ‘peeling the onion’ from the outside in.
两步方程包含影响未知数的两种不同运算，例如 2x + 3 = 11。解方程的拆解顺序至关重要：先抵消加法或减法，再抵消乘法或除法。可以将其想象为从外到内“剥洋葱”的过程。
Solve 2x + 3 = 11: Start by subtracting 3 from both sides to get 2x = 8. Then divide both sides by 2 to find x = 4. Check: 2(4) + 3 = 8 + 3 = 11, correct. This method ensures you reverse the order of operations.
求解 2x + 3 = 11：首先从两边减去 3，得到 2x = 8。然后将两边同时除以 2，得出 x = 4。验算：2(4) + 3 = 8 + 3 = 11，正确。这一方法确保你按照逆运算的顺序步骤求解。
Example: 5x – 7 = 23 → 5x = 30 → x = 6
5. Equations with Brackets | 带括号的方程
When an equation contains brackets, such as 3(x + 2) = 18, you have two strategies: either expand the brackets first or divide both sides by the coefficient in front of the bracket. Expanding gives 3x + 6 = 18, then solve: subtract 6 → 3x = 12, divide by 3 → x = 4.
当方程带有括号时，如 3(x + 2) = 18，你可以采用两种策略：要么先展开括号，要么在两边同时除以括号前的系数。展开得到 3x + 6 = 18，然后求解：减去 6 → 3x = 12，除以 3 → x = 4。
Alternatively, dividing both sides by 3 first gives x + 2 = 6, then subtract 2 to get x = 4. The second approach is often quicker when the number outside the bracket divides evenly into the right-hand side.
或者，先将两边同时除以 3，得到 x + 2 = 6，然后减去 2，得出 x = 4。当括号外的数能被等号右边的数整除时，第二种方法通常更快。
Example: 4(x – 3) = 24 → x – 3 = 6 → x = 9
6. Equations with Unknowns on Both Sides | 未知数在两边
Equations like 5x – 7 = 2x + 5 require you to collect the unknown terms on one side and constants on the other. Start by eliminating the smaller unknown term from both sides. Here, subtract 2x from both sides: 5x – 2x – 7 = 2x – 2x + 5, simplifying to 3x – 7 = 5.
像 5x – 7 = 2x + 5 这样的方程要求你将未知项集中到等号一边，常数项集中到另一边。首先从两边消去较小的未知数项。本例中，两边减去 2x：5x – 2x – 7 = 2x – 2x + 5，简化为 3x – 7 = 5。
Now add 7 to both sides: 3x = 12, then divide by 3 to find x = 4. Check by substituting back: LHS = 5(4) – 7 = 20 – 7 = 13; RHS = 2(4) + 5 = 8 + 5 = 13. The solution is correct.
现在两边加上 7：3x = 12，然后除以 3 得到 x = 4。代回原方程验算：左侧 = 5(4) – 7 = 20 – 7 = 13；右侧 = 2(4) + 5 = 8 + 5 = 13。解正确。
Example: 7x + 3 = 4x + 18 → 3x = 15 → x = 5
7. Equations Involving Fractions | 涉及分数的方程
When fractions appear, it is often helpful to eliminate them by multiplying every term by the lowest common denominator (LCD). For example, in x/2 + 1 = x/3 + 7/6, the LCD of 2, 3, and 6 is 6. Multiply each term by 6: 6·(x/2) + 6·1 = 6·(x/3) + 6·(7/6), giving 3x + 6 = 2x + 7.
当方程中出现分数时，通常采取乘以所有分母的最小公倍数 (LCD) 来消除分数。例如，方程 x/2 + 1 = x/3 + 7/6 中，2、3 和 6 的 LCD 是 6。将每一项乘以 6：6·(x/2) + 6·1 = 6·(x/3) + 6·(7/6)，得到 3x + 6 = 2x + 7。
Now solve the resulting integer equation: subtract 2x → x + 6 = 7, subtract 6 → x = 1. Always verify: LHS: 1/2 + 1 = 1.5; RHS: 1/3 + 7/6 = 2/6 + 7/6 = 9/6 = 1.5. The solution holds.
现在求解所得的整数方程：减去 2x → x + 6 = 7，减去 6 → x = 1。始终要验证：左侧：1/2 + 1 = 1.5；右侧：1/3 + 7/6 = 2/6 + 7/6 = 9/6 = 1.5。解成立。
Example: x/4 – 1 = x/6 + 5/12 → Multiply by 12: 3x – 12 = 2x + 5 → x = 17
8. Checking Your Solution | 验算你的解
Never skip the verification step. Substitute your found value back into the original equation to ensure both sides evaluate to the same number. If they do not, retrace your steps to find the algebraic error. This habit prevents silly mistakes and builds confidence in your answer.
切勿跳过验算步骤。将求出的数值代回原方程，确保等号两边的计算结果相等。如果不相等，请回溯每一个步骤以找出代数
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