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Student's Edition Grade 2 Lesson 16-2: Solving Linear Equations and Inequalities | 学生版 第2级 第16-2课：解线性方程与不等式

2026-08-21 16:24 · 本文部分内容由AI辅助创作

📚 Student's Edition Grade 2 Lesson 16-2: Solving Linear Equations and Inequalities | 学生版 第2级 第16-2课：解线性方程与不等式

Linear equations and inequalities are core algebra skills in IGCSE Mathematics. They appear in almost every paper, either as stand-alone questions or as part of larger problem-solving tasks. This lesson reviews the key techniques in a clear order: inverse operations, expanding brackets, clearing fractions, collecting like terms on both sides, translating words into equations, and finally applying the inequality rules with care - especially the rule about negative numbers.

线性方程与不等式是 IGCSE 数学中的核心代数技能。它们几乎出现在每份试卷中，要么作为独立题目，要么作为更复杂问题的一部分。本课按清晰的顺序复习关键技巧：逆运算、去括号、去分母、在等号两边合并同类项、把文字条件转化为方程，最后是谨慎使用不等式规则——尤其是关于负数的规则。

1. What Is a Linear Equation? | 什么是线性方程？

c, where a, b and c are numbers and a is not zero. The solution of a linear equation is the value of the unknown that makes both sides equal. A linear equation normally has exactly one solution, which is why IGCSE questions usually ask for a single value of x.

线性方程是未知数只出现一次幂的方程。其标准形式为 ax + b = c，其中 a、b、c 是数字且 a 不为零。线性方程的解就是使等式两边相等的未知数的值。线性方程通常恰好只有一个解，这也是 IGCSE 题目一般只要求求出一个 x 值的原因。

also linear, but x² + 4 = 20 is not linear because the highest power is 2. Knowing what is not linear matters, because a question may ask you to explain why an equation has two solutions instead of one.

例如，在方程 2x + 3 = 11 中，解为 x = 4，因为 2(4) + 3 = 11。方程 5y - 7 = 18 也是线性方程，但 x² + 4 = 20 不是线性方程，因为最高次数是 2。知道什么不是线性方程同样重要，因为题目可能要求你解释为什么某个方程有两个解而不是一个解。

Use the table below to test whether an expression is a linear equation. Two features matter: the unknown must appear to power 1 only, and the unknown must not appear inside a root, a denominator or a product of two unknowns.

可用下表判断一个式子是否为线性方程。两个特征最关键：未知数只能是一次幂，且未知数不能出现在根号内、分母中或两个未知数的乘积里。

Equation 方程 Linear? 是否线性 Reason 理由
2x + 3 = 11 Yes 是 x is to the power 1 only
5y - 7 = 18 Yes 是 y is to the power 1 only
x² + 4 = 20 No 否 Highest power is 2
6 ÷ x = 3 No 否 x is in the denominator
xy = 12 No 否 Product of two unknowns
3(x + 1) = 9 Yes 是 Expands to 3x + 3 = 9

2. Solving One-Step Equations | 解一步方程

To solve a one-step equation, apply the inverse operation to both sides. If x has been increased by a number, subtract that number from both sides. If x has been multiplied by a number, divide both sides by that number. The equation stays balanced because you do exactly the same thing to the left and to the right.

解一步方程时，对等式两边同时进行逆运算。如果 x 加了一个数，两边同时减去这个数。如果 x 乘以了一个数，两边同时除以这个数。因为你对左边和右边做了完全相同的运算，等式始终保持平衡。

x + a = b → x = b - a

ax = b → x = b ÷ a

ls automatic. Notice that the answer is always written as x = ..., never as just a number.

请把下面四个例子做到每一步都能自动写出。注意答案永远写成 x = ... 的形式，而不是只写一个数字。

Always check your answer by substituting it back into the original equation. This habit catches sign errors and arithmetic slips before they cost you marks.

始终将答案代回原方程进行检验。这个习惯可以在丢分之前帮你发现符号错误和计算失误。

3. Solving Two-Step Equations | 解两步方程

, work backwards through the order of operations (BIDMAS in reverse): brackets, then powers, then × and ÷, then + and -.

大多数 IGCSE 方程需要两步运算。先抵消加法或减法，再抵消乘法或除法。换句话说，就是按运算顺序倒着走（BIDMAS 的逆序）：先括号，再乘方，再乘除，最后加减。

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ax + b = c → ax = c - b → x = (c - b) ÷ a

Three worked examples cover the situations you will meet most often in Paper 1 and Paper 2 short-answer questions.

下面三个例题覆盖了你在 Paper 1 和 Paper 2 短答题中最常遇到的情况。

back: 5 - 2(2) = 1, which is correct. Many students lose the mark here by writing x = -2.

请仔细检查第三个例子。两边同时减 5 得到 -2x = -4，两边同时除以 -2 得到 x = 2。代回检验：5 - 2(2) = 1，正确。很多学生在这里写成 x = -2 而丢分。

4. Expanding Brackets Before Solving | 先去括号再求解

When the unknown sits inside a bracket, you have two valid routes. If the coefficient of the bracket divides the other side exactly, divide first - it is quicker. Otherwise, expand the bracket and then solve as a normal two-step equation. Both routes give the same answer, so choose the one with fewer steps.

当未知数位于括号内时，你有两条同样正确的路线。如果括号外的系数能整除等号另一边，先做除法——这样更快。否则先展开括号，再按普通两步方程求解。两条路线答案相同，选步骤更少的那条。

a(x + b) = c → x + b = c ÷ a → x = (c ÷ a) - b

-2 × (-5) = +10. Write the expanded line down before solving, and the mistake disappears.

含括号方程最常见的错误来自负的乘数。在 -2(x - 5) 中，括号内两项都要变号：-2 × x = -2x，-2 × (-5) = +10。先把展开后的式子写下来再求解，这个错误就不会发生。

5. Clearing Fractions | 去分母

If an equation contains fractions, multiply every single term on both sides by the lowest common multiple (LCM) of the denominators. This clears all the fractions in one line and turns the question into a standard linear equation. Remember that the multiplier must be applied to terms without fractions as well.

如果方程中含有分数，就把等号两边的每一个项都乘以各分母的最小公倍数（LCM）。这样一行之内就能消去所有分母，把题目变成标准线性方程。请记住，不含分数的项也必须乘。

x ÷ 2 + x ÷ 3 = 5 → ×6 → 3x + 2x = 30 → 5x = 30 → x = 6

When a fraction is subtracted, the minus sign belongs to the whole numerator. Write extra brackets such as -(2x - 1) and expand carefully. If the answer is a decimal or a fraction, leave it in an exact form unless the question asks you to round.

当分数被减号连接时，这个负号属于整个分子。请主动加上括号，例如写成 -(2x - 1)，再小心展开。如果答案是小数或分数，除非题目要求取近似值，否则应保留精确形式。

6. Unknown on Both Sides | 未知数在等号两边

When the unknown appears on both sides, collect the unknown terms on one side and the pure numbers on the other. Choose the side that keeps the coefficient of x positive, because this avoids negative division and reduces sign errors. Then solve the resulting two-step equation.

当未知数出现在等号两边时，把含未知数的项集中到一边，把纯数字集中到另一边。选择能让 x 的系数保持正数的那一边，这样可以避免负数除法并减少符号错误。然后解剩下的两步方程。