📚 7C Homework Answers: Solving Linear Equations | 7C 作业答案：解一元一次方程
This article provides complete worked solutions for Exercise 7C, which focuses on solving one-step and two-step linear equations. Each answer is explained step by step so you can check your homework, understand the method, and avoid the mistakes that commonly appear in this exercise.
本文提供练习 7C 的完整详细解答，该练习重点是一元一次方程的一步和两步求解。每个答案都逐步解释，方便你核对作业、理解解题方法，并避开本练习中常见的错误。
Exercise 7C tests whether you can isolate a variable, use inverse operations, expand brackets, collect like terms, and handle fractional or decimal coefficients. The examples below keep the same order and focus as the original homework answers.
练习 7C 考查你能否将变量单独留在方程一边、使用逆运算、去括号、合并同类项，以及处理分数或小数系数。下面的例题保持与原作业答案相同的顺序和重点。
1. General Strategy for Solving Linear Equations | 解一元一次方程的总体策略
ke x + 4 = 9, 3x = 15, 2x + 1 = 11, or 2(x − 3) = 8. The goal is always to find the value of x that makes the equation true.
一元一次方程是指变量的最高次数为 1 的方程。在练习 7C 中，变量通常是 x，方程可能形如 x + 4 = 9、3x = 15、2x + 1 = 11 或 2(x − 3) = 8。目标始终是求出使方程成立的 x 值。
To solve any linear equation, aim to isolate the variable on one side of the equation. Use inverse operations: addition and subtraction undo each other, and multiplication and division undo each other.
解任何一元一次方程，目标都是将变量单独留在方程的一边。使用逆运算：加法和减法互相抵消，乘法和除法互相抵消。
Always keep the equation balanced by performing the same operation on both sides. If you subtract 5 from the left side, you must subtract 5 from the right side as well. Finally, check your answer by substituting it back into the original equation.
始终对方程两边执行相同的运算，以保持方程平衡。如果你在左边减去 5，就必须在右边也减去 5。最后，把答案代回原方程进行检验。
The core skills tested in 7C include expanding brackets, collecting like terms, and handling fractional or decimal coefficients. The usual order is to undo addition or subtraction first, then undo multiplication or division.
7C 考查的核心技能包括去括号、合并同类项，以及处理分数或小数系数。通常的顺序是先抵消加减，再抵消乘除。
ax + b = c → x = (c − b) ÷ a, a ≠ 0
This general form summarises the whole strategy. Whatever numbers a, b, and c are, the same balanced operations lead to x.
这个一般形式概括了整个策略。无论 a、b、c 是什么数，使用相同的平衡运算都能求出 x。
2. One-Step Equations | 一步方程
ion, x − 4 = 9 needs addition, 5x = 35 needs division, and x ÷ 3 = 6 needs multiplication.
一步方程只需要一次逆运算就能把 x 单独留下。例如，x + 7 = 12 需要减法，x − 4 = 9 需要加法，5x = 35 需要除法，x ÷ 3 = 6 需要乘法。
− 7 = 12 − 7, so x = 5. For x − 4 = 9, add 4 to both sides: x − 4 + 4 = 9 + 4, so x = 13.
对于 x + 7 = 12，两边同时减去 7：x + 7 − 7 = 12 − 7，所以 x = 5。对于 x − 4 = 9，两边同时加上 4：x − 4 + 4 = 9 + 4，所以 x = 13。
35 ÷ 5, so x = 7. For x ÷ 3 = 6, multiply both sides by 3: (x ÷ 3) × 3 = 6 × 3, so x = 18.
对于 5x = 35，两边同时除以 5：5x ÷ 5 = 35 ÷ 5，所以 x = 7。对于 x ÷ 3 = 6，两边同时乘以 3：(x ÷ 3) × 3 = 6 × 3，所以 x = 18。
x + a = b → x = b − a
ax = b → x = b ÷ a, a ≠ 0
One-step equations are the foundation of 7C. If these are accurate, two-step equations become much easier.
一步方程是 7C 的基础。如果这些题目能准确完成，两步方程就会容易得多。
3. Worked Example: Solve 3x + 5 = 20 | 例题：解 3x + 5 = 20
This example corresponds to Question 2 in Exercise 7C. It is a two-step equation because two inverse operations are needed: first subtraction, then division.
这个例子对应练习 7C 的第 2 题。它是一个两步方程，因为需要两次逆运算：先减法，再除法。
Subtract 5 from both sides: 3x + 5 − 5 = 20 − 5, giving 3x = 15.
两边同时减去 5：3x + 5 − 5 = 20 − 5，得到 3x = 15。
3x + 5 − 5 = 20 − 5 → 3x = 15
Divide both sides by 3: 3x ÷ 3 = 15 ÷ 3, hence x = 5.
两边同时除以 3：3x ÷ 3 = 15 ÷ 3，因此 x = 5。
3x ÷ 3 = 15 ÷ 3 → x = 5
Answer: x = 5. Check: 3(5) + 5 = 15 + 5 = 20, so the solution is correct.
答案：x = 5。 检验：3(5) + 5 = 15 + 5 = 20，所以解是正确的。
orrect because addition and subtraction must be undone before multiplication and division.
常见错误：有些学生先除以 3，而不是先减去 5。这样会得到 x + 5 ≈ 6.67，这是错误的，因为必须先处理加减，再处理乘除。
4. Worked Example: Solve 2(x − 3) = 8 | 例题：解 2(x − 3) = 8
This example corresponds to Question 3 in Exercise 7C. The equation contains a bracket, so you may either expand the bracket first or divide both sides by 2 first. Both methods are valid if they are done correctly.
这个例子对应练习 7C 的第 3 题。方程中含有括号，所以你可以先去括号，也可以先两边同时除以 2。只要方法正确，两种做法都成立。
First expand the bracket on the left: 2(x − 3) means 2 × x − 2 × 3, so 2x − 6 = 8.
首先展开左边的括号：2(x − 3) 表示 2 × x − 2 × 3，所以 2x − 6 = 8。
2(x − 3) = 8 → 2x − 6 = 8
Add 6 to both sides: 2x − 6 + 6 = 8 + 6, so 2x = 14.
两边同时加上 6：2x − 6 + 6 = 8 + 6，得到 2x = 14。
Divide both sides by 2: 2x ÷ 2 = 14 ÷ 2, hence x = 7.
两边同时除以 2：2x ÷ 2 = 14 ÷ 2，因此 x = 7。
Answer: x = 7. Check: 2(7 − 3) = 2 × 4 = 8, so the solution is correct.
答案：x = 7。 检验：2(7 − 3) = 2 × 4 = 8，所以解是正确的。
e bracket by 2. Writing 2(x − 3) = 2x − 3 would be wrong; the correct expansion is 2x − 6.
常见错误：忘记用 2 乘括号内的每一项。把 2(x − 3) 写成 2x − 3 是错误的；正确展开是 2x − 6。
5. Equations with Brackets | 含括号方程
When an equation contains brackets, first decide whether it is faster to expand or to divide. If the bracket is already simple and the number outside divides exactly, dividing first can save time.
当方程含括号时，先判断是展开更快，还是先除法更快。如果括号内已经很简单，且外面的数能整除，先除可以节省时间。
5(x + 2) = 35. Divide both sides by 5: x + 2 = 7, so x = 5. Check: 5(5 + 2) = 5 × 7 = 35.
例 1：5(x + 2) = 35。两边同时除以 5：x + 2 = 7，所以 x = 5。检验：5(5 + 2) = 5 × 7 = 35。
Expand: 6x − 3 = 21. Add 3: 6x = 24. Divide by 6: x = 4. Check: 3(2 × 4 − 1) = 3 × 7 = 21.
例 2：3(2x − 1) = 21。展开：6x − 3 = 21。两边加 3：6x = 24。两边除以 6：x = 4。检验：3(2 × 4 − 1) = 3 × 7 = 21。
t like terms. For example, 2(x + 3) = 3(x − 1) becomes 2x + 6 = 3x − 3, which gives x = 9.
如果两边都有括号，先展开两边，再合并同类项。例如，2(x + 3) = 3(x − 1) 变为 2x + 6 = 3x − 3，解得 x = 9。
6. Equations with Fractions | 含分数方程
Fractional equations often look harder than they are. The key is to multiply every term by the denominator or by the lowest common multiple of the denominators.
分数方程通常看起来比实际难。关键是每一项都乘以分母，或乘以所有分母的最小公倍数。
x ÷ 2 + 3 = 7. Subtract 3: x ÷ 2 = 4. Multiply by 2: x = 8. Check: 8 ÷ 2 + 3 = 4 + 3 = 7.
例 1：x ÷ 2 + 3 = 7。两边减去 3：x ÷ 2 = 4。两边乘以 2：x = 8。检验：8 ÷ 2 + 3 = 4 + 3 = 7。
) ÷ 3 = 5. Multiply both sides by 3: 2x + 1 = 15. Subtract 1: 2x = 14. Divide by 2: x = 7.
例 2：(2x + 1) ÷ 3 = 5。两边乘以 3：2x + 1 = 15。两边减去 1：2x = 14。两边除以 2：x = 7。
Example 3: (2 ÷ 3)x = 8. Multiply both sides by 3: 2x = 24. Divide by 2: x = 12.
例 3：(2 ÷ 3)x = 8。两边乘以 3：2x = 24。两边除以 2：x = 12。
(2 ÷ 3)x = 8 → 2x = 24 → x = 12
When multiplying to clear fractions, multiply every term, not only the fraction term. This is one of the most common sources of lost marks in 7C.
为了消去分数而相乘时，要乘每一项，而不是只乘分数项。这是 7C 中最常见的失分点之一。
7. Equations with Decimals | 含小数方程
Decimal coefficients can be turned into integers by multiplying the whole equation by 10, 100, or another power of 10. This makes the arithmetic cleaner.
小数系数可以通过将整个方程乘以 10、100 或其他 10 的幂转化为整数。这样计算会更简洁。
0.5x + 1.2 = 3.7. Multiply by 10: 5x + 12 = 37. Subtract 12: 5x = 25. Divide by 5: x = 5.
例 1：0.5x + 1.2 = 3.7。两边乘以 10：5x + 12 = 37。两边减去 12：5x = 25。两边除以 5：x = 5。
Example 2: 0.25x = 3. Multiply by 100: 25x = 300. Divide by 25: x = 12.
例 2：0.25x = 3。两边乘以 100：25x = 300。两边除以 25：x = 12。
e 3: 1.2x − 0.6 = 3.0. Multiply by 10: 12x − 6 = 30. Add 6: 12x = 36. Divide by 12: x = 3.
例 3：1.2x − 0.6 = 3.0。两边乘以 10：12x − 6 = 30。两边加 6：12x = 36。两边除以 12：x = 3。
Always check whether the
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