📚 KS3 Cambridge: Solving Linear Equations with Brackets | 解含括号的一元一次方程
Linear equations with brackets are a key skill in KS3 Cambridge Mathematics. In this article, you will learn how to expand brackets, collect like terms, balance both sides, and check your solution step by step.
含括号的一元一次方程是 KS3 剑桥数学中的核心技能。本文将带你一步步学习如何展开括号、合并同类项、平衡等式两边，并检验你的答案。
ou will be able to solve equations such as 3(x - 2) = 12 and 5x + 4 = 2x + 19 confidently.
本主题建立在你之前学习代数式和方程的基础上。学完本文后，你将能够自信地解出像 3(x - 2) = 12 和 5x + 4 = 2x + 19 这样的方程。
1. What Is a Linear Equation? | 什么是一元一次方程？
be written in the general form ax + b = c, where a, b and c are numbers and a is not zero.
一元一次方程是指未知数的最高次数为 1 的方程。它的一般形式为 ax + b = c，其中 a、b、c 是数，且 a 不等于 0。
n exponent of 1. Quadratic equations, such as x² + 3x = 10, are not covered in this topic.
例如，2x + 6 = 14 是一元一次方程，因为变量 x 的指数是 1。二次方程（如 x² + 3x = 10）不在本主题范围内。
Linear equations graph as straight lines, which is why they are called 'linear'. Solving them means finding the value of x that makes the statement true.
一元一次方程的图像是一条直线，因此被称为“线性”方程。解这类方程意味着求出使等式成立的 x 的值。
2. The Distributive Law and Brackets | 分配律与括号
Before you can solve an equation with brackets, you must understand the distributive law. This law states that multiplying a number by a sum is the same as multiplying each addend separately and then adding the results.
在解含括号的方程之前，你必须先理解分配律。分配律指出，用一个数乘以一个和，等于分别乘以每个加数后再相加。
a(b + c) = ab + ac
mplifies to 2x + 6. You must multiply every term inside the bracket by the number outside.
在代数中，这意味着 2(x + 3) 等同于 2 × x + 2 × 3，化简后得到 2x + 6。你必须用括号外的数乘以括号内的每一项。
This step is essential because the equation cannot be solved correctly while the unknown is still trapped inside the brackets.
这一步至关重要，因为当未知数还在括号内时，方程无法被正确求解。
3. Expanding Brackets with Positive Numbers | 正数乘以括号的展开
When the number outside the bracket is positive, the signs of the terms inside the bracket stay the same after expansion. For example, 4(x + 5) becomes 4x + 20.
当括号外的数是正数时，展开后括号内各项的符号保持不变。例如，4(x + 5) 展开为 4x + 20。
+ 12 = 27. Then subtract 12 from both sides to get 3x = 15, and divide by 3 to find x = 5.
以方程 3(x + 4) = 27 为例。首先展开左边，得到 3x + 12 = 27。然后两边同时减去 12，得到 3x = 15，再除以 3，求得 x = 5。
Always show the expansion as a separate line in your working. This makes it easier to spot arithmetic mistakes and helps the examiner follow your method.
解题时一定要把展开写成单独的一行。这样可以更容易发现计算错误，也有助于阅卷人理解你的解题方法。
4. Expanding Brackets with Negative Numbers | 负数乘以括号的展开
e sign of every term inside. For instance, -3(x - 2) becomes -3x + 6 because -3 × -2 = +6.
括号外是负数时，会改变括号内每一项的符号。例如，-3(x - 2) 展开为 -3x + 6，因为 -3 × -2 = +6。
= 14. Add 10 to both sides: -2x = 24. Finally, divide both sides by -2 to obtain x = -12.
以方程 -2(x + 5) = 14 为例。展开得到 -2x - 10 = 14。两边同时加 10，得 -2x = 24。最后两边同时除以 -2，得到 x = -12。
Be especially careful with double negatives. If you see -5(x - 6), the correct expansion is -5x + 30, not -5x - 30.
要特别注意双重负号。如果你看到 -5(x - 6)，正确的展开是 -5x + 30，而不是 -5x - 30。
5. Collecting Like Terms | 合并同类项
After expanding brackets, the equation may have several x terms and several number terms. Collect the like terms on each side before you begin balancing the equation.
展开括号后，方程中可能有多个含 x 的项和多个数字项。在开始平衡等式之前，要先把两边的同类项分别合并。
For example, 4x + 5 + 3x - 2 simplifies to 7x + 3. Like terms have exactly the same variable part, so 4x and 3x can be combined, but 4x and 4 cannot.
例如，4x + 5 + 3x - 2 可以简化为 7x + 3。同类项必须含有完全相同的字母部分，所以 4x 和 3x 可以合并，但 4x 和 4 不能合并。
t like terms to get 7x + 6 = 35, then solve to find x = 29/7, which is approximately 4.14.
假设你有 2(x + 3) + 5x = 35。展开得到 2x + 6 + 5x = 35。合并同类项得到 7x + 6 = 35，然后解方程，得到 x = 29/7，约等于 4.14。
6. Using Inverse Operations | 使用逆运算
Solving a linear equation involves using inverse operations to isolate x. Addition and subtraction are inverse operations, and so are multiplication and division.
解一元一次方程需要用逆运算来分离出 x。加法和减法是互逆运算，乘法和除法也是互逆运算。
f operations is: multiply x by 2, then add 6. To undo this, subtract 6 first, then divide by 2.
对于方程 2x + 6 = 14，运算顺序是：先把 x 乘以 2，再加 6。要逆推这个过程，就要先减 6，再除以 2。
2x + 6 = 14 → 2x = 8 → x = 4
Always apply the inverse operation to both sides of the equation. This keeps the equation balanced, just like keeping both sides of a scale level.
一定要对等式两边同时进行逆运算。这能保持等式平衡，就像让天平的两端保持水平一样。
7. Unknown on Both Sides | 未知数在等号两边
When the unknown appears on both sides of the equation, the first goal is to gather all x terms on one side and all number terms on the other side. Choose to move the smaller x term to avoid negative coefficients.
当未知数出现在等号两边时，第一个目标是把所有含 x 的项移到一边，把所有数字项移到另一边。最好移动较小的 x 项，以避免出现负系数。
des to get 3x - 3 = 9. Then add 3 to both sides to get 3x = 12. Dividing by 3 gives x = 4.
对于 5x - 3 = 2x + 9，两边同时减去 2x，得到 3x - 3 = 9。然后两边同时加 3，得到 3x = 12。除以 3，得到 x = 4。
If you accidentally move the larger x term instead, you will still get the same answer, but your working may include more negative signs.
如果你不小心移动了较大的 x 项，最终答案仍然相同，但计算过程可能会出现更多负号。
8. Equations with Fractions | 含分数的方程
Fractions in equations can be cleared by multiplying
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