📚 Solving Linear Equations and Inequalities | 解一元一次方程与不等式
A linear equation is one of the most important building blocks in KS3 mathematics. It appears in algebra, problem solving and even geometry. In this article, you will learn how to solve equations step by step, including equations with brackets, fractions and unknowns on both sides. You will also learn how to solve and represent inequalities.
一元一次方程是 KS3 数学中最重要的基础内容之一。它出现在代数、问题解决甚至几何中。在本文中，你将逐步学习如何解方程，包括含有括号、分数和两边都有未知数的方程。你还将学习如何解不等式并表示其解集。
1. What is a Linear Equation? | 什么是一元一次方程？
A linear equation is an equation where the unknown variable, usually x, is only raised to the power 1. It contains an equals sign and at least one unknown term. The solution is the value of the unknown that makes the equation true.
一元一次方程是指未知数（通常用 x 表示）的指数仅为 1 的方程。它包含等号和至少一个未知项。方程的解就是使等式成立的未知数的值。
x + 4 = 9 ⇒ x = 5
e one solution, no solution, or infinitely many solutions, but at KS3 level you will mostly meet equations with one solution.
在这个例子中，5 是方程的解，因为 5 + 4 = 9。一元一次方程可以有一个解、没有解或无穷多个解，但在 KS3 阶段你遇到的大多数方程只有一个解。
2. The Balancing Method | 平衡法
Think of an equation as a balance scale. Whatever you do to one side of the equation, you must do to the other side as well. This is called the balancing method and it is the golden rule for solving equations.
把方程想象成一个天平。你对等式一边做的任何运算，必须对另一边做同样的运算。这称为平衡法，也是解方程的黄金法则。
x + 7 = 15
x + 7 − 7 = 15 − 7
x = 8
To keep the equation balanced, subtracting 7 from both sides is necessary. The same idea applies to addition, subtraction, multiplication and division.
为了保持等式平衡，必须从两边同时减去 7。同样的思想也适用于加法、减法、乘法和除法。
3. Using Inverse Operations | 运用逆运算
Addition and subtraction are inverse operations. Multiplication and division are inverse operations. To isolate the unknown, apply the inverse operation to both sides of the equation.
加法和减法互为逆运算，乘法和除法互为逆运算。为了分离未知数，需要对等式两边同时进行逆运算。
| Operation | Inverse Operation |
|---|---|
| + | − |
| − | + |
| × | ÷ |
| ÷ | × |
4x = 20
x = 20 ÷ 4
x = 5
The coefficient 4 is multiplying x, so we divide both sides by 4. This cancels the multiplication and leaves x on its own.
系数 4 正在乘以 x，所以我们将两边同时除以 4。这样就能抵消乘法，使 x 单独留在等式的一边。
4. Equations with the Unknown on Both Sides | 两边都含未知数的方程
olve these, start by moving the smaller x term to the other side using inverse operations.
有些方程两边都有 x，例如 5x + 2 = 2x + 14。解这类方程时，首先通过逆运算把较小的 x 项移到另一边。
5x + 2 = 2x + 14
5x − 2x + 2 = 14
3x + 2 = 14
3x = 12
x = 4
After moving the 2x term, the equation becomes simpler. Then use inverse operations to isolate x. Always check your answer by substituting it back into the original equation.
移动 2x 项之后，方程变得更简单。然后使用逆运算分离 x。一定要通过代入原方程来检验答案。
5. Equations with Brackets | 含有括号的方程
When an equation contains brackets, always expand them first. Multiply each term inside the bracket by the number or term outside the bracket.
当方程含有括号时，一定要先展开括号。用括号外的数或项乘以括号内的每一项。
2(x + 3) = 14
2x + 6 = 14
2x = 8
x = 4
xample, 3(2x − 1) = 2(x + 5) becomes 6x − 3 = 2x + 10, then solve to get x = 13/4 or 3.25.
对于更复杂的方程，先仔细展开括号，再合并同类项。例如，3(2x − 1) = 2(x + 5) 可化为 6x − 3 = 2x + 10，然后解得 x = 13/4 或 3.25。
6. Equations with Fractions | 含有分数的方程
To remove fractions, multiply every term on both sides by the lowest common denominator, or LCD. This makes the equation easier to solve without changing the solution.
为了去掉分数，将等式两边的每一项都乘以最小公分母（LCD）。这样可以使方程更容易求解，且不会改变方程的解。
x/3 + 1 = 2
3(x/3) + 3(1) = 3(2)
x + 3 = 6
x = 3
ts around it before multiplying. For example, (x + 4)/3 = 5 becomes x + 4 = 15, so x = 11.
如果分数的分子含有多个项，在乘法前先用括号括起来。例如，(x + 4)/3 = 5 可化为 x + 4 = 15，因此 x = 11。
7. Solving Linear Inequalities | 解一元一次不等式
You solve inequalities in the same way as equations, but remember one key rule: multiplying or dividing by a negative number reverses the inequality sign.
不等式与方程类似，但使用 <、>、≤ 和 ≥ 等符号。解不等式的方法与解方程相同，但请记住一条关键规则：乘以或除以负数时，不等式方向要改变。
3x + 1 < 10
3x < 9
x < 3
With a negative coefficient, the direction must change. For example, −2x > 6 becomes x < −3 after dividing both sides by −2 and reversing the sign.
当系数为负数时，方向必须改变。例如，−2x > 6 在两边同时除以 −2 并改变方向后变为 x < −3。
8. Representing Inequalities on a Number Line | 在数轴上表示不等式
; or >, and a closed circle for ≤ or ≥. Shade the region that satisfies the inequality.
可以在数轴上表示不等式的解集。对于 < 或 > 使用空心圆，对于 ≤ 或 ≥ 使用实心圆。涂黑满足不等式的区域。
x ≤ 3 is shown with an open circle at −1, a closed circle at 3, and shading between them.
例如，x ≥ 2 在 2 处画实心圆，并向右涂黑。不等式 −1 < x ≤ 3 在 −1 处画空心圆，在 3 处画实心圆，并在两者之间涂黑。
9. Common Mistakes and How to Avoid Them | 常见错误及避免方法
Many marks are lost in exams because of small errors. Learning to spot common mistakes will help you avoid them.
考试中很多失分都是由小错误造成的。学会识别常见错误将帮助你避免这些错误。
Mistake 1: Forgetting to do the same operation on both sides. Always apply the balancing method to both sides of the equation.
错误一：忘记对等式两边进行相同的运算。一定要对等式两边同时使用平衡法。
Mistake 2: Sign errors when expanding brackets. Remember that −(x − 3) equals −x + 3, not −x − 3.
错误二：展开括号时出现符号错误。请记住 −(x − 3) 等于 −x + 3，而不是 −x − 3。
Mistake 3: Forgetting to reverse the inequality sign when multiplying or dividing by a negative number. For example, −2x > 6 gives x < −3.
错误三：乘以或除以负数时忘记改变不等式方向。例如，−2x > 6 应得出 x < −3。
Mistake 4: Losing a term when moving it to the other side. Write each step clearly and do not skip stages.
错误四：将项移到另一边时漏掉某一项。每一步都要写清楚，不要跳步。
10. Word Problems Leading to Linear Equations | 应用题转化为方程
Many exam questions ask you to form an equation from a real-life situation. Read the question carefully, define the unknown, and translate the words into algebra.
很多考试题目要求你根据实际情境列出方程。仔细读题、设未知数，并把文字转化为代数式。
ce a number is 15. Let n be the number. The equation is 2n + 3 = 15, so 2n = 12 and n = 6.
例一：一个数的两倍再加 3 等于 15。设这个数为 n，方程为 2n + 3 = 15，因此 2n = 12，n = 6。
th is w + 3. The perimeter equation is 2(w + w + 3) = 26, giving 4w + 6 = 26, so w = 5 cm.
例二：一个长方形的周长是 26 厘米，它的长比宽多 3 厘米。设宽为 w，则长为 w + 3。周长方程为 2(w + w + 3) = 26，可得 4w + 6 = 26，所以 w = 5 厘米。
11. Practice Questions and Exam Tips | 练习题与考试技巧
Try these questions to test your understanding. Always show all steps clearly and check your final answer by substitution.
请尝试以下题目来检验你的理解。始终清楚地写出所有步骤，并通过代入检验最终答案。
1. Solve 4x + 5 = 21.
1. 解方程 4x + 5 = 21。
2. Solve 7x − 3 = 2x + 17.
2. 解方程 7x −
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