📚 KS3 Algebra Essentials: Simplifying Expressions and Solving Linear Equations | KS3 代数核心:化简表达式与解一元一次方程
Algebra is one of the most important building blocks in KS3 Mathematics. In the Cambridge Lower Secondary programme, students are expected to move from simple number patterns to working confidently with letters, symbols, equations and formulae. This article focuses on simplifying algebraic expressions and solving linear equations. These skills appear throughout the KS3 curriculum and are essential for later topics such as graphs, sequences, and real-life problem solving.
代数是 KS3 数学中最重要的基础模块之一。在剑桥初中课程中,学生需要从简单的数字规律过渡到熟练运用字母、符号、方程和公式。本文重点讲解代数表达式的化简以及一元一次方程的解法。这些技能贯穿 KS3 课程,也是后续学习图像、数列和实际问题解决的重要基础。
1. What is an algebraic expression? | 什么是代数表达式?
An algebraic expression is a combination of numbers, variables, and operation symbols. It does not contain an equals sign. For example, 3x + 5, 2a – 7b, and 4y² + 1 are all algebraic expressions. A variable is a letter that stands for an unknown value, and a coefficient is the number in front of a variable. In the term 3x, the coefficient is 3 and the variable is x.
代数表达式是由数字、变量和运算符号组成的式子,它不包含等号。例如 3x + 5、2a – 7b 和 4y² + 1 都是代数表达式。变量是表示未知值的字母,系数是变量前面的数字。在项 3x 中,系数是 3,变量是 x。
Expressions can have one term, two terms, or many terms. A term is a single number, a single variable, or a product of numbers and variables. For example, 5, x, and 2xy are each one term. When terms are separated by plus or minus signs, we can see the structure of an expression clearly.
表达式可以有一项、两项或多项。项是单个数字、单个变量,或者数字与变量的乘积。例如 5、x 和 2xy 各是一项。当项由加号或减号分隔时,我们可以清楚地看到表达式的结构。
2. Collecting like terms | 合并同类项
Like terms are terms that have exactly the same variable or variables raised to the same power. For example, 3x and 5x are like terms because they both contain x to the power of 1. However, 3x and 3x² are not like terms because the powers of x are different. Numbers without variables, such as 4 and -7, are also like terms with each other.
同类项是指含有完全相同变量且变量次数相同的项。例如 3x 和 5x 是同类项,因为它们都含有一次方的 x。但是 3x 和 3x² 不是同类项,因为 x 的次数不同。没有变量的数字,如 4 和 -7,彼此也是同类项。
To collect like terms, add or subtract the coefficients while keeping the variable part unchanged. For example, 3x + 2x = 5x and 8y – 3y = 5y. The expression 4a + 3b + 2a – b can be simplified to 6a + 2b because 4a + 2a = 6a and 3b – b = 2b.
合并同类项时,将系数相加或相减,同时保持变量部分不变。例如 3x + 2x = 5x,8y – 3y = 5y。表达式 4a + 3b + 2a – b 可以化简为 6a + 2b,因为 4a + 2a = 6a,3b – b = 2b。
5x + 4y – 2x + 3y = 3x + 7y
Always remember that you cannot combine unlike terms. For instance, 2x + 3y cannot be simplified further because x and y are different variables. Writing 2x + 3y = 5xy is a serious error and must be avoided.
始终记住不能合并不同类项。例如 2x + 3y 无法进一步化简,因为 x 和 y 是不同的变量。把 2x + 3y 写成 5xy 是错误的,必须避免。
3. Expanding brackets | 去括号展开
Expanding brackets means removing the brackets by multiplying each term inside the bracket by the term outside. This uses the distributive law: a(b + c) = ab + ac. For example, 3(x + 4) expands to 3x + 12 because 3 × x = 3x and 3 × 4 = 12.
去括号展开是指用括号外的项乘以括号内的每一项,从而去掉括号。这运用了分配律:a(b + c) = ab + ac。例如 3(x + 4) 展开后得到 3x + 12,因为 3 × x = 3x,3 × 4 = 12。
If there is a negative sign outside the bracket, it must be multiplied by every term inside. For example, -2(x – 5) = -2x + 10 because -2 × x = -2x and -2 × (-5) = +10. A common mistake is to write -2x – 10, but the second sign should be positive because a negative times a negative gives a positive.
如果括号外有负号,则必须乘以括号内的每一项。例如 -2(x – 5) = -2x + 10,因为 -2 × x = -2x,-2 × (-5) = +10。常见错误是写成 -2x – 10,但第二项的符号应为正,因为负数乘以负数得正数。
4(2x – 3) = 8x – 12
When two brackets are multiplied together, such as (x + 2)(x + 3), expand one bracket at a time. First multiply x by both terms in the second bracket, then multiply 2 by both terms: x × x = x², x × 3 = 3x, 2 × x = 2x, and 2 × 3 = 6. Combining like terms gives x² + 5x + 6.
当两个括号相乘时,例如 (x + 2)(x + 3),每次展开一个括号。先用 x 乘以第二个括号中的两项,再用 2 乘以这两项:x × x = x²,x × 3 = 3x,2 × x = 2x,2 × 3 = 6。合并同类项得到 x² + 5x + 6。
4. Factorising simple expressions | 因式分解简单表达式
Factorising is the reverse of expanding brackets. It means writing an expression as a product of factors. To factorise an expression, look for the highest common factor of all the terms. For example, in the expression 6x + 9, the highest common factor of 6 and 9 is 3. Therefore, 6x + 9 = 3(2x + 3).
因式分解是去括号展开的逆运算。它意味着把一个表达式写成因式的乘积。要对表达式进行因式分解,需要找出所有项的最高公因式。例如在表达式 6x + 9 中,6 和 9 的最高公因数是 3。因此 6x + 9 = 3(2x + 3)。
For an expression like 8xy + 4x, the common factor is 4x because both terms contain x and the largest number that divides 8 and 4 is 4. So 8xy + 4x = 4x(2y + 1). Always check your answer by expanding the brackets again.
对于 8xy + 4x 这样的表达式,公因式是 4x,因为两项都含有 x,且能整除 8 和 4 的最大数是 4。所以 8xy + 4x = 4x(2y + 1)。始终通过再次展开括号来检查答案。
3x² + 6x = 3x(x + 2)
Factorising is useful in solving equations, simplifying fractions, and understanding the structure of expressions. In KS3, you are usually expected to factorise by taking out a single common factor. More complex quadratic factorising is covered in later stages.
因式分解在解方程、化简分式以及理解表达式结构时非常有用。在 KS3 阶段,通常只要求提取单个公因式。更复杂的二次因式分解会在后续阶段学习。
5. What is a linear equation? | 什么是一元一次方程?
An equation is a mathematical statement that two expressions are equal. It contains an equals sign. A linear equation is an equation where the highest power of the variable is 1. For example, x + 5 = 12, 3x – 2 = 10, and 4(x – 1) = 8 are linear equations. The word ‘linear’ comes from the fact that the graph of such an equation is a straight line.
方程是表示两个表达式相等的数学语句,它包含等号。一元一次方程是指未知数的最高次数为 1 的方程。例如 x + 5 = 12、3x – 2 = 10 和 4(x – 1) = 8 都是一元一次方程。“一次”一词源于这类方程的图像是一条直线。
Solving an equation means finding the value of the variable that makes the equation true. Think of an equation as a balance scale: the left side and the right side must remain equal. Whatever operation you do to one side, you must do the same to the other side. This is called the balance method.
解方程意味着求出使方程成立的未知数的值。可以把方程看作一架天平:左边和右边必须保持相等。无论对一边进行什么运算,都必须对另一边进行相同的运算。这叫做平衡法。
x + 5 = 12 ⇒ x = 7
6. Solving equations with one operation | 解一步方程
Some equations can be solved in one step. If the equation is x + 3 = 10, subtract 3 from both sides to get x = 7. If the equation is x – 4 = 9, add 4 to both sides to get x = 13. The goal is to isolate the variable on one side of the equation.
有些方程只需一步就能解出。如果方程是 x + 3 = 10,两边同时减去 3,得到 x = 7。如果方程是 x – 4 = 9,两边同时加上 4,得到 x = 13。目标是使未知数单独出现在方程的一边。
For multiplication, if 5x = 30, divide both sides by 5 to get x = 6. For division, if x / 4 = 7, multiply both sides by 4 to get x = 28. Always write each step clearly and keep the equals signs aligned.
对于乘法,如果 5x = 30,两边同时除以 5,得到 x = 6。对于除法,如果 x / 4 = 7,两边同时乘以 4,得到 x = 28。始终清晰地写出每一步,并保持等号对齐。
5x = 30 ⇒ x = 6
A quick check can be done by substituting your answer back into the original equation. For example, 5 × 6 = 30, which is correct. This habit helps catch careless mistakes in tests and homework.
可以通过将答案代回原方程来快速检验。例如 5 × 6 = 30,这是正确的。这个习惯有助于在考试和作业中发现粗心错误。
7. Solving equations with two or more operations | 解两步及多步方程
Many linear equations require more than one operation to solve. For example, to solve 2x + 5 = 13, first subtract 5 from both sides: 2x = 8. Then divide both sides by 2: x = 4. Always reverse the order of operations: deal with addition or subtraction before multiplication or division.
许多一元一次方程需要不止一步运算才能解出。例如解 2x + 5 = 13,首先两边同时减去 5:2x = 8。然后两边同时除以 2:x = 4。总是按照与运算顺序相反的方向操作:先处理加减,再处理乘除。
If the equation is x/3 – 2 = 7, first add 2 to both sides to get x/3 = 9. Then multiply both sides by 3 to get x = 27. This systematic approach makes even longer equations manageable.
如果方程是 x/3 – 2 = 7,首先两边同时加上 2,得到 x/3 = 9。然后两边同时乘以 3,得到 x = 27。这种系统化的方法能使更长的方程也变得容易处理。
2x + 5 = 13 ⇒ 2x = 8 ⇒ x = 4
When writing your solution, show each step on a new line. This makes it easier for you and your teacher to follow your reasoning. In Cambridge KS3 tests, method marks are often awarded even if the final answer is wrong, as long as the steps are logical.
写解题过程时,每一步另起一行。这能让你和老师更容易理解你的推理。在剑桥 KS3 测试中,即使最终答案错误,只要步骤合乎逻辑,通常也能得到过程分。
8. Equations with unknowns on both sides | 未知数在方程两边
Some equations have the variable on both sides of the equals sign, for example 5x – 3 = 2x + 9. The first step is to collect all variable terms on one side and all constant terms on the other. Subtract 2x from both sides: 3x – 3 = 9. Then add 3 to both sides: 3x = 12. Finally divide both sides by 3: x = 4.
有些方程的等号两边都含有未知数,例如 5x – 3 = 2x + 9。第一步是让所有含未知数的项在一边,所有常数项在另一边。两边同时减去 2x:3x – 3 = 9。然后两边同时加上 3:3x = 12。最后两边同时除以 3:x = 4。
If the variable term has a negative coefficient, such as 7 – x = 2x + 1, add x to both sides to make the coefficient positive: 7 = 3x + 1. Then subtract 1 from both sides: 6 = 3x. Divide both sides by 3: x = 2.
如果变量项的系数为负,例如 7 – x = 2x + 1,两边同时加上 x,使系数变为正数:7 = 3x + 1。然后两边同时减去 1:6 = 3x。两边同时除以 3:x = 2。
5x – 3 = 2x + 9 ⇒ 3x = 12 ⇒ x = 4
Always check by substituting the value back into both sides. For x = 4, the left side is 5 × 4 – 3 = 17, and the right side is 2 × 4 + 9 = 17. Both sides are equal, so the solution is correct.
始终通过将数值代回两边来检验。当 x = 4 时,左边为 5 × 4 – 3 = 17,右边为 2 × 4 + 9 = 17。两边相等,所以解是正确的。
9. Equations involving brackets | 含括号的方程
When an equation includes brackets, the first step is usually to expand the brackets using the distributive law. For example, to solve 3(x + 2) = 18, expand to get 3x + 6 = 18. Then subtract 6 from both sides: 3x = 12. Finally divide both sides by 3: x = 4.
当方程中含有括号时,第一步通常是运用分配律展开括号。例如解 3(x + 2) = 18,展开得到 3x + 6 = 18。然后两边同时减去 6:3x = 12。最后两边同时除以 3:x = 4。
If the equation is 2(3x – 1) = 4x + 8, expand the left side first: 6x – 2 = 4x + 8. Then subtract 4x from both sides: 2x – 2 = 8. Add 2 to both sides: 2x = 10. Divide both sides by 2: x = 5. Check: left side is 2(3 × 5 – 1) = 28, right side is 4 × 5 + 8 = 28.
如果方程是 2(3x – 1) = 4x + 8,先展开左边:6x – 2 = 4x + 8。然后两边同时减去 4x:2x – 2 = 8。两边同时加上 2:2x = 10。两边同时除以 2:x = 5。检验:左边为 2(3 × 5 – 1) = 28,右边为 4 × 5 + 8 = 28。
3(x + 2) = 18 ⇒ 3x + 6 = 18 ⇒ x = 4
10. Using equations in real-life problems | 用方程解决实际问题
Linear equations often appear in word problems. To solve a word problem, follow these steps: read the question carefully, define a variable for the unknown, write an equation, solve the equation, and check that the answer makes sense in the context.
一元一次方程经常出现在文字题中。解文字题的步骤如下:仔细读题,为未知数定义一个变量,列出方程,解方程,并检查答案在情境中是否合理。
For example, a rectangle has a length that is 5 cm more than its width. The perimeter is 34 cm. Let the width be w cm. Then the length is (w + 5) cm. The perimeter equation is 2w + 2(w + 5) = 34. Expand: 2w + 2w + 10 = 34, so 4w + 10 = 34. Subtract 10: 4w = 24. Divide by 4: w = 6. Thus the width is 6 cm and the length is 11 cm.
例如一个矩形的长比宽多 5 厘米。周长为 34 厘米。设宽为 w 厘米,则长为 (w + 5) 厘米。周长方程为 2w + 2(w + 5) = 34。展开:2w + 2w + 10 = 34,所以 4w + 10 = 34。减去 10:4w = 24。除以 4:w = 6。因此宽为 6 厘米,长为 11 厘米。
Another common type is age or money problems. For example, if Ali has three times as much money as Ben, and together they have £48, then let Ben have £x. Ali has £3x. The equation is x + 3x = 48, so 4x = 48 and x = 12. Ben has £12 and Ali has £36.
另一种常见类型是年龄或金钱问题。例如 Ali 的钱是 Ben 的三倍,他们一共有 48 英镑,设 Ben 有 £x,Ali 有 £3x。方程为 x + 3x = 48,所以 4x = 48,x = 12。Ben 有 12 英镑,Ali 有
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