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Solving Linear Equations Step by Step | 逐步解线性方程

2026-08-05 07:15 · 本文部分内容由AI辅助创作

📚 Solving Linear Equations Step by Step | 逐步解线性方程

Linear equations are fundamental in mathematics, forming the backbone of algebra and problem-solving. Mastering how to solve them builds confidence for more advanced topics. In this guide, we will thoroughly explore linear equations using the balancing method, covering one-step, two-step, equations with brackets, variables on both sides, fractions, decimals, special cases, rearranging formulas, and real-world applications. Follow along and practise to sharpen your skills.

线性方程是数学的基础,构成了代数和问题解决的支柱。掌握如何解它们会为更高级的主题建立信心。在本指南中,我们将使用平衡法全面探索线性方程,涵盖一步方程、两步方程、带括号的方程、两边都有变量的方程、分数与小数方程、特殊情形、公式变形以及实际应用。跟随并练习以增强你的技能。


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

written as ax + b = c, where a, b and c are numbers (constants) and a ≠ 0. For example, 4x − 7 = 5 is a linear equation. Solving it means finding the value of x that makes the equality true.

线性方程是变量(通常是 x)的指数为 1 的方程。没有像 x² 或 x³ 这样的项。标准形式通常写作 ax + b = c,其中 a、b 和 c 是数字(常数)且 a ≠ 0。例如,4x − 7 = 5 是一个线性方程。解它意味着找到使等式成立的 x 值。

ax + b = c, a ≠ 0

3 − 7 = 12 − 7 = 5, which is correct. This simple check is an important habit in algebra.

线性方程的图像是一条直线。当我们解 4x − 7 = 5 时,两边加 7 得到 4x = 12,再除以 4 得到 x = 3。检验:4 × 3 − 7 = 12 − 7 = 5,正确。这个简单的检验是代数中的重要习惯。


2. The Balancing Method | 平衡法

The balancing method treats an equation like a balanced scale. Whatever you do to one side, you must do to the other side. You can add, subtract, multiply or divide both sides by the same non-zero number without changing the equality. The goal is to isolate the variable on one side.

平衡法把方程看作一架平衡的天平。你对一边做什么,就必须对另一边做同样的事。你可以对两边同时加、减、乘或除以同一个非零数,而不改变等式的成立。目标是让变量单独留在一边。

x + 5 = 12 → x + 5 − 5 = 12 − 5 → x = 7

We use inverse operations. Addition undoes subtraction, and multiplication undoes division. Keep your work organised, write one operation per line, and align the equals signs. This makes errors easier to spot and corrections easier to make.

我们使用逆运算。加法抵消减法,乘法抵消除法。保持书写整齐,每行写一个运算,并让等号对齐。这样更容易发现错误,也更容易改正。


3. One-Step Equations | 一步方程

One-step equations require only one inverse operation to solve. They come in four common types: addition, subtraction, multiplication and division. For addition, subtract from both sides. For subtraction, add to both sides. For multiplication, divide both sides. For division, multiply both sides.

一步方程只需要一个逆运算即可解出。它们有四种常见类型:加法、减法、乘法和除法。对于加法,两边同时减;对于减法,两边同时加;对于乘法,两边同时除;对于除法,两边同时乘。

x + 7 = 10 → x = 3

x − 4 = 9 → x = 13

6x = 42 → x = 7

x ÷ 5 = 8 → x = 40

titute x = 7: 6 × 7 = 42, correct. This takes only a few seconds and confirms your answer.

始终通过代入来检验。对于 x + 7 = 10,代入 x = 3:3 + 7 = 10,正确。对于 6x = 42,代入 x = 7:6 × 7 = 42,正确。这只需几秒钟,却能确认你的答案。


4. Two-Step Equations | 两步方程

example, solve 3x + 5 = 20. Subtract 5 from both sides: 3x = 15. Then divide by 3: x = 5.

两步方程需要两个逆运算。通常我们先抵消加法或减法,然后抵消乘法或除法。例如,解 3x + 5 = 20。两边减 5:3x = 15。然后除以 3:x = 5。

3x + 5 = 20 → 3x = 15 → x = 5

students prefer to reverse the order of operations, but the balancing method always works.

另一个例子是 2x − 7 = 11。两边加 7:2x = 18。除以 2:x = 9。检验:2 × 9 − 7 = 18 − 7 = 11,正确。有些学生喜欢逆用运算顺序,但平衡法总是有效。


5. Equations with Brackets | 带括号的方程

n. For example, 3(x + 2) = 15 becomes 3x + 6 = 15. Subtract 6: 3x = 9. Divide by 3: x = 3.

当方程含有括号时,先用分配律展开:a(b + c) = ab + ac。然后解得到的线性方程。例如,3(x + 2) = 15 变为 3x + 6 = 15。减 6:3x = 9。除以 3:x = 3。

3(x + 2) = 15 → 3x + 6 = 15 → 3x = 9 → x = 3

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and: 2x − 6 = 4x + 8. Subtract 2x: −6 = 2x + 8. Subtract 8: −14 = 2x. Divide by 2: x = −7.

你也可以先两边除以 3:x + 2 = 5,所以 x = 3。两种方法都有效。对于 2(x − 3) = 4x + 8,展开:2x − 6 = 4x + 8。减 2x:−6 = 2x + 8。减 8:−14 = 2x。除以 2:x = −7。

2(x − 3) = 4x + 8 → 2x − 6 = 4x + 8 → −14 = 2x → x = −7


6. Variables on Both Sides | 两边都有变量

4 = 2x + 11. Subtract 2x from both sides: 3x − 4 = 11. Add 4: 3x = 15. Divide by 3: x = 5.

当变量出现在两边时,把变量项集中到一边,常数项集中到另一边。例如,解 5x − 4 = 2x + 11。两边减 2x:3x − 4 = 11。加 4:3x = 15。除以 3:x = 5。

5x − 4 = 2x + 11 → 3x − 4 = 11 → 3x = 15 → x = 5

Divide by 7: x = 3. Check: 7 − 3 × 3 = 7 − 9 = −2, and 4 × 3 − 14 = 12 − 14 = −2, correct.

如果变量项是负的,就把它加到两边。例如,7 − 3x = 4x − 14。加 3x:7 = 7x − 14。加 14:21 = 7x。除以 7:x = 3。检验:7 − 3 × 3 = 7 − 9 = −2,且 4 × 3 − 14 = 12 − 14 = −2,正确。


7. Equations with Fractions and Decimals | 分数与小数方程

/3 + 1/2 = 2. Multiply by 6: 2x + 3 = 12. Subtract 3: 2x = 9. Divide by 2: x = 9/2 or 4.5.

对于含分数的方程,两边乘以分母的最小公倍数以消去分数。例如,x/3 + 1/2 = 2。乘以 6:2x + 3 = 12。减 3:2x = 9。除以 2:x = 9/2 或 4.5。

x/3 + 1/2 = 2 → 2x + 3 = 12 → 2x = 9 → x = 9/2

0: 5x + 12 = 37. Subtract 12: 5x = 25. Divide by 5: x = 5. Check: 0.5 × 5 + 1.2 = 2.5 + 1.2 = 3.7, correct.

对于含小数的方程,两边乘以 10 的幂以消去小数,或直接解。例如,0.5x + 1.2 = 3.7。乘以 10:5x + 12 = 37。减 12:5x = 25。除以 5:x = 5。检验:0.5 × 5 + 1.2 = 2.5 + 1.2 = 3.7,正确。


8. Special Cases: No Solution or Infinite Solutions | 特殊情形:无解或无穷多解

For example, 2x + 3 = 2x + 7. Subtract 2x: 3 = 7, which is false, so there is no solution.

有时变量会完全消去。如果得到的结果是假的,方程无解。例如,2x + 3 = 2x + 7。减 2x:3 = 7,这是假的,所以无解。

2x + 3 = 2x + 7 → 3 = 7 (false) → no solution

2x + 2 = 2x + 2. Subtract 2x: 2 = 2, which is true for all x. The two sides are identical.

如果得到的结果总是真的,方程有无穷多解。例如,2(x + 1) = 2x + 2。展开:2x + 2 = 2x + 2。减 2x:2 = 2,这对所有 x 都成立。两边完全相同。

2(x + 1) = 2x + 2 → 2 = 2 (true) → infinitely many solutions

On a graph, no solution means the lines are parallel. Infinitely many solutions means both equations represent the same line. Understanding these cases helps you avoid confusion in exams.

在图像上,无解意味着两条直线平行。无穷多解意味着两个方程表示同一条直线。理解这些情形有助于你在考试中避免困惑。


9. Rearranging Formulas | 公式变形

sides: y − 5 = 3x. Then divide by 3: x = (y − 5)/3. This is still a linear equation in x.

同样的平衡法可以用于公式变形。要使 x 成为 y = 3x + 5 的主项,两边减 5:y − 5 = 3x。然后除以 3:x = (y − 5)/3。这仍然是一个关于 x 的线性方程。

y = 3x + 5 → y − 5 = 3x → x = (y − 5)/3

. For speed v = s/t, make t the subject: multiply by t: vt = s, then divide by v: t = s/v.

实际公式也一样。矩形的周长是 P = 2(l + w)。要使 w 成为主项,除以 2:P/2 = l + w。减 l:w = P/2 − l。对于速度 v = s/t,要使 t 成为主项:乘以 t:vt = s,再除以 v:t = s/v。


10. Word Problems and Real-World Applications | 应用题与实际应用

ntext. For example, the sum of three consecutive integers is 48. Let the integers be n, n + 1 and n + 2. Then n + (n + 1) + (n + 2) = 48.

应用题遵循清晰的步骤:定义变量,把文字翻译成方程,解方程,并在实际情境中解释答案。

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