📚 Solving Linear Equations | 解一元一次方程
In KS3 mathematics, solving linear equations is a core algebra skill. A linear equation is an equation where the unknown, usually written as x, is only raised to the power of 1. This unit helps you use inverse operations and the balance method to find the value of the unknown.
在 KS3 数学中，解一元一次方程是代数部分的核心技能。一元一次方程是指未知数（通常写作 x）的指数仅为 1 的方程。本单元帮助你使用逆运算和天平法求出未知数的值。
1. What Is a Linear Equation? | 什么是一元一次方程？
2x - 7 are linear. The equation 3x + 5 = 20 is linear because the highest power of x is 1.
一元一次方程包含一个未知数和一个等号。未知数的指数仅为 1，因此 3x + 5 和 2x - 7 这样的表达式都是一次的。方程 3x + 5 = 20 是一元一次方程，因为 x 的最高次数为 1。
are numbers and a is not zero. The solution is the value of x that makes the equation true.
一元一次方程通常可以写成 ax + b = c 的形式，其中 a、b 和 c 是数，且 a 不等于 0。解就是使方程成立的 x 的值。
ke x² + 3 = 12 are not linear because x is squared. Quadratic equations are studied later.
像 x² + 3 = 12 这样的方程不是一次方程，因为 x 被平方了。二次方程会在以后学习。
2. The Balance Method | 天平法
The balance method treats an equation like a balanced pair of scales. Whatever you do to one side, you must do to the other side, so the equation stays true.
天平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操作，这样方程才保持成立。
If you add 4 to one side, add 4 to the other. If you divide one side by 3, divide the other side by 3. This is the most important rule when solving equations.
如果你在一边加 4，另一边也要加 4。如果你把一边除以 3，另一边也要除以 3。这是解方程时最重要的规则。
x + 4 = 10 → x + 4 − 4 = 10 − 4 → x = 6
The table below shows common operations and their inverse operations.
下表列出了常见运算及其逆运算。
| Operation in equation | Inverse operation |
| + 5 | − 5 |
| − 7 | + 7 |
| × 3 | ÷ 3 |
| ÷ 4 | × 4 |
3. One-Step Equations | 一步方程
uation such as x + 7 = 15, subtract 7 from both sides. For x − 3 = 9, add 3 to both sides.
对于 x + 7 = 15 这样的一步方程，两边同时减去 7。对于 x − 3 = 9，两边同时加上 3。
x + 7 = 15 → x = 8
x − 3 = 9 → x = 12
rse operation. If 5x = 30, divide both sides by 5. If x ÷ 4 = 6, multiply both sides by 4.
对于乘法和除法，使用逆运算。如果 5x = 30，两边同时除以 5。如果 x ÷ 4 = 6，两边同时乘以 4。
5x = 30 → x = 6
x ÷ 4 = 6 → x = 24
Always write the final answer as x = a number, not just a number on its own.
最后一定要把答案写成 x = 某个数的形式，而不是只写一个数。
4. Two-Step Equations | 两步方程
ultiplied by 2 and then 3 is added. Undo the addition first, then undo the multiplication.
两步方程中 x 带有两种运算。例如在 2x + 3 = 11 中，x 先乘以 2，再加上 3。先逆运算加法，再逆运算乘法。
Subtract 3 from both sides: 2x = 8. Then divide both sides by 2: x = 4.
两边先减去 3：2x = 8。然后两边除以 2：x = 4。
2x + 3 = 11 → 2x = 8 → x = 4
For 3x − 5 = 16, undo the subtraction first by adding 5 to both sides, then divide by 3.
对于 3x − 5 = 16，先通过两边加 5 来逆运算减法，然后除以 3。
3x − 5 = 16 → 3x = 21 → x = 7
Remember: undo addition or subtraction before undo multiplication or division. This is the reverse of the order of operations.
记住：先逆运算加法或减法，再逆运算乘法或除法。这与运算顺序相反。
5. Expanding Brackets First | 先去括号
sing the distributive law. For 3(x + 2) = 21, multiply both terms inside the bracket by 3.
如果方程含有括号，先用分配律展开括号。对于 3(x + 2) = 21，把括号内的两项都乘以 3。
3(x + 2) = 3x + 6, so 3x + 6 = 21. Then subtract 6 and divide by 3 to get x = 5.
3(x + 2) = 3x + 6，所以 3x + 6 = 21。然后减去 6 并除以 3，得到 x = 5。
3(x + 2) = 21 → 3x + 6 = 21 → x = 5
For 2(3x − 1) = 16, expand to 6x − 2 = 16. Add 2 to both sides, then divide by 6: x = 3.
对于 2(3x − 1) = 16，展开得 6x − 2 = 16。两边加 2，然后除以 6：x = 3。
Always check that every term inside the bracket is multiplied by the number outside. A common error is to multiply only the first term.
一定要检查括号内的每一项都乘以外面的数。常见的错误是只乘第一项。
6. Unknowns on Both Sides | 未知数在等号两边
s on one side first. For 5x + 2 = 2x + 14, subtract 2x from both sides to get 3x + 2 = 14.
当 x 同时出现在等号两边时，先把含 x 的项移到同一边。对于 5x + 2 = 2x + 14，两边同时减去 2x，得到 3x + 2 = 14。
Then subtract 2 from both sides: 3x = 12, so x = 4.
然后两边同时减去 2：3x = 12，所以 x = 4。
5x + 2 = 2x + 14 → 3x = 12 → x = 4
x + 11, subtract 3x from both sides to get 4x − 5 = 11. Then add 5 and divide by 4: x = 4.
对于 7x − 5 = 3x + 11，两边同时减去 3x，得到 4x − 5 = 11。然后加 5 并除以 4：x = 4。
It helps to collect the smaller x term on the side of the larger x term so the coefficient remains positive.
可以把较小的 x 项移到较大的 x 项那一边，这样系数保持为正，有助于计算。
7. Equations with Fractions | 含分数的方程
denominator to clear it. For x/3 + 1 = 5, multiply all terms by 3: x + 3 = 15, so x = 12.
如果方程含有分数，把每一项都乘以分母以去掉分母。对于 x/3 + 1 = 5，把每一项都乘以 3：x + 3 = 15，所以 x = 12。
x/3 + 1 = 5 → x + 3 = 15 → x = 12
For x/2 = 3/4, multiply both sides by 2: x = 6/4 = 3/2.
对于 x/2 = 3/4，两边乘以 2：x = 6/4 = 3/2。
, such as x/2 + x/3 = 5, multiply every term by the lowest common denominator, which is 6.
当有两个分数时，例如 x/2 + x/3 = 5，将每一项乘以最小公分母，这里是 6。
Then 3x + 2x = 30, so 5x = 30 and x = 6.
于是 3x + 2x = 30，所以 5x = 30，x = 6。
x/2 + x/3 = 5 → 3x + 2x = 30 → x = 6
8. Checking Your Solution | 检验解
Always substitute your solution back into the original equation. If the left-hand side equals the right-hand side, your answer is correct.
一定要把你的解代回原方程检验。如果左边等于右边，答案就是正确的。
For 4x − 3 = 13, the solution is x = 4. Check: 4(4) − 3 = 16 − 3 = 13, which matches.
对于 4x − 3 = 13，解是 x = 4。检验：4(4) − 3 = 16 − 3 = 13，与右边一致。
Checking helps you catch sign errors and arithmetic mistakes. It is especially important in exams because it requires only substitution.
检验可以帮助你发现符号错误和计算错误。这在考试中尤其重要，因为只需要代入即可。
cause 2(5 + 3) = 2 × 8 = 16. If the two sides do not match, go back and look for an error.
对于 2(x + 3) = 16，解 x = 5 可以检验，因为 2(5 + 3) = 2 × 8 = 16。如果两边不相等，就返回去查找错误。
9. Common Misconceptions | 常见误区
Many errors in solving equations come from not applying inverse operations correctly. The list below shows mistakes to avoid.
解方程时的许多错误来自没有正确使用逆运算。下面的列表展示了需要避免的常见错误。
Forgetting to operate on both sides of the equation.
忘记在等号两边同时进行运算。
Undoing multiplication before addition or subtraction.
在逆运算加法或减法之前就逆运算了乘法。
Multiplying only one term inside a bracket.
括号内只乘了一项。
Changing the sign incorrectly when moving a term across the equals sign.
把某一项移到等号另一边时符号改变错误。
Forgetting to check the final answer by substitution.
最后忘记用代入法检验答案。
To avoid these, write every step clearly and use the balance method for each operation.
为了避免这些问题，请清晰地写出每一步，并对每一步运算都使用天平法。
10. Word Problems and Real-Life Contexts | 应用题与实际情境
into equations. For example: The sum of a number and 9 is 15. Write x + 9 = 15, so x = 6.
把文字题翻译成方程。例如：一个数加 9 等于 15。写出 x + 9 = 15，所以 x = 6。
s length 2x + 1 cm and width 5 cm. If its perimeter is 22 cm, write 2(2x + 1) + 2(5) = 22.
例 1：一个矩形的长为 2x + 1 cm，宽为 5 cm。如果周长为 22 cm，写出 2(2x + 1) + 2(5) = 22。
Solve: 4x + 2 + 10 = 22, so 4x + 12 = 22, then 4x = 10, x = 2.5.
解：4x + 2 + 10
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