5.1

Simple Linear Equations in One Variable

We have come across number sentences such as △ + 3 = 8 previously. If we replace △ with the letter $x$ to represent the variable, we write $x + 3 = 8$.

$x + 3 = 8$ is called an equation, where the equal sign '=' indicates that the left-hand side (LHS) and the right-hand side (RHS) are equal. To solve the equation, we need to find the value of the variable $x$ such that LHS = RHS.

For what value of $x$ will LHS = RHS in the equation $x + 3 = 8$? Let us look at the following:

When $x = 1$,
LHS = $x + 3$ = $1 + 3$ = $4$
$\neq 8$
∴ LHS ≠ RHS

When $x = 5$,
LHS = $x + 3$ = $5 + 3$ = $8$
∴ LHS = RHS ✓

When we substitute $x = 5$ into the equation $x + 3 = 8$, the LHS = RHS. We say $x = 5$ satisfies the equation and it is the solution or root of the equation. However, when we substitute $x = 1$ into the equation, the LHS ≠ RHS. Therefore, $x = 1$ is not a solution of the equation.

Note that $x + 3 = 8$ is a linear equation in one variable, $x$.

An equation of the form $ax + b = c$, where $a$, $b$ and $c$ are constants, $a \neq 0$ and the variable $x$ has a power of 1, is called a linear equation in one variable, $x$.
To find the solution or root of a linear equation in one variable $x$, we need to find the value of $x$ that will result in both sides of the equation having the same value, i.e., LHS = RHS.

What happens if $a = 0$ in the equation $ax + b = c$?

💡 思考

如果 $a = 0$，方程变成 $0 \cdot x + b = c$ 即 $b = c$，这就不再是关于 $x$ 的方程了——它要么恒真（$b = c$，任何 $x$ 都满足，无穷多解）要么恒假（$b \neq c$，无解）。这就是为什么定义要求 $a \neq 0$。

⚖️The Balance Method

The equal sign '=' in an equation can be viewed as a balance between the LHS and the RHS. We can use algebra discs and the idea of a balance to help us make sense of the process of solving an equation.

Let us look at how we can solve the equation $2x + 7 = -3$.

Step	Balance (visual)	Equation
1	Place corresponding discs on the LHS and RHS of a balance.	$2x + 7 = -3$
2	Simplify by adding −7 on both sides. RECALL: the −7 disc on LHS forms a zero pair with the +7 disc.	$2x + 7 \boxed{- 7} = -3 \boxed{- 7}$
3	Group the numbers on the right-hand side.	$2x = -10$
4	Divide both sides by 2.	$\dfrac{2x}{2} = \dfrac{-10}{2}$ $x = -5$

The solution is $\mathbf{x = -5}$.

We observe that, at each step in the simplification, the equation produced is equivalent to the previous one. That is, $2x + 7 = -3$, $2x = -10$ and $x = -5$ are equivalent.

Can you form three equivalent equations that all have the solution $x = 2$? Show your working clearly.

💡 思路

从 $x = 2$ 出发，两边做同样的操作：
• 加 5：$x + 5 = 7$
• 乘 3：$3x = 6$
• 先乘 4 再减 1：$4x - 1 = 7$
这些都是 $x = 2$ 的等价方程。

🧪 Activity 1 · Solve linear equations

Objective: To solve linear equations in one variable.

Use algebra discs (or the balance idea) to help you solve the following equations.

(a) $x - 2 = 5$ → $x = $ 检查
(b) $3x + 4 = 10$ → $x = $ 检查
(c) $2x - 5 = -9$ → $x = $ 检查
(d) $2x = x + 7$ → $x = $ 检查
(e) $3x - 1 = -x + 7$ → $x = $ 检查
(f) $x - 7 = -4x + 8$ → $x = $ 检查

To summarise, we can always 'maintain the balance' when we (a) add, (b) subtract, (c) multiply, (d) divide by the same number on both sides of the equation. Hence the expressions on both sides of the equation are equivalent.

Given $ac = bc$, can we conclude $a = b$? Why?

💡 思考

不一定！ 如果 $c \neq 0$，那 $ac = bc \Rightarrow a = b$ ✓。
但如果 $c = 0$，$0 = 0$ 对任何 $a, b$ 都成立——所以推不出 $a = b$。
这就是为什么"除以 $c$"时，必须保证 $c \neq 0$。

📖 Worked Example 1 · Add to both sides

Solve the equation $x - 5 = 11$.

SOLUTION

$x - 5 = 11$
$x - 5 \boxed{+ 5} = 11 \boxed{+ 5}$ (Add 5 to both sides.)
∴ $x = \mathbf{16}$

SPOTLIGHT. If $a = b$, then $a + c = b + c$.

✍️ Try It Yourself 1

(a) $x - 2 = 7$ → $x = $ 检查
(b) $x - 12 = -7$ → $x = $ 检查
(c) $-8 + x = 24$ → $x = $ 检查

📖 Worked Example 2 · Subtract from both sides

Solve the equation $x + 6 = 13$.

SOLUTION

$x + 6 = 13$
$x + 6 \boxed{- 6} = 13 \boxed{- 6}$ (Subtract 6 from both sides.)
∴ $x = \mathbf{7}$

SPOTLIGHT. If $a = b$, then $a - c = b - c$.

✍️ Try It Yourself 2

(a) $x + 11 = 5$ → $x = $ 检查
(b) $x + 9 = -13$ → $x = $ 检查
(c) $25 + x = 5$ → $x = $ 检查

📖 Worked Example 3 · Multiply both sides

Solve the equation $\dfrac{x}{4} = 7$.

SOLUTION

$\dfrac{x}{4} = 7$
$\dfrac{x}{4} \boxed{\times 4} = 7 \boxed{\times 4}$ (Multiply both sides by 4.)
∴ $x = \mathbf{28}$

SPOTLIGHT. If $a = b$, then $ac = bc$.

✍️ Try It Yourself 3

(a) $\dfrac{x}{3} = -5$ → $x = $ 检查
(b) $-\dfrac{x}{4} = 7$ → $x = $ 检查
(c) $\dfrac{1}{5}x = 2.4$ → $x = $ 检查

📖 Worked Example 4 · Divide both sides

Solve the equation $-6x = 8$.

SOLUTION

$-6x = 8$
$\dfrac{-6x}{\boxed{-6}} = \dfrac{8}{\boxed{-6}}$ (Divide both sides by −6.)
$x = -\dfrac{4}{3}$
∴ $x = \mathbf{-1\dfrac{1}{3}}$

SPOTLIGHT.
If $a = b$, then $\dfrac{a}{c} = \dfrac{b}{c}$, provided $c \neq 0$.
$-x$ is not necessarily a negative number (it's negative iff $x > 0$).

✍️ Try It Yourself 4

(a) $-10x = 35$ → $x = $ 检查
(b) $4x = -1.6$ → $x = $ 检查
(c) $-9x = -27$ → $x = $ 检查

📖 Worked Example 5 · Variables on both sides

Solve the equation $4x + 7 = 9x - 6$.

SOLUTION

$4x + 7 = 9x - 6$
$4x + 7 - 9x = 9x - 6 - 9x$ (Subtract $9x$ from both sides.)
$-5x + 7 = -6$
$-5x + 7 - 7 = -6 - 7$ (Subtract 7 from both sides.)
$-5x = -13$
$\dfrac{-5x}{-5} = \dfrac{-13}{-5}$ (Divide both sides by −5.)
$x = \dfrac{13}{5}$
∴ $x = \mathbf{2\dfrac{3}{5}}$

✍️ Try It Yourself 5

(a) $10x + 21 = -25 - 13x$ → $x = $ 检查
💡 思路
$10x + 13x = -25 - 21$；$23x = -46$；$x = -2$。
(b) $5 - 7x = 3x + 29$ → $x = $ 检查
💡 思路
$-7x - 3x = 29 - 5$；$-10x = 24$；$x = -12/5$。

📖 Worked Example 6 · Formulate + Solve

(a) The sum of three times $y$ and −5 is 9. Write an equation in $y$ and solve it.
(b) Given that $(2x + 1)$ has the same value as $(-x + 3)$, write an equation in $x$ and solve it.

SOLUTION

(a) Three times a number $y$ is $3y$. The sum of $3y$ and $−5$ is $3y + (−5)$.
∴ $3y + (-5) = 9$
$3y - 5 = 9$
$3y - 5 + 5 = 9 + 5$ (Add 5 to both sides.)
$3y = 14$
$\dfrac{3y}{3} = \dfrac{14}{3}$ (Divide both sides by 3.)
∴ $y = \mathbf{\dfrac{14}{3}} = \mathbf{4\dfrac{2}{3}}$

(b) $2x + 1 = -x + 3$
$2x + 1 + x = -x + 3 + x$ (Add $x$ to both sides.)
$3x + 1 = 3$
$3x + 1 - 1 = 3 - 1$ (Subtract 1 from both sides.)
$3x = 2$
$\dfrac{3x}{3} = \dfrac{2}{3}$ (Divide both sides by 3.)
∴ $x = \mathbf{\dfrac{2}{3}}$

In Worked Example 6(a), after obtaining $3y - 5 = 9$, we can also divide both sides of the equation by 3 to simplify the $y$ term in the equation as shown below.
$3y - 5 = 9$ → $\dfrac{3y - 5}{3} = \dfrac{9}{3}$ → $y - \dfrac{5}{3} = 3$ → $y = 3 + \dfrac{5}{3} = \dfrac{14}{3}$ → ∴ $y = 4\dfrac{2}{3}$.
Which way of simplification do you prefer? Why?

💡 思考

多数情况下先加再除更直观（保持系数为整数）；但如果是 $\dfrac{3y - 5}{3} = 9$ 这种"整体除"，先除可能更快。Both arrive at the same answer.

✍️ Try It Yourself 6

Write an equation for each of the following and solve it.

(a) The product of $\dfrac{m}{2}$ and 8 is $-20$. → $m = $ 检查
💡 思路
$\dfrac{m}{2} \times 8 = -20$；$4m = -20$；$m = -5$。
(b) It is given that $c = 4x - 5$, $d = -x - 2$ and $c = d$. → $x = $ 检查
💡 思路
$4x - 5 = -x - 2$；$5x = 3$；$x = 3/5$。

PPractice Exercise 5.1

BASIC MASTERY

Q1. Solve the following equations.
(a) $x + 8 = 17$ → $x = $ 检查
(b) $x + 36 = -40$ → $x = $ 检查
(c) $x - 7 = 25$ → $x = $ 检查
(d) $x - 22 = -15$ → $x = $ 检查
Q2. Solve the following equations.
(a) $\dfrac{x}{10} = 3.6$ → $x = $ 检查
(b) $\dfrac{1}{5}x = -40$ → $x = $ 检查
(c) $4x = 5.6$ → $x = $ 检查
(d) $-9x = 21$ → $x = $ 检查
(e) $-\dfrac{1}{8}x = 6$ → $x = $ 检查
(f) $-24x = -84$ → $x = $ 检查
Q3. Solve the following equations.
(a) $2x + 16 = 7$ → $x = $ 检查
(b) $3x + 8 = -1$ → $x = $ 检查
(c) $-5x + 4.5 = -3$ → $x = $ 检查
(d) $\dfrac{x}{3} - 12 = 0$ → $x = $ 检查
(e) $\dfrac{1}{4}x + 19 = 6$ → $x = $ 检查
(f) $1 - \dfrac{1}{7}x = -8$ → $x = $ 检查

INTERMEDIATE

Q4. Solve the following equations.
(a) $3x = x + 6 + 10x$ → $x = $ 检查
(b) $2x - 5 = -14 - x$ → $x = $ 检查
(c) $9 - 5y = y - 1$ → $y = $ 检查
(d) $-3y + 4 = -5y + 0.8$ → $y = $ 检查
(e) $3a + 6 = -4a - 8$ → $a = $ 检查
(f) $-1 - 7b = -3 + 2b$ → $b = $ 检查
(g) $2w + 7w - 3 = 5w + 4$ → $w = $ 检查
(h) $19 - 3z - 6z = 7 - 14z$ → $z = $ 检查
(i) $\dfrac{1}{3}m - 2 = -1\dfrac{1}{4} - \dfrac{1}{6}m$ → $m = $ 检查
(j) $5 + \dfrac{p}{2} = \dfrac{p}{4} - 5$ → $p = $ 检查
Q5. Write an equation for each of the following and solve it.
(a) The sum of one-sixth of $x$ and −9 is −25. → $x = $ 检查
(b) Twice a number $y$ decreased by 5 is −7. → $y = $ 检查
(c) The product of $6x$ and −12 is 72. → $x = $ 检查
(d) The quotient of $m$ divided by −2 is −11. → $m = $ 检查
(e) It is given that $a = \dfrac{1}{2}x$, $b = -3x + 18$ and $2a = b$. → $x = $ 检查
Q6. The cost $C of screen printing $x$ T-shirts is given by the formula $C = 100 + 12x$.
(a) Find the cost of printing 50 T-shirts. → $ 检查
(b) If the budget is $976, how many T-shirts can be printed? → 检查
Q7. In the diagram, a spring is hung from a ceiling with a mass $m$ kg attached at one end. The length, $L$ cm, of the spring is given by the formula $L = 70 + 5m$. Find
(a) the length of the spring when $m = 6$ → $L = $ cm 检查
(b) the value of $m$ when $L = 85$ → $m = $ 检查

ADVANCED

Q8. Given that $x = 6$ is the solution of the equation $3x - 2a = 8$, find the value of $a$. → $a = $ 检查
💡 思路
代入 $x = 6$：$18 - 2a = 8 \Rightarrow 2a = 10 \Rightarrow a = 5$。
Q9. Leo solved the equation $6x - 12 = 5x - 12$ as follows:
$6x - 12 = 5x - 12$
$6x - 12 + 12 = 5x - 12 + 12$
$6x = 5x$
$\dfrac{6x}{x} = \dfrac{5x}{x}$
$6 = 5$
Identify the mistake in Leo's solution and help him to solve the equation.
Correct answer: $x = $ 检查
💡 错误诊断

Leo 把两边除以了 $x$——但他不知道 $x$ 是否等于 0！如果 $x = 0$ 就不能除以 $x$。这一步隐含假设 $x \neq 0$，但实际上恰好就是 $x = 0$。
正确做法：$6x = 5x \Rightarrow 6x - 5x = 0 \Rightarrow x = 0$。验证：左 $= 6(0) - 12 = -12$，右 $= 5(0) - 12 = -12$ ✓。
Q10. 🟧 and 🟧 (orange discs) represent a number each. From the bar diagrams (top: $2x = 4$; bottom: $x + y = 2y - 5$ with $x$ already known), find the values of $x$ and $y$.
$x = $ 检查 ; $y = $ 检查
💡 思路

(Top bar) $2x = 4 \Rightarrow x = 2$.
(Bottom bar) $x + y = 2y - 5$；代入 $x = 2$：$2 + y = 2y - 5 \Rightarrow y = 7$.
Q11. The distance, $d$ km, of a train from a station after $t$ hours is given by the formula $d = 240 - 80t$. Find the time $t$ when
(a) $d = 40$ → $t = $ hours 检查
(b) the train reaches the station ($d = 0$) → $t = $ hours 检查
Q12. (OPEN) Create an equation of the form $ax + b = c$, where $a$, $b$ and $c$ are constants, such that the solution of the equation is $x = 4$.
Your equation: 检查
💡 提示

开放题。最简单：$x = 4$ 本身就是 $1 \cdot x + 0 = 4$。其他例子：$2x - 3 = 5$、$x + 1 = 5$、$3x = 12$、$\dfrac{x}{2} = 2$ 等等。只要把任意 $a, c$ 选定，$b$ 也能反推确定。
Q13. Can a linear equation in one variable have
(a) more than one solution? 检查
(b) no solution? 检查
If your answer is yes, provide an example. If your answer is no, explain your answer.

💡 思路

(a) 在 $ax + b = c$ ($a \neq 0$) 中只有一解 $x = (c-b)/a$，所以不会有"多于一个解"（除非允许 $a = 0$，但定义要求 $a \neq 0$）。
严格说："多于一个但有限个" 不可能；恒等式 $0 = 0$（无穷多解）也违背定义。
(b) 有可能无解：化简后变成矛盾，比如 $2x + 1 = 2x + 5$ → $1 = 5$，无解。
Q14. (a) Let $x = 0.\dot{2}\dot{7}$ (即 27 循环, 0.272727...).
(i) $100x = $ 检查
(ii) Hence express $x$ as a fraction in its simplest form. → $x = $ 检查
💡 思路

$x = 0.272727...$；$100x = 27.272727... = 27 + x$；$\Rightarrow 99x = 27 \Rightarrow x = \tfrac{27}{99} = \tfrac{3}{11}$.

(b) Extending the technique in (a), express each of the following as a fraction in its simplest form.
(i) $0.\dot{4}$ (= 0.4444...) → 检查
(ii) $0.\dot{2}1\dot{6}$ (= 0.216216216...) → 检查
💡 思路

(b)(i) $y = 0.4\overline{4}$；$10y - y = 4$；$y = 4/9$.
(b)(ii) $z = 0.216\overline{216}$（3 位循环）；$1000z - z = 216$；$z = 216/999 = 8/37$（约分：$\gcd(216, 999) = 27$；$216/27 = 8, 999/27 = 37$）。