🏠 首页 CHAPTER 5 · SIMPLE EQUATIONS IN ONE VARIABLE 5.2

5.2

Linear Equations Involving Brackets and Fractions

💡 5.1 学了"平衡法" 解 $ax + b = c$。5.2 升级两个常见难点：
1️⃣ 带括号：先用分配律展开（第 4.3 节学过），再解。
2️⃣ 带分数：两边同时乘分母的 LCM，把方程"清干净"成纯整数系数的。

AEquations Involving Brackets

To solve linear equations involving brackets, we can apply the distributive law to remove the brackets when expanding the expression.

$\stackrel{\frown}{a(x + y)} = ax + ay$
📖 Worked Example 7

Solve the equation $9(x + 1) = 2(3x + 8)$.

SOLUTION

  • $9(x + 1) = 2(3x + 8)$
  • $9x + 9 = 6x + 16$  (Expand both sides by applying the distributive law.)
  • $9x + 9 - 6x = 6x + 16 - 6x$  (Subtract $6x$ from both sides.)
  • $3x + 9 = 16$
  • $3x + 9 - 9 = 16 - 9$  (Subtract 9 from both sides.)
  • $3x = 7$
  • $\dfrac{3x}{3} = \dfrac{7}{3}$  (Divide both sides by 3.)
  • $x = \dfrac{7}{3}$
  • ∴ $x = \mathbf{2\dfrac{1}{3}}$
Show how you would verify that $x = 2\dfrac{1}{3}$ is the solution of $9(x + 1) = 2(3x + 8)$ in Worked Example 7.
💡 思路

把 $x = \tfrac{7}{3}$ 代入两边算：
LHS = $9(\tfrac{7}{3} + 1) = 9 \cdot \tfrac{10}{3} = 30$；
RHS = $2(3 \cdot \tfrac{7}{3} + 8) = 2(7 + 8) = 30$ ✓

✍️ Try It Yourself 7
  1. (a) $7(2 - x) = -2(3x + 5)$ → $x = $ 检查
    💡 思路

    $14 - 7x = -6x - 10 \Rightarrow -7x + 6x = -10 - 14 \Rightarrow -x = -24 \Rightarrow x = 24$.

  2. (b) $x - 4(x - 3) = 3(5 + 2x)$ → $x = $ 检查
    💡 思路

    $x - 4x + 12 = 15 + 6x \Rightarrow -3x + 12 = 15 + 6x \Rightarrow -9x = 3 \Rightarrow x = -\tfrac{1}{3}$.

BEquations Involving Fractions

When we have an equation involving fractions, we can simplify the equation by multiplying the LHS and RHS of the equation by the LCM of the denominators. The following examples illustrate how this is done.

📖 Worked Example 8

Solve the equation $\dfrac{3x + 2}{5} = \dfrac{4x - 7}{6}$.

SOLUTION

  • $\dfrac{3x + 2}{5} = \dfrac{4x - 7}{6}$
  • $30 \left(\dfrac{3x + 2}{5}\right) = 30 \left(\dfrac{4x - 7}{6}\right)$  (Multiply both sides by 30, which is the LCM of 5 and 6.)
  • $6(3x + 2) = 5(4x - 7)$
  • $18x + 12 = 20x - 35$  (Expand both sides by applying the distributive law.)
  • $18x + 12 - 20x = 20x - 35 - 20x$  (Subtract $20x$ from both sides.)
  • $-2x + 12 = -35$
  • $-2x + 12 - 12 = -35 - 12$  (Subtract 12 from both sides.)
  • $-2x = -47$
  • $\dfrac{-2x}{-2} = \dfrac{-47}{-2}$  (Divide both sides by $-2$.)
  • ∴ $x = \mathbf{\dfrac{47}{2}} = \mathbf{23\dfrac{1}{2}}$
Why do we multiply both sides of the equation in Worked Example 8 by the LCM of 5 and 6?
💡 思考

LCM 是同时能整除两个分母的最小数。乘 LCM 后两个分数同时变成整数系数，最简洁。也可以乘 5 × 6 = 30（就是 LCM），或者甚至 60 也行，只是数字变大而已（参见下面 THINK²）。

In Worked Example 8, can we multiply both sides by 60? Explain your answer.
💡 思考

可以。$60 = 2 \times 30$ 也是 5 和 6 的公倍数。乘以 60 得 $12(3x+2) = 10(4x-7)$，最后结果一样。但通常选 LCM 因为数字最小、计算最快。

✍️ Try It Yourself 8
  1. (a) $\dfrac{8m + 3}{7} = \dfrac{5m - 3}{4}$ → $m = $ 检查
    💡 思路

    × 28 (LCM 7, 4): $4(8m+3) = 7(5m-3)$；$32m + 12 = 35m - 21$；$33 = 3m$；$m = 11$.

  2. (b) $\dfrac{2(t + 6)}{3} = \dfrac{-t + 1}{9}$ → $t = $ 检查
    💡 思路

    × 9: $6(t+6) = -t + 1$；$6t + 36 = -t + 1$；$7t = -35$；$t = -5$.

📖 Worked Example 9

Solve the equation $\dfrac{y}{5} - 6 = \dfrac{y - 3}{2}$.

SOLUTION

  • $\dfrac{y}{5} - 6 = \dfrac{y - 3}{2}$
  • $10\left(\dfrac{y}{5} - 6\right) = 10\left(\dfrac{y - 3}{2}\right)$  (Multiply both sides by 10, which is the LCM of 2 and 5.)
  • $10\left(\dfrac{y}{5}\right) - 10 \times 6 = 10\left(\dfrac{y - 3}{2}\right)$  (Expand LHS by applying the distributive law.)
  • $2y - 60 = 5(y - 3)$
  • $2y - 60 = 5y - 15$  (Expand RHS by applying the distributive law.)
  • $-3y = 45$
  • $\dfrac{-3y}{-3} = \dfrac{45}{-3}$  (Divide both sides by $-3$.)
  • ∴ $y = \mathbf{-15}$
✍️ Try It Yourself 9
  1. (a) $\dfrac{y}{3} + \dfrac{y - 2}{4} = 3$ → $y = $ 检查
    💡 思路

    × 12: $4y + 3(y-2) = 36$；$7y - 6 = 36$；$7y = 42$；$y = 6$.

  2. (b) $\dfrac{x - 3}{6} = \dfrac{1}{5} + \dfrac{3(x - 1)}{2}$ → $x = $ 检查
    💡 思路

    LCM(6, 5, 2) = 30. × 30: $5(x-3) = 6 + 45(x-1)$；$5x - 15 = 6 + 45x - 45$；$5x - 15 = 45x - 39$；$-40x = -24$；$x = \tfrac{3}{5}$.
    验证：LHS = $\tfrac{3/5 - 3}{6} = \tfrac{-12/5}{6} = -\tfrac{2}{5}$；RHS = $\tfrac{1}{5} + \tfrac{3(-2/5)}{2} = \tfrac{1}{5} - \tfrac{3}{5} = -\tfrac{2}{5}$ ✓.

📖 Worked Example 10 · Solve for one unknown in a formula

Four quantities $s$, $u$, $v$ and $t$ are related by the formula $s = \dfrac{1}{2}(u + v)t$. Find the value of $v$ given that $u = 10$, $s = 15$ and $t = 3$.

SOLUTION

  • Substituting $u = 10$, $s = 15$ and $t = 3$ into $s = \dfrac{1}{2}(u + v)t$,
  • $15 = \dfrac{1}{2}(10 + v)(3)$  ($v$ is the unknown.)
  • $15 \times 2 = \dfrac{3}{2}(10 + v) \times 2$  (Multiply both sides by 2.)
  • $30 = 3(10 + v)$
  • $\dfrac{30}{3} = \dfrac{3}{3}(10 + v)$  (Divide both sides by 3.)
  • $10 = 10 + v$
  • $10 + v - 10 = 10 - 10$  (Subtract 10 from both sides.)
  • ∴ $v = \mathbf{0}$
Jane suggests we can simplify the equation $15 = \dfrac{3}{2}(10 + v)$ by multiplying both sides by $\dfrac{2}{3}$ directly. Do you agree? Explain your answer.
💡 思考

同意。乘以 $\tfrac{2}{3}$ 就是 "乘 2 再除以 3" 的合并操作，效果一样：$15 \times \tfrac{2}{3} = 10 + v \Rightarrow 10 = 10 + v \Rightarrow v = 0$。两种做法等价。

✍️ Try It Yourself 10

Given the formula $A = \dfrac{1}{2}(a + b)h$, find the value of $a$ when $b = 13$, $h = 9$ and $A = 90$.

$a = $ 检查

💡 思路

$90 = \tfrac{1}{2}(a+13)(9) \Rightarrow 180 = 9(a+13) \Rightarrow 20 = a + 13 \Rightarrow a = 7$.

In general,

  • for equations involving brackets, we apply the distributive law to remove the brackets when expanding the expressions.
  • for equations involving fractions, we multiply both sides of the equation by the LCM of the denominators.

PPractice Exercise 5.2

BASIC MASTERY
  1. Q1. Solve the following equations.
    (a) $3x + 4 = 2(2x + 7)$ → $x = $ 检查
    (b) $3(5x + 8) = 3x - 2$ → $x = $ 检查
    (c) $3x - 4 = 2(-3x + 7)$ → $x = $ 检查
    (d) $3(-5x + 8) = -42 - 9x$ → $x = $ 检查
    (e) $2(8x + 5) = 4(3x + 1)$ → $x = $ 检查
    (f) $3(4x - 1) = 7(2x - 5)$ → $x = $ 检查
  2. Q2. Solve the following equations.
    (a) $5(x + 3) - 4(2x - 9) = 0$ → $x = $ 检查
    (b) $3(3x - 1) - 4(5 - 2x) = -10$ → $x = $ 检查
    (c) $9x - 2(x + 8) = 5x - 11(2 - x)$ → $x = $ 检查
    (d) $1 - 4(2x + 3) = 5(x - 2) - 3(x - 1)$ → $x = $ 检查
  3. Q3. Solve the following equations.
    (a) $12 + \dfrac{2}{5}x = 16$ → $x = $ 检查
    (b) $9 - \dfrac{4}{3}x = -7$ → $x = $ 检查
    (c) $\dfrac{2x + 9}{5} = 5$ → $x = $ 检查
    (d) $\dfrac{3x - 11}{7} - 2 = 0$ → $x = $ 检查
INTERMEDIATE
  1. Q4. Solve the following equations.
    (a) $\dfrac{7x - 2}{2} = \dfrac{5x - 3}{3}$ → $x = $ 检查
    (b) $\dfrac{2x + 3}{3} = \dfrac{3x - 15}{11}$ → $x = $ 检查
    (c) $-x + \dfrac{x}{4} = 15$ → $x = $ 检查
    (d) $\dfrac{x}{2} - \dfrac{x}{3} = 7$ → $x = $ 检查
    (e) $\dfrac{1}{4}y + \dfrac{7}{2} = -2y + \dfrac{4}{5}$ → $y = $ 检查
    (f) $\dfrac{2y}{3} + \dfrac{5y}{4} = \dfrac{y}{6} - 8$ → $y = $ 检查
    (g) $\dfrac{3(5x - 6)}{4} + 2 = 4x$ → $x = $ 检查
    (h) $\dfrac{2(1 - 4x)}{5} - 9 = 3(2 - x)$ → $x = $ 检查
  2. Q5. Solve the following equations.
    (a) $\dfrac{t}{5} - \dfrac{3(t + 2)}{7} = 2$ → $t = $ 检查
    (b) $\dfrac{2t - 1}{3} + \dfrac{3t - 4}{5} = t$ → $t = $ 检查
    (c) $\dfrac{y + 9}{2} = \dfrac{y - 3}{4} - \dfrac{y}{3}$ → $y = $ 检查
    (d) $2y - \dfrac{5(y - 2)}{6} = \dfrac{2(1 + 3y)}{3}$ → $y = $ 检查
    (e) $\dfrac{z - 7}{3} - \dfrac{z - 5}{12} = \dfrac{2z - 27}{15}$ → $z = $ 检查
    (f) $\dfrac{4z + 3}{5} - \dfrac{7z - 1}{3} = \dfrac{2 - 19z}{10}$ → $z = $ 检查
ADVANCED
  1. Q6. The conversion formula between $f$ degree Fahrenheit (°F) and $c$ degree Celsius (°C) is $c = \dfrac{5}{9}(f - 32)$.
    (a) The boiling point of water is 100 °C. Find the boiling point of water in °F. → $f = $ °F 检查
    (b) The body temperature of John is 37 °C. Find his body temperature in °F. → $f = $ °F 检查
  2. Q7. Let $m$ denote the average of two numbers, $a$ and $b$.
    (a) Express $m$ in terms of $a$ and $b$. → $m = $ 检查
    (b) Given that $a = 7$ and $m = 10$, find the value of $b$. → $b = $ 检查
  3. Q8. The kinetic energy, $E$ joules, of an object of mass $m$ kg moving with velocity $v$ m/s is given by $E = \dfrac{1}{2}mv^2$. Find the value of $m$ when $E = 160$ and $v = 5$. → $m = $ kg 检查
  4. Q9. In the linear equation $3L + R(2x + 3y) = 5z$, which variables should be given in order to find the value of $R$?
    💡 思路

    要解出 $R = \dfrac{5z - 3L}{2x + 3y}$，需要给定 $L, x, y, z$ 四个变量（且 $2x + 3y \neq 0$，否则无法除）。

  5. Q10. Ivan solved the following equation to find the value of $x$, where $a$ is a constant.
    $\dfrac{x + 1}{2} + \dfrac{x - a}{3} = 1$
    $3(x + 1) + 2(x - a) = 1$
    $3x + 3 + 2x - a = 1$
    $5x = a - 2$
    $x = \dfrac{a - 2}{5}$
    (a) Identify the errors and show the correct working.

    (b) If the solution of the equation is $x = 2$, find the value of $a$. → $a = $ 检查
    💡 错误诊断

    LCM(2, 3) = 6. 两边乘 6 应该得到：
    $6 \cdot \dfrac{x+1}{2} + 6 \cdot \dfrac{x-a}{3} = 6 \cdot 1$
    → $3(x+1) + 2(x-a) = \mathbf{6}$ (不是 1)
    Ivan 忘了把右边的 1 也乘 6。
    正确：$3x + 3 + 2x - 2a = 6 \Rightarrow 5x = 2a + 3 \Rightarrow x = \dfrac{2a + 3}{5}$.
    (b) $x = 2$: $10 = 2a + 3 \Rightarrow a = \tfrac{7}{2}$.

  6. Q11. (OPEN) Create an equation in variable $x$ of the form $\dfrac{x - a}{b} = c$, where $a$, $b$ and $c$ are constants, such that the solution of the equation is $x = -1$.
    Your equation: 检查
    💡 提示

    开放题。代入 $x = -1$，方程必须成立：$\dfrac{-1 - a}{b} = c \Leftrightarrow -1 - a = bc \Leftrightarrow a = -1 - bc$。
    示例：选 $b = 2, c = -1$ → $a = -1 - (-2) = 1$，方程 $\dfrac{x - 1}{2} = -1$。
    其他：$\dfrac{x - 2}{3} = -1$（$a=2, b=3, c=-1$）、$\dfrac{x + 3}{2} = 1$（$a=-3, b=2, c=1$）等。