5.3
Simple Fractional Equations
When the variable of an equation is in the denominator of a term, the equation is called a fractional equation. Examples of fractional equations are
$\dfrac{6}{x - 2} = 3 \qquad$ and $\qquad \dfrac{1}{x + 3} = \dfrac{2}{x}$
We can use multiplication to transform simple fractional equations into linear equations. In solving fractional equations, it is important to check the solutions, which cannot include those values that will make a denominator of the original equation zero.
Solve the equation $\dfrac{6}{x - 2} = 3$.
SOLUTION
- $\dfrac{6}{x - 2} = 3$
- $(x - 2)\left(\dfrac{6}{x - 2}\right) = (x - 2)(3)$ (Multiply both sides by $(x - 2)$.)
- $6 = 3(x - 2)$ (The equation is transformed into a linear equation.)
- $6 = 3x - 6$
- $12 = 3x$
- $x = \dfrac{12}{3}$
- ∴ $x = \mathbf{4}$
When $x = 4$, the denominator of $x - 2 = 4 - 2 = 2 \neq 0$ ✓.
Substituting $x = 4$ into the LHS, we have LHS $= \dfrac{6}{4 - 2} = \dfrac{6}{2} = 3 = $ RHS ✓.
Since LHS = RHS, the solution $x = 4$ is correct.
💡 思考
因为除以 0 没有意义。如果我们乘 $(x-2)$ 然后解得 $x = 2$（恰好让分母变零），那就是假解——原方程 $\tfrac{6}{x-2}$ 在 $x = 2$ 处根本就没定义。必须排除。在本题 $x = 4$ 没问题，所以解有效。
- (a) $\dfrac{8}{x + 1} = 4$ → $x = $ 检查
💡 思路
× $(x+1)$: $8 = 4(x+1)$；$8 = 4x + 4$；$x = 1$. 检查：$x = 1 \neq -1$ ✓。
- (b) $\dfrac{-2}{y - 6} - 3 = 0$ → $y = $ 检查
💡 思路
$\dfrac{-2}{y-6} = 3$；× $(y-6)$: $-2 = 3(y-6)$；$-2 = 3y - 18$；$3y = 16$；$y = \tfrac{16}{3}$. 检查 $y \neq 6$ ✓。
Solve each of the following equations.
- (a) $\dfrac{t}{t + 3} = 1\dfrac{1}{4}$
- (b) $\dfrac{2}{x - 1} = \dfrac{3}{x + 1}$
- (c) $\dfrac{1}{2x + 1} = \dfrac{4}{3x - 6}$
SOLUTION
(a)
$\dfrac{t}{t + 3} = 1\tfrac{1}{4} = \dfrac{5}{4}$
$4(t + 3) \cdot \dfrac{t}{t + 3} = 4(t + 3) \cdot \dfrac{5}{4}$ (Multiply both sides by $4(t + 3)$.)
$4t = 5(t + 3)$ (Linear equation.)
$4t = 5t + 15$
$-t = 15$
∴ $t = \mathbf{-15}$ ($t + 3 = -12 \neq 0$ ✓)
(b)
$\dfrac{2}{x - 1} = \dfrac{3}{x + 1}$
$(x - 1)(x + 1) \cdot \dfrac{2}{x - 1} = (x - 1)(x + 1) \cdot \dfrac{3}{x + 1}$ (Multiply both sides by $(x-1)(x+1)$.)
$2(x + 1) = 3(x - 1)$
$2x + 2 = 3x - 3$
$-x = -5$
∴ $x = \mathbf{5}$ (check: $x-1 = 4 \neq 0$, $x+1 = 6 \neq 0$ ✓)
(c)
$\dfrac{1}{2x + 1} = \dfrac{4}{3x - 6}$
$(2x + 1)(3x - 6) \cdot \dfrac{1}{2x + 1} = (2x + 1)(3x - 6) \cdot \dfrac{4}{3x - 6}$
$3x - 6 = 4(2x + 1)$
$3x - 6 = 8x + 4$
$-5x = 10$
$\dfrac{-5x}{-5} = \dfrac{10}{-5}$
∴ $x = \mathbf{-2}$ (check: $2x+1 = -3 \neq 0$, $3x-6 = -12 \neq 0$ ✓)
- (a) $\dfrac{-y}{y + 4} = 2\dfrac{1}{3}$ → $y = $ 检查
💡 思路
$\dfrac{-y}{y+4} = \dfrac{7}{3}$；× $3(y+4)$：$-3y = 7(y+4)$；$-3y = 7y + 28$；$-10y = 28$；$y = -\tfrac{14}{5}$.
- (b) $\dfrac{-1}{m - 7} = \dfrac{6}{m}$ → $m = $ 检查
💡 思路
× $m(m-7)$: $-m = 6(m-7) = 6m - 42$；$-7m = -42$；$m = 6$. ($m-7 = -1 \neq 0$, $m \neq 0$ ✓)
- (c) $\dfrac{3}{2a - 1} = \dfrac{2}{4 - a}$ → $a = $ 检查
💡 思路
× $(2a-1)(4-a)$: $3(4-a) = 2(2a-1)$；$12 - 3a = 4a - 2$；$14 = 7a$；$a = 2$.
PPractice Exercise 5.3
- Q1. Solve the following equations.
(a) $\dfrac{15}{x} = 3$ → $x = $ 检查
(b) $\dfrac{7}{2x} = -2$ → $x = $ 检查
(c) $\dfrac{1}{3x} + 4 = 0$ → $x = $ 检查
(d) $9 - \dfrac{6}{x} = 0$ → $x = $ 检查
(e) $\dfrac{2}{x} + 3 = \dfrac{1}{2}$ → $x = $ 检查
(f) $\dfrac{2}{x + 3} = 4$ → $x = $ 检查
(g) $\dfrac{7}{2x + 5} = -3$ → $x = $ 检查
(h) $\dfrac{7}{x - 9} - 5 = 0$ → $x = $ 检查
- Q2. Solve the following equations.
(a) $\dfrac{3t + 5}{t} = \dfrac{3}{2}$ → $t = $ 检查
(b) $\dfrac{7 - z}{3z} = -\dfrac{1}{6}$ → $z = $ 检查
(c) $\dfrac{3y + 2}{2y - 7} = 4$ → $y = $ 检查
(d) $\dfrac{4v - 1}{15 + 2v} + 3 = 0$ → $v = $ 检查
(e) $\dfrac{n}{2(n + 3)} = \dfrac{7}{5}$ → $n = $ 检查
(f) $\dfrac{-2u}{u + 4} = 2\dfrac{2}{3}$ → $u = $ 检查 - Q3. Solve the following equations.
(a) $\dfrac{5}{x + 1} = \dfrac{7}{2x}$ → $x = $ 检查
(b) $\dfrac{3}{2x - 5} = \dfrac{1}{x + 1}$ → $x = $ 检查
(c) $\dfrac{4}{2x - 1} = \dfrac{-3}{x + 1}$ → $x = $ 检查
(d) $\dfrac{3}{5 - x} = \dfrac{-2}{1 - 2x}$ → $x = $ 检查 - Q4. Huimin solved an equation as follows:
$\dfrac{x}{x - 4} + 1 = \dfrac{8}{x - 4}$Did she solve it correctly? If not, identify the error and show the correct working.
$(x - 4)\dfrac{x}{x - 4} + 1 = (x - 4)\dfrac{8}{x - 4}$
$x + 1 = 8$
$x = 7$
Correct answer: $x = $ 检查💡 错误诊断
Huimin 忘记把 所有项 都乘 $(x-4)$——她只把第一项和右边乘了 $(x-4)$，但把"+1"原封不动留着了。
正确：$(x-4) \cdot \dfrac{x}{x-4} + (x-4) \cdot 1 = (x-4) \cdot \dfrac{8}{x-4}$
$\Rightarrow x + (x-4) = 8$
$\Rightarrow 2x - 4 = 8$
$\Rightarrow 2x = 12$
$\Rightarrow x = 6$. 验证：$\tfrac{6}{2} + 1 = 3 + 1 = 4$；$\tfrac{8}{2} = 4$ ✓. - Q5. Write an equation for each of the following and solve it.
(a) The quotient when 7 is divided by the sum of $2x$ and 5 equals 3. → $x = $ 检查💡 思路
$\dfrac{7}{2x + 5} = 3$；$7 = 3(2x+5)$；$7 = 6x + 15$；$x = -\tfrac{8}{6} = -\tfrac{4}{3}$.
(b) The sum of $\dfrac{2}{x}$ and $\dfrac{3}{4}$ is the same as the difference when $\dfrac{1}{x}$ is subtracted from 2. → $x = $ 检查💡 思路
$\dfrac{2}{x} + \dfrac{3}{4} = 2 - \dfrac{1}{x}$；$\dfrac{3}{x} = 2 - \dfrac{3}{4} = \dfrac{5}{4}$；$x = \dfrac{12}{5}$.
- Q6. The time $t$ seconds for a car to increase its speed from $u$ m/s to $v$ m/s is given by $t = \dfrac{2d}{u + v}$, where $d$ m is the distance covered by the car. If $d = 40$, $t = 5$ and $u = 4$, find the value of $v$.
$v = $ m/s 检查💡 思路
$5 = \dfrac{80}{4 + v}$；$5(4 + v) = 80$；$4 + v = 16$；$v = 12$.
- Q7. In the diagram, a block of mass $m$ kg on a smooth table is pulled by another block of mass $n$ kg hanging over a smooth pulley. The acceleration, $a$ m/s², of the $m$ kg mass is given by $a = \dfrac{ng}{n + m}$. If $a = 3\dfrac{1}{3}$, $g = 10$ and $m = 8$, find the value of $n$.
$n = $ 检查💡 思路
$\tfrac{10}{3} = \dfrac{10n}{n + 8}$；$10(n+8) = 30n$；$10n + 80 = 30n$；$20n = 80$；$n = 4$.
- Q8. The object distance, $u$ cm, and the image distance, $v$ cm, of a lens is related by the formula $\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$, where $f$ cm is the focal length of the lens. If $f = 20$ and $u = 30$, find the value of $v$.
$v = $ cm 检查💡 思路
$\dfrac{1}{v} = \dfrac{1}{20} - \dfrac{1}{30} = \dfrac{3 - 2}{60} = \dfrac{1}{60}$；所以 $v = 60$.