Inequalities · 不等式
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| inequality/ɪniːˈkwɒlɪti/ | 不等式 | bù děng shì |
| intervals/ˈɪntəvlz/ | 区间 | qū jiān |
| number line/ˈnʌmbə laɪn/ | 数轴 | shù zhóu |
| region/ˈriːdʒn/ | 区域 | qū yù |
Speed limits and number ranges
- A road sign says "Speed ≤ 30 mph". That's an inequality 不等式 — a range of allowed values, not a single answer.
- Inequalities describe intervals 区间: a set of values, not one value. Solving them is like solving equations, with one crucial twist.
速度限制和数字范围
- 一个路标说"速度 ≤ 30 mph"。那是一个不等式(inequality)——一个允许值的范围,不是一个单一的答案。
- 不等式描述区间(intervals):一组值,不是一个值。解它们就像解方程,带一个关键的转折。
Solving inequalities (Extended) (Extended)
- Solve exactly like an equation — but reverse the sign ($<$ becomes $>$, and vice versa) if you multiply or divide by a negative number.
Solve $3x + 1 < 10$: subtract 1 → $3x < 9$ → divide by 3 → $x < 3$.
The sign flip. Solve $-2x > 6$: divide by $-2$ and flip → $x < -3$ (not $x > -3$). Forgetting to flip is the most common mistake.
$-\tfrac13 \leqslant x < 3$: a closed circle includes the end value, an open circle excludes it
解不等式（拓展）
- 完全像一个方程那样解——但如果你乘或除以一个负数,翻转符号($<$ 变成 $>$,反之亦然)。
解 $3x + 1 < 10$:减 1 → $3x < 9$ → 除以 3 → $x < 3$。
符号翻转。 解 $-2x > 6$:除以 $-2$ 并翻转 → $x < -3$(不是 $x > -3$)。忘记翻转是最常见的错误。
$-\tfrac13 \leqslant x < 3$:一个实心圆包括端值,一个空心圆排除它
Inequalities · 不等式
y = ax + b
An inequality asks where the line is above or below a value. · 一个不等式询问直线在哪里高于或低于一个值。
You must reverse the inequality sign when you: · 你必须翻转不等号当你:
Multiplying or dividing by a negative flips < to > (and vice versa). · 乘或除以负数时，不等号< to >方向翻转（反之亦然）。
Solve 3x + 1 < 10. What is the upper bound for x? · 解 3x + 1 < 10。x 的上界是多少?
3x < 9 → x < 3. The upper bound is 3 (not included). · 3x < 9 → x < 3。上界是 3(不被包括)。
Solving −2x > 6 gives x > −3. · 解 −2x > 6 给出 x > −3。
Dividing by −2 and flipping gives x < −3, not x > −3. · 两边同除−2并翻转符号，得 x < −3, not x > −3。
Solving double inequalities (Extended)
- $-3 \leq 3x - 2 < 7$: apply the same operation to all three parts.
- Add 2: $-1 \leq 3x < 9$. Divide by 3: $-\dfrac{1}{3} \leq x < 3$.
解双重不等式（拓展）
- $-3 \leq 3x - 2 < 7$:对所有三个部分应用相同的运算。
- 加 2:$-1 \leq 3x < 9$。除以 3:$-\dfrac{1}{3} \leq x < 3$。
Solve −3 ≤ 3x − 2 < 7. The solution is −1/3 ≤ x < ? What is the upper bound? · 解 −3 ≤ 3x − 2 < 7。解是 −1/3 ≤ x < ? 上界是多少?
Add 2: −1 ≤ 3x < 9; divide by 3: −1/3 ≤ x < 3. · 加 2:−1 ≤ 3x < 9;除以 3:−1/3 ≤ x < 3。
On a number line 数轴
- Open circle (○) for $<$ or $>$ — the end value is not included.
- Closed circle (●) for $\leq$ or $\geq$ — the end value is included.
Closed circle at $-\frac{1}{3}$ (included), open circle at $3$ (excluded). The shaded region 区域 is the solution.
在一条数轴上
- 空心圆(○)用于 $<$ 或 $>$——端值不被包括。
- 实心圆(●)用于 $\leq$ 或 $\geq$——端值被包括。
在 $-\frac{1}{3}$ 处实心圆(包括),在 $3$ 处空心圆(排除)。阴影区域是解。
On a number line, the symbol ≤ is drawn with a: · 在一条数轴上,符号 ≤ 用以下画出:
≤ and ≥ include the end value, shown by a closed circle; < and > use an open circle. · ≤ 和 ≥ 包含端点值，用实心圆表示；< and > 使用空心圆。
Match each inequality symbol to its number-line drawing. · 把每个不等号匹配到它的数轴画法。
< and > use open circles (end value excluded); ≤ and ≥ use closed circles (included). · < and > 使用空心圆（不包含端点）；≤ 和 ≥ 使用实心圆（包含端点）。
Inequality regions (Extended)
- An inequality in two variables ($y > 2x + 1$) shades a region on the coordinate plane.
- Broken (dashed) boundary line for strict inequalities ($<$ or $>$).
- Solid boundary line for $\leq$ or $\geq$.
不等式区域(扩展)
- 一个两变量的不等式($y > 2x + 1$)在坐标平面上给一个区域上色。
- 严格不等式($<$ 或 $>$)用虚线边界线。
- $\leq$ 或 $\geq$ 用实线边界线。
Core: interpret before solving
- For $-3\le x<2$, mark a filled circle at $-3$, an open circle at 2 and shade between. The integer values are $-3,-2,-1,0,1$.
- A parcel rule $2\le m<5$ kg includes 2 kg but excludes 5 kg. Core questions ask you to interpret and represent such ranges; solving algebraic inequalities belongs to Extended.
核心：求解前先解读
- 对于 $-3\le x<2$，在 $-3$ 处标实心圆点，在 2 处标空心圆点，并在两者之间涂色。整数解为 $-3,-2,-1,0,1$。
- 包裹重量规则 $2\le m<5$ kg 包含 2 kg 但不包含 5 kg。核心题目要求你解读并表达此类范围；代数不等式的求解属于拓展内容。
How many integers satisfy −3 ≤ x < 2? · 有多少个整数满足 −3 ≤ x < 2？
The integers are −3, −2, −1, 0 and 1. · 这些整数为 −3, −2, −1, 0 和 1。
You've got it
- solve like an equation, but flip the sign when multiplying/dividing by a negative
- $-3 \leq 3x - 2 < 7 \Rightarrow -\dfrac{1}{3} \leq x < 3$
- number line: open circle excludes, closed circle includes the end
- 2-variable inequalities shade a region (dashed boundary for strict, solid otherwise)
你掌握了
- 像方程那样解,但在乘/除以一个负数时翻转符号
- $-3 \leq 3x - 2 < 7 \Rightarrow -\dfrac{1}{3} \leq x < 3$
- 数轴:空心圆排除,实心圆包括端
- 2 变量不等式表示一个区域（严格不等式用虚线边界，否则用实线）