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Linear Functions and Systems · Solutions一次函数与方程组 · 详解

Companion to the Practice Set · Mark-by-mark walkthroughs · SAT / AP-Feeder / ON / BC styles练习题配套答案 · 逐分讲解 · SAT / AP 衔接 / 安 / 卑诗省考风格

EASY易 MEDIUM中 HARD难 🇺🇸 US美 🇨🇦 ON安 🇨🇦 BC卑 SAT-style MCQSAT 风格选择题 AP-feeder FRQAP 衔接简答题 ON Provincial-style安大略省考风格 BC Provincial-style卑诗省考风格 Honors荣誉级
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  短答题 · 详解SAT MCQ + ON/BC short answer · 18 marksSAT 选择题 + 安/卑省考短答 · 共 18 分

Section A · Worked SolutionsA 部分 · 详细解答

Q1EASY易 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §1 Slope斜率 · HSF-IF.B.6 [3 marks][3 分]

Slope of the line through $(-4, 7)$ and $(2, -5)$.求过点 $(-4, 7)$ 与 $(2, -5)$ 的直线斜率。

Answer:答案：  (A)  $m = -2$

(a) Apply slope formula套用斜率公式 M1·A1·A1

$$ m \;=\; \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \;=\; \frac{-5 - 7}{2 - (-4)} \;=\; \frac{-12}{6} \;=\; -2. $$ So the slope is $-2$, matching option (A).故斜率为 $-2$，对应选项 (A)。
Why the distractors fail.干扰项分析。
(B) $-\tfrac{1}{2}$: the reciprocal, comes from swapping numerator and denominator in the slope formula.是真值的倒数，来源是在斜率公式中把分子分母颠倒。
(C) $\tfrac{1}{2}$: reciprocal with the sign also dropped, two errors at once.既颠倒又丢掉负号，一步走错两次。
(D) $2$: dropped the negative sign from $-12 / 6$.在算 $-12 / 6$ 时丢掉了负号。
Pin down "rise over run", direction-locked.把"纵向变化除以横向变化"锁定方向。 The sign of slope is fixed the moment you pick which point is $(x_{1}, y_{1})$, but only if you keep that choice consistent in both numerator and denominator. The common slip is writing $\dfrac{y_{2} - y_{1}}{x_{1} - x_{2}}$, which flips the sign. Either lock the order before you compute, or use the magnitude $\left|\dfrac{\Delta y}{\Delta x}\right|$ and read the sign off the geometry: as $x$ increases from $-4$ to $2$, $y$ falls from $7$ to $-5$, so $m < 0$.一旦选定哪个点是 $(x_{1}, y_{1})$，斜率符号就定了——前提是分子与分母都保持同一选择。常见的错是写成 $\dfrac{y_{2} - y_{1}}{x_{1} - x_{2}}$，这会把符号翻反。要么在动笔前锁定顺序，要么先算绝对值 $\left|\dfrac{\Delta y}{\Delta x}\right|$，再从几何上读出符号：$x$ 从 $-4$ 增到 $2$ 时，$y$ 从 $7$ 降到 $-5$，所以 $m < 0$。
Q2EASY易 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §2 Slope-Intercept斜截式 · HSA-REI.B.3 [3 marks][3 分]

Rewrite $4x - 3y = 12$ in slope-intercept form.将 $4x - 3y = 12$ 化为斜截式。

Answer:答案：  (B)  $y = \tfrac{4}{3}\,x - 4$

(a) Isolate $y$分离 $y$ M1·A1

Start from $4x - 3y = 12$. Subtract $4x$ from both sides:从 $4x - 3y = 12$ 出发，两边减去 $4x$： $$ -3y \;=\; -4x + 12. $$ Divide every term by $-3$, flipping every sign:两边同除以 $-3$，每一项都变号： $$ y \;=\; \tfrac{-4}{-3}\,x + \tfrac{12}{-3} \;=\; \tfrac{4}{3}\,x - 4. $$

(b) Read $(m, b)$ and match读出 $(m, b)$ 并比对 A1

Slope $m = \tfrac{4}{3}$, $y$-intercept $b = -4$. Hence option (B).斜率 $m = \tfrac{4}{3}$，$y$ 轴截距 $b = -4$，故选 (B)。
Why the distractors fail.干扰项分析。
(A) $y = -\tfrac{4}{3} x + 4$: sign error on both the slope and the intercept, divided by $+3$ instead of $-3$.斜率与截距均符号反，是把分母写成 $+3$ 而非 $-3$ 所致。
(C) $y = 4x - 12$: "moved" the $-3y$ across the equals sign without dividing the rest of the equation.把 $-3y$ "搬"过等号，却没把右边一并除以 $-3$。
(D) $y = \tfrac{3}{4} x - 3$: read off the equation as if it were already $y = mx + b$, treating the $-3$ multiplier as a constant.当作方程已是 $y = mx + b$ 形式直接读取，把 $-3$ 当成常数处理。
Read slope only from $y = mx + b$ form.只能从 $y = mx + b$ 形式读出斜率。 A line in standard form $Ax + By = C$ has slope $-A/B$ and $y$-intercept $C/B$, not $A$ and $C$. Many students glance at $4x - 3y = 12$ and answer "slope 4". On the SAT the only safe move is to isolate $y$ before reading anything off the equation. The second trap is sign-handling: dividing by a negative coefficient flips every term, both the $x$-coefficient and the constant.标准式 $Ax + By = C$ 的斜率是 $-A/B$、$y$ 轴截距是 $C/B$，而不是 $A$ 和 $C$。不少同学瞥一眼 $4x - 3y = 12$ 就答"斜率 4"。在 SAT 上唯一稳妥的做法是先分离 $y$，再从方程读取信息。第二个陷阱是符号处理：除以负系数会让每一项都变号，$x$ 的系数与常数项都得跟着翻。
Q3EASY易 🇨🇦 ON安 ON Provincial-style安大略省考风格 §1 Rate of Change变化率 · MPM1D Linear Relations线性关系 [4 marks][4 分]

Pool drains linearly: $V(3) = 4200$ L, $V(11) = 2600$ L. (a) Rate. (b) Interpret sign. (c) Restriction on $t$.泳池线性排水：$V(3) = 4200$ 升，$V(11) = 2600$ 升。(a) 求变化率。(b) 解释符号含义。(c) 写出 $t$ 的限制。

Answer:答案：  (a) $\dfrac{\Delta V}{\Delta t} = -200$ L/min升/分  ·  (b) negative $\Rightarrow$ water is leaving负值 $\Rightarrow$ 水在流失  ·  (c) $0 \le t \le 24$ min分

(a) Compute the rate, with units计算变化率（带单位） M1·A1

Using two data points $(t, V) = (3, 4200)$ and $(11, 2600)$:取两组数据 $(t, V) = (3, 4200)$ 与 $(11, 2600)$： $$ \frac{\Delta V}{\Delta t} \;=\; \frac{2600 - 4200}{11 - 3} \;=\; \frac{-1600}{8} \;=\; -200 \;\; \text{L/min}. $$

(b) Interpret the sign解释符号含义 A1

The negative sign means $V$ is decreasing over time: the pool is losing $200$ litres of water every minute. This is consistent with "the pool is being drained."负号表示 $V$ 随时间递减：泳池每分钟流失 $200$ 升水，与题干"泳池正在排水"一致。

(c) Domain restriction定义域限制 A1

The linear model is $V(t) = -200 t + b$. Using $V(3) = 4200$: $b = 4200 + 600 = 4800$, so $V(t) = 4800 - 200 t$. Physically the volume cannot go below zero: $V(t) \ge 0 \;\Leftrightarrow\; t \le 24$. Time also cannot run backwards from the start of the drain: $t \ge 0$. Hence $0 \le t \le 24$ minutes.线性模型为 $V(t) = -200 t + b$。代入 $V(3) = 4200$：$b = 4200 + 600 = 4800$，故 $V(t) = 4800 - 200 t$。物理上水量不能为负：$V(t) \ge 0 \;\Leftrightarrow\; t \le 24$。同时排水开始后时间不能倒流：$t \ge 0$。综合得 $0 \le t \le 24$ 分钟。
"Restriction on $t$" wants the physical boundary, not just $t \ge 0$."对 $t$ 的限制"要的是物理边界，不仅仅是 $t \ge 0$。 Provincial markers reward students who recognise that a linear model of a physical quantity has two domain boundaries: a start time (often $t = 0$) and a depletion time (set the modelled quantity to zero and solve). Writing only "$t \ge 0$" leaves the upper-bound mark on the table, the pool can't keep draining once it is empty at $t = 24$. State both endpoints explicitly with units.省考阅卷人乐于奖励能看出"物理量的线性模型有两个定义域边界"的学生：一个起始时刻（通常 $t = 0$）和一个耗尽时刻（令被建模量为零再求解）。只写 "$t \ge 0$" 会丢掉上界这一分，因为 $t = 24$ 时池子已空，不可能继续排水。两个端点都要明确写出并带上单位。
Q4MEDIUM中 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §2 Vertical Line竖直线 · HSF-IF.C.7a [3 marks][3 分]

Equation of the line through $(5, -8)$ and $(5, 17)$.求过点 $(5, -8)$ 与 $(5, 17)$ 的直线方程。

Answer:答案：  (C)  $x = 5$

(a) Recognise the vertical-line pattern识别竖直线特征 M1·A1

Both given points share the same $x$-coordinate, $x = 5$. Attempting the slope formula gives两点的 $x$ 坐标相同，都是 $x = 5$。若硬套斜率公式： $$ m \;=\; \frac{17 - (-8)}{5 - 5} \;=\; \frac{25}{0}, $$ which is undefined. A line with undefined slope is vertical, and a vertical line consists of every point whose $x$-coordinate equals the common value.分母为零、斜率不存在。斜率不存在的直线为竖直线，由所有 $x$ 坐标都等于该共同值的点组成。

(b) Write the equation写出方程 A1

Hence $x = 5$, option (C).故 $x = 5$，选 (C)。
Why the distractors fail.干扰项分析。
(A) $y = 5$: a horizontal line at height 5, students confuse "constant $x$" with "constant $y$".这是过 $y = 5$ 的水平线；学生把"$x$ 恒等"误作"$y$ 恒等"。
(B) $y = -8$: uses only one point and writes a horizontal line through it.只用了一个点，写出过该点的水平线。
(D) $y = \tfrac{25}{2} x$: reads $\Delta y / \Delta x = 25 / 2$ from the two points, but $\Delta x = 0$, not $2$. This distractor invites the student who tries to "force" the slope formula.从两点读出 $\Delta y / \Delta x = 25 / 2$，但 $\Delta x = 0$ 而非 $2$。此干扰项专门"诱惑"那些硬套斜率公式的学生。
A vertical line is not a function.竖直线不是函数。 The single most-tested vertical-line trap on the SAT: students see two points and reach for $y = mx + b$, which fails because $m$ is undefined. The honest move is to inspect the $x$-coordinates first: if they match, the line is $x = (\text{that value})$ and you are done. The same diagnostic works for horizontal lines: matching $y$-coordinates $\Rightarrow$ $y = (\text{that value})$. Train the eye to compare coordinates before computing.SAT 上最常考的竖直线陷阱：学生一看到两点就伸手去写 $y = mx + b$，结果因 $m$ 不存在而翻车。正确做法是先比较 $x$ 坐标：若相同，直线即 $x = (\text{该值})$，秒答。同一招也用于水平线：$y$ 坐标相同 $\Rightarrow$ $y = (\text{该值})$。养成下笔前先比坐标的习惯。
Q5MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §3 Graphing作图 · FMPC10 linear functions一次函数 [5 marks][5 分]

$3x + 4y = 12$. (a) Intercepts. (b) Sketch with intercepts and one lattice point. (c) Slope and sign.$3x + 4y = 12$。(a) 求截距。(b) 画图，标出截距与另一格点。(c) 写出斜率及其符号。

Answer:答案：  (a) $x$-int $(4, 0)$, $y$-int $(0, 3)$$x$ 轴截距 $(4, 0)$，$y$ 轴截距 $(0, 3)$  ·  (b) line through $(4,0), (0,3), (-4, 6)$过 $(4,0)$、$(0,3)$、$(-4, 6)$ 的直线  ·  (c) $m = -\tfrac{3}{4}$, negative$m = -\tfrac{3}{4}$，为负

(a) Intercepts截距 A1·A1

$y$-intercept: set $x = 0 \Rightarrow 4y = 12 \Rightarrow y = 3$. So $(0, 3)$.$y$ 轴截距：令 $x = 0 \Rightarrow 4y = 12 \Rightarrow y = 3$，得 $(0, 3)$。
$x$-intercept: set $y = 0 \Rightarrow 3x = 12 \Rightarrow x = 4$. So $(4, 0)$.$x$ 轴截距：令 $y = 0 \Rightarrow 3x = 12 \Rightarrow x = 4$，得 $(4, 0)$。

(b) Sketch with an extra lattice point作图并补一个格点 M1·A1

A convenient lattice point: pick $x = -4$, giving $3(-4) + 4y = 12 \Rightarrow 4y = 24 \Rightarrow y = 6$. So $(-4, 6)$ lies on the line. Plot all three points $(4, 0)$, $(0, 3)$, $(-4, 6)$, label them, and draw a straight line through them. The line falls from upper-left to lower-right.方便的格点：取 $x = -4$，则 $3(-4) + 4y = 12 \Rightarrow 4y = 24 \Rightarrow y = 6$，故 $(-4, 6)$ 在直线上。画出三点 $(4, 0)$、$(0, 3)$、$(-4, 6)$，标注后连成直线。该线从左上向右下倾斜。

(c) Slope and sign斜率与符号 A1

Using the two intercepts:用两截距计算： $$ m \;=\; \frac{3 - 0}{0 - 4} \;=\; -\tfrac{3}{4}. $$ Slope is negative. (Cross-check: rearrange $3x + 4y = 12 \Rightarrow y = -\tfrac{3}{4}\,x + 3$, same slope.)斜率为负。（交叉验证：把 $3x + 4y = 12$ 化为 $y = -\tfrac{3}{4}\,x + 3$，斜率一致。）
Intercept method beats slope-intercept for sketching.作图时，截距法胜过斜截式。 When a line is in standard form $Ax + By = C$, the fastest route to a sketch is the intercept method: set the other variable to zero twice. This gives you two well-defined lattice points (if $A$, $B$, $C$ behave) and a line in three pen strokes. Converting to $y = mx + b$ first is wasted work for a sketch. The third lattice point is a check, if it doesn't fall on the line drawn through the intercepts, your arithmetic is wrong. Provincial markers look for that third point as evidence the student didn't just trace two dots.当直线为标准式 $Ax + By = C$ 时，最快的作图法就是截距法：分别令另一变量为零两次。若 $A$、$B$、$C$ 取整，立即得到两个清晰的格点，三笔画完。先转 $y = mx + b$ 反而是绕远路。第三个格点是核对工具——若它不落在由两截距连成的直线上，说明算错了。省考阅卷人正是把这第三个点视作"学生不是只描了两个点"的证据。
PART II  ·  EXTENDED RESPONSE · SOLUTIONS第二部分  ·  简答题 · 详解AP-feeder FRQ + honors · 35 marksAP 衔接简答题 + 荣誉级 · 共 35 分

Section B · Worked SolutionsB 部分 · 详细解答

Q6MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §4 Parallel & Perpendicular平行与垂直 · HSF-LE.A.2 [7 marks][7 分]

$\ell_{0}: 2x + 5y = 10$. (a) Slope-intercept form and $m_{0}$. (b) Parallel line through $(-3, 1)$. (c) Perpendicular line through $(4, -2)$.$\ell_{0}: 2x + 5y = 10$。(a) 化为斜截式并求 $m_{0}$。(b) 求过 $(-3, 1)$ 的平行线。(c) 求过 $(4, -2)$ 的垂直线。

Answer:答案：  (a) $y = -\tfrac{2}{5}x + 2$, $m_{0} = -\tfrac{2}{5}$  ·  (b) $y = -\tfrac{2}{5}x - \tfrac{1}{5}$  ·  (c) $y = \tfrac{5}{2}x - 12$

(a) Slope-intercept form of $\ell_{0}$$\ell_{0}$ 的斜截式 M1·A1

From $2x + 5y = 10$, isolate $y$:由 $2x + 5y = 10$ 分离 $y$： $$ 5y \;=\; -2x + 10 \;\Longrightarrow\; y \;=\; -\tfrac{2}{5}\,x + 2. $$ So $m_{0} = -\tfrac{2}{5}$ and the $y$-intercept is $2$.故 $m_{0} = -\tfrac{2}{5}$，$y$ 轴截距为 $2$。

(b) Parallel through $(-3, 1)$过 $(-3, 1)$ 的平行线 M1·A1

Parallel $\Rightarrow$ same slope: $m = -\tfrac{2}{5}$. Use point-slope through $(-3, 1)$:平行 $\Rightarrow$ 斜率相同：$m = -\tfrac{2}{5}$。用点斜式过 $(-3, 1)$： $$ y - 1 \;=\; -\tfrac{2}{5}\,(x - (-3)) \;=\; -\tfrac{2}{5}\,x - \tfrac{6}{5}. $$ Hence $y = -\tfrac{2}{5}\,x - \tfrac{6}{5} + 1 = -\tfrac{2}{5}\,x - \tfrac{1}{5}$.故 $y = -\tfrac{2}{5}\,x - \tfrac{6}{5} + 1 = -\tfrac{2}{5}\,x - \tfrac{1}{5}$。

(c) Perpendicular through $(4, -2)$过 $(4, -2)$ 的垂直线 M1·A1·A1

Perpendicular slope is the negative reciprocal of $m_{0}$:垂直方向的斜率为 $m_{0}$ 的负倒数： $$ m_{\perp} \;=\; -\frac{1}{m_{0}} \;=\; -\frac{1}{-2/5} \;=\; \tfrac{5}{2}. $$ Point-slope through $(4, -2)$:用点斜式过 $(4, -2)$： $$ y - (-2) \;=\; \tfrac{5}{2}\,(x - 4) \;\Longrightarrow\; y + 2 \;=\; \tfrac{5}{2}\,x - 10. $$ So $y = \tfrac{5}{2}\,x - 12$.故 $y = \tfrac{5}{2}\,x - 12$。
Convert before reading; reciprocate and flip sign for perpendiculars.先转化再读数；垂直方向要同时取倒数与变号。 Two locked moves on every parallel/perpendicular question: (1) Reading the slope of $2x + 5y = 10$ as "$2$" is the most-missed-by-one-mark error in HSF-LE.A.2; only $y = mx + b$ form gives the slope directly. (2) "Negative reciprocal" means both sign flips and reciprocal, slope $-\tfrac{2}{5}$ has perpendicular slope $+\tfrac{5}{2}$, not $\tfrac{2}{5}$ and not $-\tfrac{5}{2}$. Getting the sign right separates the A1 from the wrong-line-family error.每道平行/垂直题都要锁住两步：(1) 把 $2x + 5y = 10$ 的斜率读成 "$2$" 是 HSF-LE.A.2 中最常丢的一分；只有 $y = mx + b$ 形式才能直接读出斜率。(2) "负倒数"是既变号又取倒数：斜率 $-\tfrac{2}{5}$ 对应的垂直斜率是 $+\tfrac{5}{2}$，而不是 $\tfrac{2}{5}$ 也不是 $-\tfrac{5}{2}$。把符号弄对，是拿到 A1 与"答错直线族"之间的分水岭。
Q7MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §4 Equations from a Table由数值表写方程 · MCR3U A1.2 / A1.7 [8 marks][8 分]

Table $(x, y)$: $(-2, 11), (1, 5), (4, -1), (7, -7), (10, -13)$. (a) First differences. (b) $y = mx + b$. (c) $f(20)$. (d) $f^{-1}(x)$.数值表 $(x, y)$：$(-2, 11), (1, 5), (4, -1), (7, -7), (10, -13)$。(a) 一阶差分。(b) $y = mx + b$。(c) $f(20)$。(d) $f^{-1}(x)$。

Answer:答案：  (a) $\Delta y = -6$ constant for $\Delta x = 3$$\Delta x = 3$ 时 $\Delta y = -6$ 恒定  ·  (b) $y = -2x + 7$  ·  (c) $f(20) = -33$  ·  (d) $f^{-1}(x) = -\tfrac{1}{2}\,x + \tfrac{7}{2}$

(a) First differences一阶差分 M1·A1

With $\Delta x = 3$ between consecutive entries, the $y$-differences are相邻 $x$ 间隔均为 $\Delta x = 3$，对应 $y$ 的差分为 $$ 5 - 11 = -6, \quad -1 - 5 = -6, \quad -7 - (-1) = -6, \quad -13 - (-7) = -6. $$ The first differences are constant at $-6$, so the relation is linear.一阶差分恒为 $-6$，故关系为线性。

(b) Equation in $y = mx + b$ form写成 $y = mx + b$ 形式 M1·A1·A1

Slope $m = \dfrac{\Delta y}{\Delta x} = \dfrac{-6}{3} = -2$. To find $b$, substitute any point, e.g. $(-2, 11)$:斜率 $m = \dfrac{\Delta y}{\Delta x} = \dfrac{-6}{3} = -2$。代入任一点如 $(-2, 11)$ 求 $b$： $$ 11 \;=\; -2(-2) + b \;=\; 4 + b \;\Longrightarrow\; b \;=\; 7. $$ Hence $f(x) = -2x + 7$. (Verify against another point: $f(10) = -20 + 7 = -13$. $\checkmark$)故 $f(x) = -2x + 7$。（用另一点验证：$f(10) = -20 + 7 = -13$，$\checkmark$）

(c) Evaluate $f(20)$求 $f(20)$ A1

$f(20) = -2(20) + 7 = -40 + 7 = -33$.

(d) Inverse function反函数 M1·A1

Start from $y = -2x + 7$. Swap $x \leftrightarrow y$: $\;x = -2y + 7$. Solve for $y$:由 $y = -2x + 7$ 出发，交换 $x \leftrightarrow y$：$\;x = -2y + 7$，解出 $y$： $$ 2y \;=\; 7 - x \;\Longrightarrow\; y \;=\; \tfrac{7 - x}{2} \;=\; -\tfrac{1}{2}\,x + \tfrac{7}{2}. $$ So $f^{-1}(x) = -\tfrac{1}{2}\,x + \tfrac{7}{2}$.故 $f^{-1}(x) = -\tfrac{1}{2}\,x + \tfrac{7}{2}$。
Slope of the inverse $=$ reciprocal of the slope, sign-preserved.反函数的斜率 $=$ 原斜率的倒数，符号不变。 A line $f$ with slope $m$ has inverse $f^{-1}$ with slope $1/m$ (not $-1/m$). Here $m = -2 \Rightarrow$ inverse slope $-\tfrac{1}{2}$, both negative. The negative-reciprocal rule is for perpendicular lines (Q6), not inverses. A line and its inverse are reflections in $y = x$, so a line of slope $-2$ reflects to a line of slope $-\tfrac{1}{2}$, same sign, reciprocal magnitude. Verify by checking that $f(f^{-1}(x)) = -2 \cdot (-\tfrac{1}{2}x + \tfrac{7}{2}) + 7 = x - 7 + 7 = x$. $\checkmark$斜率为 $m$ 的直线 $f$，其反函数 $f^{-1}$ 的斜率为 $1/m$（而非 $-1/m$）。此处 $m = -2 \Rightarrow$ 反函数斜率为 $-\tfrac{1}{2}$，二者同号。负倒数法则适用于垂直直线（见 Q6），并非反函数。一条直线与其反函数关于 $y = x$ 对称，所以斜率 $-2$ 的直线反射后斜率为 $-\tfrac{1}{2}$，符号相同、绝对值互为倒数。验证：$f(f^{-1}(x)) = -2 \cdot (-\tfrac{1}{2}x + \tfrac{7}{2}) + 7 = x - 7 + 7 = x$，$\checkmark$。
Q8HARD难 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §5 Linear Systems线性方程组 · HSA-REI.C.5 / C.6 [9 marks][9 分]

System $3x + 2y = 16$, $5x - 4y = 1$. (a) Elimination. (b) Substitution check. (c) Why row operations preserve solutions. (d) Geometric meaning.方程组 $3x + 2y = 16$、$5x - 4y = 1$。(a) 消元法。(b) 代入法验证。(c) 为何行变换保持解集。(d) 几何意义。

Answer:答案：  (a), (b) $(x, y) = \left(3, \tfrac{7}{2}\right)$  ·  (d) two lines intersect at a single point两条直线相交于一点

(a) Elimination消元法 M1·A1·A1

Multiply the first equation by $2$ to align the $y$-coefficient:将第一个方程乘以 $2$，使 $y$ 的系数对齐： $$ R_{1} \to 2 R_{1}: \quad 6x + 4y \;=\; 32. $$ Add to $R_{2}$ (the $y$-terms cancel):与 $R_{2}$ 相加（$y$ 项相消）： $$ R_{2} + (2R_{1}): \quad (5x - 4y) + (6x + 4y) \;=\; 1 + 32 \;\Longrightarrow\; 11x \;=\; 33 \;\Longrightarrow\; x \;=\; 3. $$ Back-substitute into $3x + 2y = 16$: $9 + 2y = 16 \Rightarrow 2y = 7 \Rightarrow y = \tfrac{7}{2}$. So $(x, y) = (3, \tfrac{7}{2})$.回代入 $3x + 2y = 16$：$9 + 2y = 16 \Rightarrow 2y = 7 \Rightarrow y = \tfrac{7}{2}$。故 $(x, y) = (3, \tfrac{7}{2})$。

(b) Substitution cross-check代入法交叉验证 M1·A1·A1

From $3x + 2y = 16$, isolate $y$: $y = \dfrac{16 - 3x}{2}$. Substitute into $5x - 4y = 1$:由 $3x + 2y = 16$ 解出 $y = \dfrac{16 - 3x}{2}$，代入 $5x - 4y = 1$： $$ 5x - 4 \cdot \frac{16 - 3x}{2} \;=\; 1 \;\Longrightarrow\; 5x - 2(16 - 3x) \;=\; 1 \;\Longrightarrow\; 5x - 32 + 6x \;=\; 1. $$ Hence $11x = 33$, $x = 3$, $y = \dfrac{16 - 9}{2} = \tfrac{7}{2}$. Both methods give the same ordered pair $(3, \tfrac{7}{2})$.故 $11x = 33$，$x = 3$，$y = \dfrac{16 - 9}{2} = \tfrac{7}{2}$。两法所得有序对相同，均为 $(3, \tfrac{7}{2})$。

(c) Why row operations preserve the solution set行变换为何保持解集不变 M1·R1

An elementary row operation of the form $R_{i} \to R_{i} + k R_{j}$ replaces equation $i$ with the sum of equation $i$ and $k$ times equation $j$. If $(x, y)$ satisfies both original equations, it satisfies the new equation $i$ (a true equality plus $k$ times another true equality is still true). The operation is reversible: applying $R_{i} \to R_{i} - k R_{j}$ to the new system returns the original, so any solution of the new system also solves the original. Hence the two systems have identical solution sets, this is exactly the substance of HSA-REI.C.5.形如 $R_{i} \to R_{i} + k R_{j}$ 的初等行变换，把方程 $i$ 替换为方程 $i$ 加上 $k$ 倍方程 $j$。若 $(x, y)$ 同时满足两个原方程，则它也满足新方程 $i$（一个成立的等式加上另一个成立等式的 $k$ 倍，仍然成立）。该操作可逆：对新方程组施加 $R_{i} \to R_{i} - k R_{j}$ 即可还原。故新系统的任一解也满足原系统。两系统解集相同——这正是 HSA-REI.C.5 的实质。

(d) Geometric meaning几何意义 A1

The two equations are lines in the $xy$-plane with different slopes ($-\tfrac{3}{2}$ and $\tfrac{5}{4}$), so they are not parallel and meet at exactly one point, namely $(3, \tfrac{7}{2})$. The system is consistent and independent.两方程是 $xy$ 平面上的两条直线，斜率不同（分别为 $-\tfrac{3}{2}$ 和 $\tfrac{5}{4}$），故不平行，恰交于一点 $(3, \tfrac{7}{2})$。该方程组相容且独立（有唯一解）。
Picking the elimination multiplier: look at the LCM of the column you want to kill.挑选消元乘数：盯准要消去那一列系数的最小公倍数。 The $y$-column has coefficients $+2$ and $-4$; their LCM is $4$. Multiplying $R_{1}$ by $2$ gives $+4$, which cancels $-4$ in one addition, no second multiplication needed, no sign hunt. Less obvious systems benefit from explicitly recording the row operation (e.g., $R_{2} \to R_{2} + 2 R_{1}$) so partial credit survives an arithmetic slip later. AP graders give an M1 for the row operation even when the A1 for the final answer goes missing.$y$ 列系数为 $+2$ 与 $-4$，最小公倍数为 $4$。$R_{1}$ 乘以 $2$ 得 $+4$，一次加法即与 $-4$ 相消，无需第二步相乘，也免去搜索符号的麻烦。在结构不那么直观的方程组里，明确记录所用行变换（如 $R_{2} \to R_{2} + 2 R_{1}$），后续算错时还能保住部分分数。AP 阅卷人即便扣掉最终答案的 A1，也会给行变换那一个 M1。
Q9HARD难Honors荣誉级 🇺🇸 US美 🇨🇦 BC卑 🇨🇦 ON安 AP-feeder FRQAP 衔接简答题 §7 Matrix Row-Reduction矩阵行简化 · HSA-REI.C.8/9 (+) / BC PC12 / ON MCV4U [11 marks][11 分]

$3 \times 3$ system: $x + 2y + z = 9$, $2x - y + z = 3$, $x + y - z = 0$. (a) Augmented matrix. (b) Row-reduce. (c) Back-substitute. (d) Verify.$3 \times 3$ 方程组：$x + 2y + z = 9$、$2x - y + z = 3$、$x + y - z = 0$。(a) 增广矩阵。(b) 行简化。(c) 回代。(d) 验证。

Answer:答案：  (a) $\left[\begin{smallmatrix}1 & 2 & 1 & \big| & 9 \\ 2 & -1 & 1 & \big| & 3 \\ 1 & 1 & -1 & \big| & 0 \end{smallmatrix}\right]$  ·  (c) $(x, y, z) = \left(1, \tfrac{7}{3}, \tfrac{10}{3}\right)$

(a) Augmented matrix增广矩阵 A1

$$ [A \,|\, \mathbf{b}] \;=\; \left[\begin{array}{ccc|c} 1 & 2 & 1 & 9 \\ 2 & -1 & 1 & 3 \\ 1 & 1 & -1 & 0 \end{array}\right]. $$

(b) Row-reduce to row-echelon form行简化为行阶梯形 M1·A1·A1·A1·R1

Step 1.第一步。 Clear column $1$ below the pivot using $R_{1}$:用 $R_{1}$ 清除第 $1$ 列主元下方的元素： $$ R_{2} \to R_{2} - 2 R_{1}: \quad (2, -1, 1, 3) - 2(1, 2, 1, 9) \;=\; (0, -5, -1, -15). $$ $$ R_{3} \to R_{3} - R_{1}: \quad (1, 1, -1, 0) - (1, 2, 1, 9) \;=\; (0, -1, -2, -9). $$ State of the matrix:此时矩阵： $$ \left[\begin{array}{ccc|c} 1 & 2 & 1 & 9 \\ 0 & -5 & -1 & -15 \\ 0 & -1 & -2 & -9 \end{array}\right]. $$ Step 2.第二步。 Clear column $2$ below the pivot in $R_{2}$ (pivot is $-5$). Use $R_{3} \to R_{3} - \tfrac{1}{5} R_{2}$:清除 $R_{2}$ 主元（即 $-5$）下方第 $2$ 列。施加 $R_{3} \to R_{3} - \tfrac{1}{5} R_{2}$： $$ R_{3} \to R_{3} - \tfrac{1}{5} R_{2}: \quad (0, -1, -2, -9) - \tfrac{1}{5}(0, -5, -1, -15). $$ Component-by-component: $-1 - \tfrac{1}{5}(-5) = -1 + 1 = 0$; $-2 - \tfrac{1}{5}(-1) = -2 + \tfrac{1}{5} = -\tfrac{9}{5}$; $-9 - \tfrac{1}{5}(-15) = -9 + 3 = -6$. So逐项计算：$-1 - \tfrac{1}{5}(-5) = -1 + 1 = 0$；$-2 - \tfrac{1}{5}(-1) = -2 + \tfrac{1}{5} = -\tfrac{9}{5}$；$-9 - \tfrac{1}{5}(-15) = -9 + 3 = -6$。故 $$ R_{3} \;=\; \bigl(0, 0, -\tfrac{9}{5}, -6\bigr). $$ Step 3.第三步。 (Optional) scale $R_{3}$ to make the pivot $1$: $R_{3} \to -\tfrac{5}{9} R_{3} = (0, 0, 1, \tfrac{10}{3})$. Row-echelon form:（可选）将 $R_{3}$ 缩放，使主元为 $1$：$R_{3} \to -\tfrac{5}{9} R_{3} = (0, 0, 1, \tfrac{10}{3})$。行阶梯形： $$ \left[\begin{array}{ccc|c} 1 & 2 & 1 & 9 \\ 0 & -5 & -1 & -15 \\ 0 & 0 & 1 & \tfrac{10}{3} \end{array}\right]. \quad \text{AG} $$

(c) Back-substitute回代 M1·A1·A1

From $R_{3}$: $\;z = \tfrac{10}{3}$.由 $R_{3}$：$\;z = \tfrac{10}{3}$。
From $R_{2}$: $-5y - z = -15 \Rightarrow -5y = -15 + \tfrac{10}{3} = -\tfrac{45}{3} + \tfrac{10}{3} = -\tfrac{35}{3} \Rightarrow y = \tfrac{7}{3}$.由 $R_{2}$：$-5y - z = -15 \Rightarrow -5y = -15 + \tfrac{10}{3} = -\tfrac{45}{3} + \tfrac{10}{3} = -\tfrac{35}{3} \Rightarrow y = \tfrac{7}{3}$。
From $R_{1}$: $x + 2y + z = 9 \Rightarrow x = 9 - 2 \cdot \tfrac{7}{3} - \tfrac{10}{3} = 9 - \tfrac{14}{3} - \tfrac{10}{3} = 9 - \tfrac{24}{3} = 9 - 8 = 1$.由 $R_{1}$：$x + 2y + z = 9 \Rightarrow x = 9 - 2 \cdot \tfrac{7}{3} - \tfrac{10}{3} = 9 - \tfrac{14}{3} - \tfrac{10}{3} = 9 - \tfrac{24}{3} = 9 - 8 = 1$。
Hence $(x, y, z) = \left(1, \tfrac{7}{3}, \tfrac{10}{3}\right)$.故 $(x, y, z) = \left(1, \tfrac{7}{3}, \tfrac{10}{3}\right)$。

(d) Verify in each original equation代回三个原方程验证 A1·A1

  • $x + 2y + z = 1 + \tfrac{14}{3} + \tfrac{10}{3} = 1 + \tfrac{24}{3} = 1 + 8 = 9$. $\checkmark$
  • $2x - y + z = 2 - \tfrac{7}{3} + \tfrac{10}{3} = 2 + \tfrac{3}{3} = 2 + 1 = 3$. $\checkmark$
  • $x + y - z = 1 + \tfrac{7}{3} - \tfrac{10}{3} = 1 - \tfrac{3}{3} = 1 - 1 = 0$. $\checkmark$
Reading a $3 \times 3$ system off the row-reduced matrix.从行简化矩阵读出 $3 \times 3$ 方程组的几何含义。 A $3 \times 3$ linear system has exactly four geometric outcomes, all visible in the rref of the augmented matrix. (1) Unique point, rref is the $3 \times 3$ identity with a column of constants on the right, three non-zero pivots, no zero rows; three planes meet at a single point (this question). (2) Line of solutions, exactly one zero row $[0\,0\,0\,|\,0]$, two pivots; the three planes share a common line (one free parameter). (3) Plane of solutions, two zero rows $[0\,0\,0\,|\,0]$, one pivot; the three "planes" are actually the same plane up to scaling. (4) No solution, a row of the form $[0\,0\,0\,|\,k]$ with $k \ne 0$; some pair of planes is parallel or the three planes form a triangular prism with no common point. Row-reduce first, then classify. The classification rule is robust to which order you perform the operations in, because rref is unique. This is the bridge from BC PC12 / ON MCV4U honors work into first-year linear algebra (rank-nullity).$3 \times 3$ 线性方程组恰有四种几何结果，全部可从增广矩阵的最简行阶梯形（rref）读出。(1) 唯一点：rref 是 $3 \times 3$ 单位阵加一列常数，三个非零主元、无零行；三个平面交于一点（本题情形）。(2) 解为一条直线：恰有一个零行 $[0\,0\,0\,|\,0]$，两个主元；三个平面共有一条直线（一个自由参数）。(3) 解为一个平面：两个零行 $[0\,0\,0\,|\,0]$，一个主元；三个"平面"实为同一平面的伸缩。(4) 无解：出现形如 $[0\,0\,0\,|\,k]$（$k \ne 0$）的行；某对平面平行，或三平面围成三棱柱而无公共点。先做行简化，再分类。无论以何种顺序作行变换，因 rref 唯一，分类法则始终稳健。这是 BC PC12 / ON MCV4U 荣誉级内容通往大一线性代数（秩-零度定理）的桥梁。
PART III  ·  MODELING / APPLIED · SOLUTIONS第三部分  ·  建模与应用 · 详解Universal · 28 marks通用题型 · 共 28 分

Section C · Modeling and ApplicationsC 部分 · 建模与应用

Q10MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §6 Break-even盈亏平衡 · HSF-LE.B.5 [8 marks][8 分]

T-shirt sale. Cost $C(n) = 90 + 6.50 n$; revenue $R(n) = 12 n$. (a) Set up. (b) Interpret slope and intercept of $C$. (c) Break-even $n^{*}$. (d) Smallest whole-shirt profit and profit value.T 恤销售。成本 $C(n) = 90 + 6.50 n$；收入 $R(n) = 12 n$。(a) 列式。(b) 解释 $C$ 的斜率与截距。(c) 求盈亏平衡 $n^{*}$。(d) 最少盈利件数及该件数下的利润。

Answer:答案：  (a) $C(n) = 90 + 6.50 n$, $R(n) = 12 n$  ·  (c) $n^{*} = \tfrac{180}{11} \approx 16.36$  ·  (d) $17$ shirts $\Rightarrow$ profit $\$3.50$$17$ 件 $\Rightarrow$ 利润 $\$3.50$

(a) Set up the cost and revenue functions建立成本与收入函数 A1·A1

Cost has a fixed setup of $\$90$ plus $\$6.50$ per shirt: $C(n) = 90 + 6.50 n$.成本含 $\$90$ 一次性开版费加每件 $\$6.50$：$C(n) = 90 + 6.50 n$。
Revenue: each shirt sells for $\$12$, so $R(n) = 12 n$. Both have domain $n \in \mathbb{Z}_{\ge 0}$ (whole shirts).收入：每件售价 $\$12$，故 $R(n) = 12 n$。两函数的定义域均为 $n \in \mathbb{Z}_{\ge 0}$（整件数）。

(b) Interpret slope and intercept of $C(n)$解释 $C(n)$ 的斜率与截距 A1·A1

Slope $6.50$:斜率 $6.50$： the marginal cost, the additional cost of producing one more shirt, in US dollars per shirt.即边际成本，每多生产一件 T 恤所增加的成本，单位为美元/件。
Intercept $90$:截距 $90$： the fixed cost, the cost when $n = 0$, i.e. the one-time setup fee owed to the printer even if zero shirts are produced. This is exactly the interpretation Common Core HSF-LE.B.5 rewards.即固定成本，是 $n = 0$ 时的成本，亦即即使一件也不印仍要支付给印刷商的一次性开版费。这正是共同核心 HSF-LE.B.5 所奖励的解读方式。

(c) Break-even $n^{*}$盈亏平衡 $n^{*}$ M1·A1

Set $R(n) = C(n)$:令 $R(n) = C(n)$： $$ 12 n \;=\; 90 + 6.50 n \;\Longrightarrow\; 5.50 n \;=\; 90 \;\Longrightarrow\; n^{*} \;=\; \frac{90}{5.50} \;=\; \frac{180}{11} \;\approx\; 16.36. $$

(d) Smallest whole-shirt profit最少盈利件数 M1·A1

Profit requires $R(n) > C(n)$, i.e. $n > 16.36$. The smallest whole number satisfying this is $n = 17$. Profit at $n = 17$:盈利要求 $R(n) > C(n)$，即 $n > 16.36$。满足条件的最小整数为 $n = 17$。此时利润： $$ P(17) \;=\; R(17) - C(17) \;=\; 12(17) - \bigl(90 + 6.50 \cdot 17\bigr) \;=\; 204 - (90 + 110.50) \;=\; 204 - 200.50 \;=\; \$3.50. $$
"Break-even" is a continuous answer; "smallest profitable" is a discrete answer."盈亏平衡"是连续答案；"最少盈利件数"是离散答案。 The break-even point $n^{*} = 16.36$ is the mathematically clean threshold, but no club can sell $0.36$ of a shirt. AP-feeder graders deduct when students answer "$n^{*} = 16$" or round break-even down, at $n = 16$, revenue is $\$192$ and cost is $\$194$, so the club is still losing $\$2$. The discrete answer is always $\lceil n^{*} \rceil$, the ceiling, not the floor. Get this dichotomy locked: the fractional break-even is the modelling output; the whole-shirt threshold is the contextual answer.盈亏平衡点 $n^{*} = 16.36$ 是数学上干净的临界值，但社团没法卖 $0.36$ 件 T 恤。AP 衔接阅卷人会扣分给"$n^{*} = 16$"或对盈亏平衡向下取整的答案——$n = 16$ 时，收入 $\$192$、成本 $\$194$，社团仍然亏 $\$2$。离散答案永远是 $\lceil n^{*} \rceil$（向上取整），而不是向下取整。把这一对要牢牢锁住：小数盈亏平衡是建模输出；整件阈值才是结合情境的答案。
Q11HARD难 🇨🇦 ON安 ON Provincial-style安大略省考风格 §6 Mixture System混合方程组 · MPM2D Analytic Geometry解析几何 [10 marks][10 分]

Mixture: $24$ L of $35\%$ alcohol from $20\%$ ($x$ L) and $50\%$ ($y$ L) stock. (a) Volume eq. (b) Alcohol eq. (c) Solve. (d) Verify percentage. (e) One-sentence answer.混合题：$24$ 升 $35\%$ 酒精由 $20\%$（$x$ 升）与 $50\%$（$y$ 升）库存配成。(a) 体积方程。(b) 酒精方程。(c) 求解。(d) 验证百分比。(e) 用一句话作答。

Answer:答案：  (a) $x + y = 24$  ·  (b) $0.20 x + 0.50 y = 8.4$  ·  (c) $(x, y) = (12, 12)$  ·  (d) mixture is $35\%$混合液为 $35\%$

(a) Volume constraint体积约束 A1

Total litres in the mixed solution equal the sum of the litres of each stock used:混合液总升数等于所用两种库存升数之和： $$ x + y \;=\; 24. $$

(b) Alcohol constraint酒精约束 M1·A1

Litres of alcohol in $20\%$ stock: $0.20\,x$. Litres of alcohol in $50\%$ stock: $0.50\,y$. The target mixture contains $0.35 \times 24 = 8.4$ litres of alcohol. So$20\%$ 库存中的纯酒精：$0.20\,x$ 升。$50\%$ 库存中的纯酒精：$0.50\,y$ 升。目标混合液含纯酒精 $0.35 \times 24 = 8.4$ 升。故 $$ 0.20\,x + 0.50\,y \;=\; 8.4. $$

(c) Solve the system解方程组 M1·A1·A1·A1

From (a), $x = 24 - y$. Substitute into (b):由 (a)，$x = 24 - y$，代入 (b)： $$ 0.20\,(24 - y) + 0.50\,y \;=\; 8.4 \;\Longrightarrow\; 4.8 - 0.20\,y + 0.50\,y \;=\; 8.4. $$ Combine $y$-terms: $0.30\,y = 8.4 - 4.8 = 3.6$, so合并 $y$ 项：$0.30\,y = 8.4 - 4.8 = 3.6$，故 $$ y \;=\; \frac{3.6}{0.30} \;=\; 12 \;\;\text{L}. $$ Back-substitute: $x = 24 - 12 = 12$ L. So $(x, y) = (12, 12)$.回代：$x = 24 - 12 = 12$ 升。故 $(x, y) = (12, 12)$。

(d) Verify the percentage验证百分比 M1·A1

Total alcohol $= 0.20(12) + 0.50(12) = 2.4 + 6.0 = 8.4$ L. Percentage:总酒精量 $= 0.20(12) + 0.50(12) = 2.4 + 6.0 = 8.4$ 升。浓度： $$ \frac{8.4}{24} \;=\; 0.35 \;=\; 35\%. \;\;\checkmark $$

(e) Concluding sentence总结句 A1

The technician must combine $12$ litres of the $20\%$ alcohol solution with $12$ litres of the $50\%$ alcohol solution to obtain $24$ litres of $35\%$ alcohol mixture.实验员需将 $12$ 升 $20\%$ 酒精溶液与 $12$ 升 $50\%$ 酒精溶液混合，即得 $24$ 升 $35\%$ 酒精混合液。
The "weighted-average shortcut", and why the algebra still matters."加权平均捷径"，以及为何代数过程仍然不可省。 The target $35\%$ sits exactly midway between $20\%$ and $50\%$: $35 = \tfrac{20 + 50}{2}$. By the linear weighted-average principle, when the target is the midpoint of the two stocks, the answer is always a $50 / 50$ split, half of each. So one could write the answer $x = y = 12$ in a single line. But provincial markers split the marks across setting up the system, solving it, and verifying the percentage; shortcut answers without the system earn 1 mark out of 10. The shortcut is a checking tool, not an answer. The deeper takeaway: in a two-stock mixture with target $t$, the ratio of stocks is $\dfrac{t - p_{2}}{p_{1} - t}$ where $p_{1}, p_{2}$ are the stock percentages, this is the "alligation rule" and generalises directly to weighted averages and centre-of-mass.目标 $35\%$ 恰位于 $20\%$ 与 $50\%$ 的中点：$35 = \tfrac{20 + 50}{2}$。由线性加权平均原理，当目标值是两库存的中点时，答案永远是 $50 / 50$ 对半，各取一半。所以一行就能写出 $x = y = 12$。但省考阅卷人把 10 分拆给列式、求解、验证百分比三段；只写捷径而不列方程组，10 分里只能拿 1 分。捷径是检验工具，不是答案。更深一层：两库存混合、目标值 $t$ 时，两库存的比例为 $\dfrac{t - p_{2}}{p_{1} - t}$（$p_{1}, p_{2}$ 为各库存百分比），这就是"十字交叉法"，可直接推广到加权平均与质心问题。
Q12MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §5 / §6 Rate System速率方程组 · FMPC10 systems方程组 [10 marks][10 分]

Two cyclists from the same trailhead. Maya: $18$ km/h from $t = 0$. Jin: $24$ km/h from $t = 0.5$. (a) $d_{M}(t)$. (b) $d_{J}(t)$ + restriction. (c) Catch-up time, $t$ and clock. (d) Sketch.两骑行者从同一入口出发。Maya：自 $t = 0$ 起 $18$ km/h。Jin：自 $t = 0.5$ 起 $24$ km/h。(a) $d_{M}(t)$。(b) $d_{J}(t)$ 及定义域限制。(c) 追上时刻（$t$ 与钟表时间）。(d) 作图。

Answer:答案：  (a) $d_{M}(t) = 18\,t$  ·  (b) $d_{J}(t) = 24\,t - 12$ for $t \ge 0.5$当 $t \ge 0.5$  ·  (c) $t = 2$ h, i.e. $11{:}00$ a.m.小时，即上午 $11{:}00$

(a) Maya's distance equationMaya 的距离方程 A1·A1

Maya leaves the trailhead at $t = 0$ at a constant $18$ km/h. With $d$ in km and $t$ in hours from $9{:}00$ a.m.:Maya 在 $t = 0$ 离开入口，恒速 $18$ km/h。设 $d$ 单位为公里，$t$ 单位为小时（自上午 $9{:}00$ 起算）： $$ d_{M}(t) \;=\; 18\,t, \quad t \ge 0. $$

(b) Jin's distance equation and restrictionJin 的距离方程与定义域限制 M1·A1·A1

Jin starts $30$ minutes ($0.5$ h) later. He has been riding for $(t - 0.5)$ hours at $24$ km/h, soJin 晚 $30$ 分钟（即 $0.5$ 小时）出发。他已骑行 $(t - 0.5)$ 小时，速度 $24$ km/h，故 $$ d_{J}(t) \;=\; 24\,(t - 0.5) \;=\; 24\,t - 12, \quad t \ge 0.5. $$ Restriction.定义域限制。 The equation only models Jin's position from the moment he leaves; for $t < 0.5$ he is still at the trailhead ($d = 0$), so the linear model $24t - 12$ (which gives negative distances) does not apply.该方程仅建模 Jin 自出发起的位置；当 $t < 0.5$ 时他仍在入口（$d = 0$），线性模型 $24t - 12$ 会给出负距离，不适用。

(c) Catch-up time追上时刻 M1·A1·A1

Catch-up means $d_{M}(t) = d_{J}(t)$:追上意味着 $d_{M}(t) = d_{J}(t)$： $$ 18\,t \;=\; 24\,t - 12 \;\Longrightarrow\; 12 \;=\; 6\,t \;\Longrightarrow\; t \;=\; 2 \;\;\text{hours}. $$ At $t = 2$, Maya has ridden $18 \cdot 2 = 36$ km and Jin has ridden $24 \cdot 1.5 = 36$ km. $\checkmark$$t = 2$ 时，Maya 骑行 $18 \cdot 2 = 36$ km，Jin 骑行 $24 \cdot 1.5 = 36$ km，$\checkmark$。
Clock time: $9{:}00$ a.m. $+ \;2$ h $= \mathbf{11{:}00}$ a.m.钟表时间：上午 $9{:}00$ $+ \;2$ 小时 $= \mathbf{11{:}00}$ 上午。

(d) Sketch on $(t, d)$ axes, $0 \le t \le 3$在 $(t, d)$ 坐标系上作图，$0 \le t \le 3$ M1·A1

Maya's line: through $(0, 0)$ and $(3, 54)$, slope $18$, drawn for all $t \in [0, 3]$.Maya 的直线：过 $(0, 0)$ 与 $(3, 54)$，斜率 $18$，区间 $t \in [0, 3]$ 全画。
Jin's line: through $(0.5, 0)$ and $(3, 60)$, slope $24$, drawn only for $t \in [0.5, 3]$.Jin 的直线：过 $(0.5, 0)$ 与 $(3, 60)$，斜率 $24$，仅画 $t \in [0.5, 3]$ 一段。
Mark the $t$-intercept of Jin's line at $(0.5, 0)$ (his start time) and the intersection point $(2, 36)$. Label each line "$d_{M}$" and "$d_{J}$". Jin's line is steeper and crosses Maya's at $t = 2$, then climbs above.标出 Jin 直线的 $t$ 轴截距 $(0.5, 0)$（他的出发时刻）以及两线交点 $(2, 36)$。两条线分别标注 "$d_{M}$" 与 "$d_{J}$"。Jin 的直线更陡，在 $t = 2$ 处与 Maya 的直线相交，之后高出。
The "head-start" structural pattern."先行一段"的结构性套路。 Any "B leaves later and goes faster, when does B catch A?" problem is a $2 \times 2$ system with two linear functions of the same $t$: $d_{A}(t) = v_{A}\,t$ and $d_{B}(t) = v_{B}\,(t - t_{0})$. Setting them equal gives a single linear equation in $t$ with closed form $t^{*} = \dfrac{v_{B} \, t_{0}}{v_{B} - v_{A}}$. For this question: $t^{*} = \dfrac{24 \cdot 0.5}{24 - 18} = \dfrac{12}{6} = 2$. This same template solves the train-relay, boat-current, plane-vs-jet, and the "second runner with a head start" SAT problems. The domain restriction on $d_{J}$, easy to forget, is the BC-provincial marker's favourite trap: writing $d_{J}(t) = 24t - 12$ without "$t \ge 0.5$" surrenders a clean A1.凡是"B 晚出发但更快，何时追上 A？"的题，都是同一 $t$ 的两个一次函数构成的 $2 \times 2$ 方程组：$d_{A}(t) = v_{A}\,t$ 与 $d_{B}(t) = v_{B}\,(t - t_{0})$。令两式相等，化为一元一次方程，闭式解为 $t^{*} = \dfrac{v_{B} \, t_{0}}{v_{B} - v_{A}}$。本题：$t^{*} = \dfrac{24 \cdot 0.5}{24 - 18} = \dfrac{12}{6} = 2$。同一模板可解火车接力、船与水流、飞机对飞机以及 SAT "第二名跑者先行一段" 类问题。$d_{J}$ 的定义域限制极易遗漏，是卑诗省考阅卷人最爱布置的陷阱：只写 $d_{J}(t) = 24t - 12$ 而不写 "$t \ge 0.5$"，干净的一个 A1 就此送出。