第 7 單元，第 1 課

Graphing Systems of Equations用圖解方程組

這課一眼看

Definitions定義

system of equations方程組（聯立方程式）: Two or more equations considered together; here two linear equations in x and y.
solution of a system方程組的解: An ordered pair that makes every equation true; on a graph, the point where the lines meet.
three cases: Lines cross: one solution; parallel: no solution; same line: infinitely many solutions.
classifying: Consistent = at least one solution (independent: one; dependent: infinitely many); inconsistent = none.

Formulas and Rules公式與性質

graphing method: Graph both lines, read the intersection, check it in both equations.
estimates估算: When lines cross between grid points, a graph gives only an estimate; the check shows it.
predicting from slopes: Different slopes: one solution; same slope, different intercepts: none; same slope and intercept: infinitely many.

Problem Types題型範例

1Definitions定義

ASystems of equations and their solutions方程組與它的解

A system of equations is a set of two or more equations that are considered together. In this module each system has two linear equations in $x$ and $y$, written with a brace, such as $\begin{cases} x + y = 6 \\ 2x - y = 3 \end{cases}$. A solution of a system is an ordered pair that makes every equation in the system true at the same time. For this system, $(3, 3)$ is a solution: $3 + 3 = 6$ and $2(3) - 3 = 3$.方程組是把兩個或更多個方程式放在一起考慮。本單元的方程組都是兩個 $x$、$y$ 的一次方程式，用大括號寫在一起，例如 $\begin{cases} x + y = 6 \\ 2x - y = 3 \end{cases}$。方程組的解是同時讓每一個方程式都成立的有序數對。以這個方程組來說，$(3, 3)$ 是解：$3 + 3 = 6$，而且 $2(3) - 3 = 3$。
Each linear equation has a line as its graph, and every point on that line is a solution of that one equation. A solution of the system must lie on both lines, so on a graph it is the point where the two lines meet.每個一次方程式的圖形都是一條直線，線上每一點都是那一個方程式的解。方程組的解必須同時落在兩條直線上，所以在圖上，它就是兩條直線的交點。

Checking ordered pairs in $x + y = 6$ and $2x - y = 3$把有序數對代入 $x + y = 6$ 與 $2x - y = 3$

Ordered pair有序數對	$x + y$$x + y$	$2x - y$$2x - y$	Solution of the system?是方程組的解嗎？
$(3, 3)$$(3, 3)$	$6$ ✓$6$ ✓	$3$ ✓$3$ ✓	yes是
$(4, 2)$$(4, 2)$	$6$ ✓$6$ ✓	$6$ ✗$6$ ✗	no不是
$(1, -1)$$(1, -1)$	$0$ ✗$0$ ✗	$3$ ✓$3$ ✓	no不是
$(0, 6)$$(0, 6)$	$6$ ✓$6$ ✓	$-6$ ✗$-6$ ✗	no不是

注意

Checking only one equation is not enough. $(4, 2)$ makes $x + y = 6$ true, but $2(4) - 2 = 6$, not $3$, so it is not a solution of the system. Substitute into every equation, and write the answer as an ordered pair with $x$ first.只檢查一個方程式是不夠的。$(4, 2)$ 會讓 $x + y = 6$ 成立，但 $2(4) - 2 = 6$，不是 $3$，所以它不是方程組的解。每一個方程式都要代入，答案寫成 $x$ 在前的有序數對。

BNumber of solutions and classifying systems解的個數與方程組的分類

Two lines in a plane can meet in three ways, so a system of two linear equations has one solution, no solution, or infinitely many solutions. If the lines cross at one point, there is exactly one solution. If the lines are parallel lines, they never meet and there is no solution. If the two equations describe the same line, every point on that line is a solution, so there are infinitely many.平面上的兩條直線只有三種相遇的方式，所以兩個一次方程式組成的方程組不是恰好一解、就是無解或無限多解。兩條線交於一點時，恰好一解；兩條線平行時，永遠不相交，無解；兩個方程式其實是同一條直線時，線上每一點都是解，有無限多解。
These cases have names. A consistent system has at least one solution; an inconsistent system has no solution. A consistent system with exactly one solution is an independent system: the two equations give different lines. A consistent system with infinitely many solutions is a dependent system: one equation can be turned into the other, so they describe the same line.這幾種情形各有名稱。至少有一個解的叫相容方程組；無解的叫不相容方程組。相容且恰好一解的叫獨立方程組，兩個方程式代表不同的直線；相容且有無限多解的叫相依方程組，其中一式可以化成另一式，所以它們代表同一條直線。

Three kinds of systems方程組的三種情形

Graph圖形	Number of solutions解的個數	Consistent?相容嗎？	Independent or dependent獨立或相依	Example例子
lines cross at one point兩線交於一點	one一個	consistent相容	independent獨立	$y = x + 1$, $y = -x + 5$: solution $(2, 3)$$y = x + 1$、$y = -x + 5$：解 $(2, 3)$
parallel lines兩線平行	none沒有	inconsistent不相容	——	$y = 2x + 1$, $y = 2x - 3$$y = 2x + 1$、$y = 2x - 3$
the same line同一條直線	infinitely many無限多個	consistent相容	dependent相依	$y = 3x - 2$, $6x - 2y = 4$$y = 3x - 2$、$6x - 2y = 4$

注意

"Dependent" does not mean "no solution"; it means the equations depend on each other, so the solutions never run out. Answers such as "infinitely many solutions" should still say which line: every point on $y = 3x - 2$.「相依」不是「無解」，而是兩個方程式彼此相依，所以解多到數不完。回答「無限多解」時，最好也說明是哪一條直線上的點，例如「$y = 3x - 2$ 上的每一點」。

2Formulas and Rules公式與性質

ASteps for solving by graphing用圖解方程組的步驟

To solve a system by graphing, graph both lines on the same coordinate plane, using slope-intercept form or the intercepts. Find the point where they cross and read its coordinates. Then substitute that point into both original equations; if both are true, it is the solution.用圖解方程組時，把兩條直線畫在同一個坐標平面上，可以用斜截式或截距來畫。找出兩線的交點並讀出坐標，再把這一點代回兩個原方程式；兩式都成立，它就是解。
A graph is only as exact as the drawing. When the lines cross between grid points, the graph gives an estimate, and the check will show that the estimate is not exact. Exact answers for such systems come from algebra, which the next lessons develop.圖解的準確度取決於畫得多準。兩條線交在格點之間時，圖只能給出估計值，代回驗算就會發現估計值不完全正確。這類方程組的精確解要靠代數方法，接下來幾課會介紹。

Solving $y = 2x - 1$ and $x + y = 5$ by graphing用圖解 $y = 2x - 1$ 與 $x + y = 5$

Step步驟	What to do做什麼	Work過程
1. Graph the first line1. 畫第一條線	Use slope-intercept form.用斜截式。	y-intercept $(0, -1)$, slope $2$y 截距 $(0, -1)$，斜率 $2$
2. Graph the second line2. 畫第二條線	Rewrite as $y = -x + 5$.改寫成 $y = -x + 5$。	y-intercept $(0, 5)$, slope $-1$y 截距 $(0, 5)$，斜率 $-1$
3. Find the intersection3. 找交點	Read where the lines cross.讀出兩線交在哪裡。	$(2, 3)$$(2, 3)$
4. Check4. 驗算	Substitute into both equations.代入兩個方程式。	$3 = 2(2) - 1$ ✓; $2 + 3 = 5$ ✓$3 = 2(2) - 1$ ✓；$2 + 3 = 5$ ✓
5. Answer5. 作答	Write the ordered pair.寫出有序數對。	$(2, 3)$$(2, 3)$

注意

Never skip the check. A small drawing error can make the lines appear to cross at the wrong grid point, and only substituting into both equations catches it.驗算絕對不能省。圖畫得稍微不準，兩條線看起來就可能交在錯的格點上，只有代回兩個方程式才抓得出來。

BPredicting the number of solutions from slopes and y-intercepts由斜率與 y 截距判斷解的個數

You can tell how many solutions a system has without graphing. Rewrite both equations in slope-intercept form $y = mx + b$ and compare. Different slope values: the lines cross once, so there is one solution. The same slope with different y-intercept values: the lines are parallel, so there is no solution. The same slope and the same y-intercept: the equations describe one line, so there are infinitely many solutions.不畫圖也能知道方程組有幾個解。把兩個方程式都改寫成斜截式 $y = mx + b$ 再比較：斜率不同，兩線交於一點，恰好一解；斜率相同、y 截距不同，兩線平行，無解；斜率和 y 截距都相同，兩式是同一條直線，有無限多解。

Predicting the number of solutions預測解的個數

System方程組	Slope-intercept forms斜截式	Slopes斜率	y-interceptsy 截距	Solutions解
$3x + y = 4$; $y = x - 2$$3x + y = 4$；$y = x - 2$	$y = -3x + 4$; $y = x - 2$$y = -3x + 4$；$y = x - 2$	$-3$, $1$$-3$、$1$	$4$, $-2$$4$、$-2$	one一個
$2x - y = 5$; $y = 2x + 1$$2x - y = 5$；$y = 2x + 1$	$y = 2x - 5$; $y = 2x + 1$$y = 2x - 5$；$y = 2x + 1$	$2$, $2$$2$、$2$	$-5$, $1$$-5$、$1$	none沒有
$x - 2y = 6$; $2x - 4y = 12$$x - 2y = 6$；$2x - 4y = 12$	$y = \frac{1}{2}x - 3$ for both兩式都是 $y = \frac{1}{2}x - 3$	$\frac{1}{2}$, $\frac{1}{2}$$\frac{1}{2}$、$\frac{1}{2}$	$-3$, $-3$$-3$、$-3$	infinitely many無限多個

注意

Compare slopes first. If the slopes are different, you are done: there is one solution, whatever the y-intercepts are. Vertical lines need their own check: $x = 2$ and $x = 5$ are parallel (no solution), while $x = 2$ and $y = 3$ cross at $(2, 3)$.先比較斜率。斜率不同就可以停了：不管 y 截距是多少，都恰好一解。鉛直線要另外看：$x = 2$ 和 $x = 5$ 互相平行（無解），而 $x = 2$ 和 $y = 3$ 交於 $(2, 3)$。

3Problem Types題型範例

ACheck whether an ordered pair is a solution判斷有序數對是不是方程組的解

Use this type when a question gives a system and one or more ordered pairs and asks which pairs are solutions. Substitute $x$ and $y$ into every equation. The pair is a solution only if every equation is true.當題目給一個方程組和一個或幾個有序數對，問哪些是解時，就是這種題型。把 $x$、$y$ 代入每一個方程式，每一式都成立才是解。

例題

Is $(-1, 4)$ a solution of $\begin{cases} 3x + y = 1 \\ x - 2y = -9 \end{cases}$?$(-1, 4)$ 是 $\begin{cases} 3x + y = 1 \\ x - 2y = -9 \end{cases}$ 的解嗎？

觀念提示

Substitute $x = -1$ and $y = 4$ into both equations, keeping the parentheses around the negative number.把 $x = -1$、$y = 4$ 代入兩個方程式，負數記得加括號。

關鍵算式

$3(-1) + 4 = 1$; $-1 - 2(4) = -9$$3(-1) + 4 = 1$；$-1 - 2(4) = -9$

完整詳解

First equation: $3(-1) + 4 = -3 + 4 = 1$ ✓.
Second equation: $-1 - 2(4) = -1 - 8 = -9$ ✓.
Both equations are true, so $(-1, 4)$ is a solution of the system.

第一式：$3(-1) + 4 = -3 + 4 = 1$ ✓。
第二式：$-1 - 2(4) = -1 - 8 = -9$ ✓。
兩式都成立，所以 $(-1, 4)$ 是方程組的解。

例題

Is $(5, 2)$ a solution of $\begin{cases} x - y = 3 \\ 2x + 3y = 14 \end{cases}$?$(5, 2)$ 是 $\begin{cases} x - y = 3 \\ 2x + 3y = 14 \end{cases}$ 的解嗎？

觀念提示

The first equation works, but that is only half the test. One false equation is enough to say no.第一式成立只是通過一半。只要有一式不成立，答案就是「不是」。

關鍵算式

$2(5) + 3(2) = 16 \ne 14$$2(5) + 3(2) = 16 \ne 14$

完整詳解

First equation: $5 - 2 = 3$ ✓.
Second equation: $2(5) + 3(2) = 10 + 6 = 16$, but the right side is $14$ ✗.
The second equation is false, so $(5, 2)$ is not a solution of the system. It lies on the first line only.

第一式：$5 - 2 = 3$ ✓。
第二式：$2(5) + 3(2) = 10 + 6 = 16$，但右邊是 $14$ ✗。
第二式不成立，所以 $(5, 2)$ 不是方程組的解，它只在第一條直線上。

自我檢查

Which of $(2, 1)$, $(1, 3)$, $(0, 5)$ is a solution of $\begin{cases} y = -2x + 5 \\ y = x + 2 \end{cases}$?$(2, 1)$、$(1, 3)$、$(0, 5)$ 哪一個是 $\begin{cases} y = -2x + 5 \\ y = x + 2 \end{cases}$ 的解？

看答案

$(2, 1)$: $-4 + 5 = 1$ ✓, but $2 + 2 = 4 \ne 1$ ✗.
$(1, 3)$: $-2 + 5 = 3$ ✓ and $1 + 2 = 3$ ✓, so it is the solution.
$(0, 5)$: $0 + 5 = 5$ ✓, but $0 + 2 = 2 \ne 5$ ✗.

$(2, 1)$：$-4 + 5 = 1$ ✓，但 $2 + 2 = 4 \ne 1$ ✗。
$(1, 3)$：$-2 + 5 = 3$ ✓，$1 + 2 = 3$ ✓，所以它是解。
$(0, 5)$：$0 + 5 = 5$ ✓，但 $0 + 2 = 2 \ne 5$ ✗。

BSolve a system by graphing用圖解方程組

Use this type when a question asks you to solve a system by graphing. Graph each line from its slope and y-intercept (or from its intercepts), read the intersection, and check it in both equations. Describe the graph by naming points on each line and the crossing point.當題目要你用圖解方程組時，就是這種題型。用斜率與 y 截距（或用兩個截距）畫出每條直線，讀出交點，再代回兩式驗算。描述圖形時，說出每條線經過的點和交點。

例題

Solve by graphing: $\begin{cases} y = x + 2 \\ y = -2x + 8 \end{cases}$.用圖解：$\begin{cases} y = x + 2 \\ y = -2x + 8 \end{cases}$。

觀念提示

Both equations are already in slope-intercept form. Step along each line from its y-intercept until the two lists of points share a point.兩式都已經是斜截式。從各自的 y 截距沿著斜率走，直到兩串點出現同一個點。

關鍵算式

$(0, 2) \to (1, 3) \to (2, 4)$; $(0, 8) \to (1, 6) \to (2, 4)$$(0, 2) \to (1, 3) \to (2, 4)$；$(0, 8) \to (1, 6) \to (2, 4)$

完整詳解

First line: y-intercept $(0, 2)$, slope $1$; it passes through $(1, 3)$ and $(2, 4)$.
Second line: y-intercept $(0, 8)$, slope $-2$; it passes through $(1, 6)$ and $(2, 4)$.
The lines cross at $(2, 4)$.
Check: $4 = 2 + 2$ ✓ and $4 = -2(2) + 8$ ✓. The solution is $(2, 4)$.

第一條線：y 截距 $(0, 2)$，斜率 $1$，經過 $(1, 3)$、$(2, 4)$。
第二條線：y 截距 $(0, 8)$，斜率 $-2$，經過 $(1, 6)$、$(2, 4)$。
兩線交於 $(2, 4)$。
驗算：$4 = 2 + 2$ ✓，$4 = -2(2) + 8$ ✓。解是 $(2, 4)$。

例題

Solve by graphing: $\begin{cases} x + 2y = 4 \\ x - y = 1 \end{cases}$.用圖解：$\begin{cases} x + 2y = 4 \\ x - y = 1 \end{cases}$。

觀念提示

In standard form, the intercepts are the quickest points: set $y = 0$ for the x-intercept and $x = 0$ for the y-intercept.標準式用截距最快：令 $y = 0$ 求 x 截距，令 $x = 0$ 求 y 截距。

關鍵算式

$x + 2y = 4$: $(4, 0), (0, 2)$; $x - y = 1$: $(1, 0), (0, -1)$$x + 2y = 4$：$(4, 0), (0, 2)$；$x - y = 1$：$(1, 0), (0, -1)$

完整詳解

First line (standard form, use intercepts): $(4, 0)$ and $(0, 2)$.
Second line: $(1, 0)$ and $(0, -1)$.
Drawn on one grid, the lines cross at $(2, 1)$.
Check: $2 + 2(1) = 4$ ✓ and $2 - 1 = 1$ ✓. The solution is $(2, 1)$.

第一條線（標準式，用截距畫）：$(4, 0)$ 和 $(0, 2)$。
第二條線：$(1, 0)$ 和 $(0, -1)$。
畫在同一張方格上，兩線交於 $(2, 1)$。
驗算：$2 + 2(1) = 4$ ✓，$2 - 1 = 1$ ✓。解是 $(2, 1)$。

自我檢查

Solve by graphing: $\begin{cases} y = 3x - 4 \\ y = -x + 4 \end{cases}$.用圖解：$\begin{cases} y = 3x - 4 \\ y = -x + 4 \end{cases}$。

看答案

First line: $(0, -4)$, $(1, -1)$, $(2, 2)$. Second line: $(0, 4)$, $(1, 3)$, $(2, 2)$.
They cross at $(2, 2)$. Check: $3(2) - 4 = 2$ ✓, $-2 + 4 = 2$ ✓.

第一條線：$(0, -4)$、$(1, -1)$、$(2, 2)$。第二條線：$(0, 4)$、$(1, 3)$、$(2, 2)$。
兩線交於 $(2, 2)$。驗算：$3(2) - 4 = 2$ ✓，$-2 + 4 = 2$ ✓。

CClassify a system without graphing不畫圖判斷方程組的類型

Use this type when a question asks how many solutions a system has, or whether it is consistent, inconsistent, independent or dependent. Rewrite each equation in slope-intercept form, compare slopes and y-intercepts, and then name the type.當題目問方程組有幾個解，或問它是相容、不相容、獨立還是相依時，就是這種題型。把每個方程式改寫成斜截式，比較斜率和 y 截距，再說出類型。

例題

Classify $\begin{cases} 4x - 2y = 6 \\ y = 2x - 3 \end{cases}$ and state the number of solutions.判斷 $\begin{cases} 4x - 2y = 6 \\ y = 2x - 3 \end{cases}$ 的類型，並說出解的個數。

觀念提示

Divide every term by $-2$, including the constant. Then compare the two equations side by side.每一項都要除以 $-2$，常數也是。然後把兩式並排比較。

關鍵算式

$4x - 2y = 6 \;\to\; y = 2x - 3$$4x - 2y = 6 \;\to\; y = 2x - 3$

完整詳解

Rewrite the first equation: $-2y = -4x + 6$, so $y = 2x - 3$.
Both equations are $y = 2x - 3$: same slope $2$, same y-intercept $-3$.
They describe the same line, so there are infinitely many solutions: every point on $y = 2x - 3$.
The system is consistent and dependent.

改寫第一式：$-2y = -4x + 6$，所以 $y = 2x - 3$。
兩式都是 $y = 2x - 3$：斜率都是 $2$，y 截距都是 $-3$。
兩式是同一條直線，所以有無限多解：$y = 2x - 3$ 上的每一點。
這是相容且相依的方程組。

例題

Classify $\begin{cases} x + 3y = 9 \\ y = -\frac{1}{3}x - 2 \end{cases}$ and state the number of solutions.判斷 $\begin{cases} x + 3y = 9 \\ y = -\frac{1}{3}x - 2 \end{cases}$ 的類型，並說出解的個數。

觀念提示

Same slope means the lines are either parallel or the same line. The y-intercepts decide which.斜率相同表示兩條線不是平行就是重合，由 y 截距決定是哪一種。

關鍵算式

$m = -\frac{1}{3}$ for both; $b = 3$ vs. $b = -2$兩式 $m = -\frac{1}{3}$；$b = 3$ 和 $b = -2$

完整詳解

Rewrite the first equation: $3y = -x + 9$, so $y = -\frac{1}{3}x + 3$.
Slopes: $-\frac{1}{3}$ and $-\frac{1}{3}$, the same. y-intercepts: $3$ and $-2$, different.
The lines are parallel, so there is no solution.
The system is inconsistent.

改寫第一式：$3y = -x + 9$，所以 $y = -\frac{1}{3}x + 3$。
斜率：$-\frac{1}{3}$ 和 $-\frac{1}{3}$，相同；y 截距：$3$ 和 $-2$，不同。
兩線平行，所以無解。
這是不相容方程組。

自我檢查

Classify each system. (a) $\begin{cases} y = 4x + 1 \\ y = -x + 1 \end{cases}$ (b) $\begin{cases} 2x + y = 3 \\ 6x + 3y = 9 \end{cases}$ (c) $\begin{cases} x = 4 \\ x = -1 \end{cases}$判斷每個方程組的類型。(a) $\begin{cases} y = 4x + 1 \\ y = -x + 1 \end{cases}$　(b) $\begin{cases} 2x + y = 3 \\ 6x + 3y = 9 \end{cases}$　(c) $\begin{cases} x = 4 \\ x = -1 \end{cases}$

看答案

(a) Slopes $4$ and $-1$ differ: one solution, consistent and independent (the solution is $(0, 1)$).
(b) Both become $y = -2x + 3$: infinitely many solutions, consistent and dependent.
(c) Two different vertical lines are parallel: no solution, inconsistent.

(a) 斜率 $4$ 和 $-1$ 不同：恰好一解，相容且獨立（解是 $(0, 1)$）。
(b) 兩式都化成 $y = -2x + 3$：無限多解，相容且相依。
(c) 兩條不同的鉛直線互相平行：無解，不相容。

DEstimate a solution from a graph or table and check it由圖形或表格估計解並驗算

Use this type when the lines cross between grid points, or when the equations are given as tables of values. On a graph, read an approximate crossing point and check it. In a table, look for an $x$ where the two $y$-values are equal; if the larger $y$-value switches from one equation to the other between two rows, the solution lies between those rows.當兩條線交在格點之間，或方程式用數值表給出時，就是這種題型。在圖上讀出大約的交點並驗算；在表格中找兩個 $y$ 值相等的 $x$，如果兩列之間「哪一式的 $y$ 比較大」換了過來，解就落在這兩列之間。

例題

The lines $y = 2x + 1$ and $y = -x + 3$ are graphed. They appear to cross near $(0.7, 2.3)$. Use a table to locate the crossing, then check the estimate and find the exact solution.畫出 $y = 2x + 1$ 和 $y = -x + 3$，兩線看起來交在 $(0.7, 2.3)$ 附近。用表格找出交點的位置，再驗算估計值並求出精確解。

觀念提示

A graph can only suggest a decimal estimate. When both equations are solved for $y$, setting the right sides equal gives the exact $x$.圖只能給出小數的估計值。兩式都已解出 $y$ 時，令兩個右邊相等就能得到精確的 $x$。

關鍵算式

$2x + 1 = -x + 3 \;\to\; x = \frac{2}{3}$$2x + 1 = -x + 3 \;\to\; x = \frac{2}{3}$

完整詳解

Table: at $x = 0$, the values are $1$ and $3$ (second is larger); at $x = 1$, they are $3$ and $2$ (first is larger). The crossing is between $x = 0$ and $x = 1$.
Check $(0.7, 2.3)$: $2(0.7) + 1 = 2.4 \ne 2.3$, so the estimate is close but not exact.
Setting the $y$-values equal: $2x + 1 = -x + 3$, so $3x = 2$, $x = \frac{2}{3}$, and $y = 2\left(\frac{2}{3}\right) + 1 = \frac{7}{3}$.
Check: $-\frac{2}{3} + 3 = \frac{7}{3}$ ✓. The exact solution is $\left(\frac{2}{3}, \frac{7}{3}\right)$.

表格：$x = 0$ 時兩值是 $1$ 和 $3$（第二式較大）；$x = 1$ 時是 $3$ 和 $2$（第一式較大）。交點在 $x = 0$ 與 $x = 1$ 之間。
驗算 $(0.7, 2.3)$：$2(0.7) + 1 = 2.4 \ne 2.3$，估計值很接近但不精確。
令兩個 $y$ 值相等：$2x + 1 = -x + 3$，$3x = 2$，$x = \frac{2}{3}$，$y = 2\left(\frac{2}{3}\right) + 1 = \frac{7}{3}$。
驗算：$-\frac{2}{3} + 3 = \frac{7}{3}$ ✓。精確解是 $\left(\frac{2}{3}, \frac{7}{3}\right)$。

例題

Use the table to solve $\begin{cases} y = 3x - 5 \\ y = x + 1 \end{cases}$. $x = 1$: $-2$ and $2$; $x = 2$: $1$ and $3$; $x = 3$: $4$ and $4$; $x = 4$: $7$ and $5$.用表格解 $\begin{cases} y = 3x - 5 \\ y = x + 1 \end{cases}$。 $x = 1$：$-2$ 和 $2$；$x = 2$：$1$ 和 $3$；$x = 3$：$4$ 和 $4$；$x = 4$：$7$ 和 $5$。

觀念提示

Look for the row where the two outputs match. The switch in which output is larger confirms you found the crossing.找兩個輸出相同的那一列。「哪一個比較大」在這裡換過來，也確認了這就是交點。

關鍵算式

$x = 3$: $3(3) - 5 = 4 = 3 + 1$$x = 3$：$3(3) - 5 = 4 = 3 + 1$

完整詳解

At $x = 3$, both equations give $y = 4$.
Check: $3(3) - 5 = 4$ ✓ and $3 + 1 = 4$ ✓.
The solution is $(3, 4)$. Before $x = 3$ the second line is higher; after it, the first line is higher.

$x = 3$ 時，兩式都得 $y = 4$。
驗算：$3(3) - 5 = 4$ ✓，$3 + 1 = 4$ ✓。
解是 $(3, 4)$。$x = 3$ 以前第二條線比較高，之後第一條線比較高。

自我檢查

For $\begin{cases} y = 4x - 2 \\ y = -2x + 6 \end{cases}$, a table gives $2$ and $4$ at $x = 1$, and $6$ and $2$ at $x = 2$. Between which $x$-values is the solution? Find it exactly.對 $\begin{cases} y = 4x - 2 \\ y = -2x + 6 \end{cases}$，表格中 $x = 1$ 時是 $2$ 和 $4$，$x = 2$ 時是 $6$ 和 $2$。解落在哪兩個 $x$ 值之間？求出精確解。

看答案

The larger value switches between $x = 1$ and $x = 2$, so the solution is between them.
$4x - 2 = -2x + 6$ gives $6x = 8$, $x = \frac{4}{3}$; $y = 4\left(\frac{4}{3}\right) - 2 = \frac{10}{3}$. Check: $-\frac{8}{3} + 6 = \frac{10}{3}$ ✓. Solution $\left(\frac{4}{3}, \frac{10}{3}\right)$.

較大的值在 $x = 1$ 與 $x = 2$ 之間換過來，所以解在這兩者之間。
$4x - 2 = -2x + 6$ 得 $6x = 8$，$x = \frac{4}{3}$；$y = 4\left(\frac{4}{3}\right) - 2 = \frac{10}{3}$。驗算：$-\frac{8}{3} + 6 = \frac{10}{3}$ ✓。解是 $\left(\frac{4}{3}, \frac{10}{3}\right)$。

ESolve a problem by graphing a system用圖解方程組解應用題

Use this type when a situation compares two quantities that change at steady rates and asks when they are equal. Write one linear equation for each quantity using the same variables, graph both, and read the intersection. State what the coordinates mean, with units, and say what happens on each side of the crossing.當情境比較兩個以固定速率變化的量，問什麼時候相等時，就是這種題型。用同樣的變數為每個量各寫一個一次方程式，兩條都畫出來，讀出交點。用完整句子帶單位說明坐標的意義，也說明交點兩側的情形。

例題

A streaming service offers Plan A at $\$6$ per month plus $\$2$ per movie, and Plan B at $\$10$ per month plus $\$1$ per movie. For how many movies a month do the plans cost the same?某串流服務有兩種方案：方案 A 每月 $\$6$，每部電影另加 $\$2$；方案 B 每月 $\$10$，每部電影另加 $\$1$。每月看幾部電影時，兩種方案的費用一樣？

觀念提示

The monthly fee is the y-intercept and the price per movie is the slope. "Cost the same" means the intersection.月費是 y 截距，每部電影的價錢是斜率。「費用一樣」就是交點。

關鍵算式

$\begin{cases} C = 2m + 6 \\ C = m + 10 \end{cases}$$\begin{cases} C = 2m + 6 \\ C = m + 10 \end{cases}$

完整詳解

Let $m$ = movies and $C$ = monthly cost. Plan A: $C = 2m + 6$. Plan B: $C = m + 10$.
Graph A from $(0, 6)$ with slope $2$: $(2, 10)$, $(4, 14)$. Graph B from $(0, 10)$ with slope $1$: $(2, 12)$, $(4, 14)$.
The lines cross at $(4, 14)$. Check: $2(4) + 6 = 14$ ✓, $4 + 10 = 14$ ✓.
With $4$ movies both plans cost $\$14$. With fewer movies Plan A is cheaper; with more, Plan B is cheaper.

設 $m$ 為電影數、$C$ 為每月費用。方案 A：$C = 2m + 6$；方案 B：$C = m + 10$。
方案 A 從 $(0, 6)$ 開始、斜率 $2$：$(2, 10)$、$(4, 14)$。方案 B 從 $(0, 10)$ 開始、斜率 $1$：$(2, 12)$、$(4, 14)$。
兩線交於 $(4, 14)$。驗算：$2(4) + 6 = 14$ ✓，$4 + 10 = 14$ ✓。
看 $4$ 部電影時兩方案都是 $\$14$；看得比較少時方案 A 較便宜，看得比較多時方案 B 較便宜。

例題

Candle P is $20$ cm tall and burns down $2$ cm per hour. Candle Q is $14$ cm tall and burns down $1$ cm per hour. Both are lit at the same time. When are they the same height?蠟燭 P 高 $20$ 公分，每小時燒短 $2$ 公分；蠟燭 Q 高 $14$ 公分，每小時燒短 $1$ 公分。兩根同時點燃，什麼時候一樣高？

觀念提示

Burning down means a negative slope. The taller candle burns faster, so the two lines must meet.燒短表示斜率是負的。比較高的那根燒得比較快，所以兩條線一定會相交。

關鍵算式

$\begin{cases} h = 20 - 2t \\ h = 14 - t \end{cases}$$\begin{cases} h = 20 - 2t \\ h = 14 - t \end{cases}$

完整詳解

Let $t$ = hours and $h$ = height in cm. P: $h = 20 - 2t$. Q: $h = 14 - t$.
P passes through $(0, 20)$, $(3, 14)$, $(6, 8)$; Q passes through $(0, 14)$, $(3, 11)$, $(6, 8)$.
The lines cross at $(6, 8)$. Check: $20 - 12 = 8$ ✓, $14 - 6 = 8$ ✓.
After $6$ hours both candles are $8$ cm tall. The model only makes sense for $0 \le t \le 10$, because P is gone at $t = 10$.

設 $t$ 為小時、$h$ 為高度（公分）。P：$h = 20 - 2t$；Q：$h = 14 - t$。
P 經過 $(0, 20)$、$(3, 14)$、$(6, 8)$；Q 經過 $(0, 14)$、$(3, 11)$、$(6, 8)$。
兩線交於 $(6, 8)$。驗算：$20 - 12 = 8$ ✓，$14 - 6 = 8$ ✓。
$6$ 小時後兩根蠟燭都是 $8$ 公分高。這個模型只在 $0 \le t \le 10$ 有意義，因為 $t = 10$ 時 P 已經燒完。

自我檢查

Tank A holds $50$ L and is filled at $5$ L per minute. Tank B holds $110$ L and is drained at $10$ L per minute. Write a system and find when they hold the same amount.水槽 A 有 $50$ 公升，每分鐘注入 $5$ 公升；水槽 B 有 $110$ 公升，每分鐘排出 $10$ 公升。寫出方程組，並求兩槽何時水量相同。

看答案

$\begin{cases} V = 50 + 5t \\ V = 110 - 10t \end{cases}$. The lines cross at $(4, 70)$.
Check: $50 + 20 = 70$ ✓, $110 - 40 = 70$ ✓. After $4$ minutes each tank holds $70$ L.

$\begin{cases} V = 50 + 5t \\ V = 110 - 10t \end{cases}$。兩線交於 $(4, 70)$。
驗算：$50 + 20 = 70$ ✓，$110 - 40 = 70$ ✓。$4$ 分鐘後兩槽都有 $70$ 公升。

下一課： Substitution代入消去法

這課的重點字

system of equations方程組（聯立方程式）
solution of a system方程組的解
consistent system相容方程組
inconsistent system不相容方程組
independent system獨立方程組
dependent system相依方程組
infinitely many solutions無限多解

背這課的 7 個字

申請試用一個月

Ordered pair有序數對	\(x + y\)\(x + y\)	\(2x - y\)\(2x - y\)	Solution of the system?是方程組的解嗎？
\((3, 3)\)\((3, 3)\)	\(6\) ✓\(6\) ✓	\(3\) ✓\(3\) ✓	yes是
\((4, 2)\)\((4, 2)\)	\(6\) ✓\(6\) ✓	\(6\) ✗\(6\) ✗	no不是
\((1, -1)\)\((1, -1)\)	\(0\) ✗\(0\) ✗	\(3\) ✓\(3\) ✓	no不是
\((0, 6)\)\((0, 6)\)	\(6\) ✓\(6\) ✓	\(-6\) ✗\(-6\) ✗	no不是

System方程組	Slope-intercept forms斜截式	Slopes斜率	y-interceptsy 截距	Solutions解
\(3x + y = 4\); \(y = x - 2\)\(3x + y = 4\)；\(y = x - 2\)	\(y = -3x + 4\); \(y = x - 2\)\(y = -3x + 4\)；\(y = x - 2\)	\(-3\), \(1\)\(-3\)、\(1\)	\(4\), \(-2\)\(4\)、\(-2\)	one一個
\(2x - y = 5\); \(y = 2x + 1\)\(2x - y = 5\)；\(y = 2x + 1\)	\(y = 2x - 5\); \(y = 2x + 1\)\(y = 2x - 5\)；\(y = 2x + 1\)	\(2\), \(2\)\(2\)、\(2\)	\(-5\), \(1\)\(-5\)、\(1\)	none沒有
\(x - 2y = 6\); \(2x - 4y = 12\)\(x - 2y = 6\)；\(2x - 4y = 12\)	\(y = \frac{1}{2}x - 3\) for both兩式都是 \(y = \frac{1}{2}x - 3\)	\(\frac{1}{2}\), \(\frac{1}{2}\)\(\frac{1}{2}\)、\(\frac{1}{2}\)	\(-3\), \(-3\)\(-3\)、\(-3\)	infinitely many無限多個

Step步驟	What to do做什麼	Work過程
1. Graph the first line1. 畫第一條線	Use slope-intercept form.用斜截式。	y-intercept \((0, -1)\), slope \(2\)y 截距 \((0, -1)\)，斜率 \(2\)
2. Graph the second line2. 畫第二條線	Rewrite as \(y = -x + 5\).改寫成 \(y = -x + 5\)。	y-intercept \((0, 5)\), slope \(-1\)y 截距 \((0, 5)\)，斜率 \(-1\)
3. Find the intersection3. 找交點	Read where the lines cross.讀出兩線交在哪裡。	\((2, 3)\)\((2, 3)\)
4. Check4. 驗算	Substitute into both equations.代入兩個方程式。	\(3 = 2(2) - 1\) ✓; \(2 + 3 = 5\) ✓\(3 = 2(2) - 1\) ✓；\(2 + 3 = 5\) ✓
5. Answer5. 作答	Write the ordered pair.寫出有序數對。	\((2, 3)\)\((2, 3)\)