例題 1 Example 1：兩條一次方程 Two linear equations
解 2x + y = 7 及 x − y = 2。Solve 2x + y = 7 and x − y = 2.
- 兩式相加消去 y：
3x = 9，所以x = 3Add the equations to eliminate y:3x = 9, sox = 3 3 − y = 2, soy = 1
答案： Answer: x = 3, y = 1
【聯立方程：消元法、代入法及圖像】Simultaneous Equations: Elimination, Substitution and Graphs
SketchMath 數學團隊 SketchMath Maths Team ｜ 更新 Updated 2026-10-20
解含 x、y 的兩條方程，可用消元法（先使係數相同，再相加或相減）或代入法。若其中一條是二次方程，便把一次方程化成 y 作主項再代入。每個解都是兩條圖像的交點。To solve two equations in x and y, eliminate one unknown (add or subtract after matching coefficients) or substitute one equation into the other. When one equation is a quadratic, make y the subject of the linear equation and substitute. Each solution is a point where the two graphs meet.
解法一覽 Methods at a Glance
消元法Elimination把方程乘以適當數，使一個係數相同，再相加或相減Multiply to match one coefficient, then add or subtract
代入法Substitution把一個未知數化為主項，再代入另一條方程Make one unknown the subject and substitute into the other equation
一次與二次Linear and quadratic把直線代入二次方程，得
Ax² + Bx + C = 0Substitute the line into the quadratic: get Ax² + Bx + C = 0解的數目Number of solutions
Δ > 0: 2, Δ = 0: 1 (tangent), Δ < 0: none聯立方程例題詳解 Worked Examples
例題 2 Example 2：直線與圓 A line and a circle
解 y = x + 1 及 x² + y² = 25。Solve y = x + 1 and x² + y² = 25.
x² + (x + 1)² = 25, so2x² + 2x − 24 = 0x² + x − 12 = 0, so(x + 4)(x − 3) = 0- 把每個 x 代回直線：
y = x + 1Put each x back into the line:y = x + 1
答案： Answer: (3, 4) and (−4, −3)
例題 3 Example 3：直線何時與曲線相切？ When does a line touch a curve?
求 c，使 y = 2x + c 與 y = x² 相切。Find c so that y = 2x + c touches y = x².
x² = 2x + c, sox² − 2x − c = 0- 相切即只有一個解：
Δ = 4 + 4c = 0Tangent means one solution:Δ = 4 + 4c = 0
答案： Answer: c = −1 (touching at (1, 1))
💡 應試陷阱與秘訣 Exam Traps & Secrets
⚠️ 要代入一次方程 Substitute into the linear equation由二次方程求得 x 後，要用一次方程求 y。代入二次方程會得出多餘的錯誤答案。After finding x from a quadratic, find y from the linear equation. Using the quadratic gives extra, false pairs.
⚠️ 兩個未知數都要寫 Write both unknowns解是一對數值。只寫出 x 會損失最後的分數。A solution is a pair. Stopping at the values of x loses the final marks.
✏️ 自我測試：互動練習 Self-Test: Interactive Practice
溫習方法：點擊以下任何題目，即時在坐標網格載入並核對計算步驟。How to study: click any question to load it on the coordinate grid and check your steps.
(消元法 Elimination)解 3x + 2y = 12 及 5x − 2y = 4。Solve 3x + 2y = 12 and 5x − 2y = 4.
(直線與圓 Line and circle)解 y = 2x − 1 及 x² + y² = 13。Solve y = 2x − 1 and x² + y² = 13.
(相切條件 Tangent condition)求 k 的值，使 y = x + k 與 x² + y² = 8 只有一個交點。Find the values of k for which y = x + k meets x² + y² = 8 at exactly one point.
聯立方程練習題（DSE 模擬題） Exercises: DSE-Style Practice
SketchMath 題庫內的 DSE 風格練習，選擇題及長題目均附解答。 DSE-style practice from the SketchMath question bank, with solutions for every MC and long question.
二元一次方程 · 中二
31 條選擇題 31 MC · 27 條長題目 27 long questions
續方程 · 中四
128 條選擇題 128 MC · 35 條長題目 35 long questions
常見問題 Frequently Asked Questions
若其中一條方程已有一個未知數作主項，用代入法；兩條都是一次方程時，通常消元法較快。
Use substitution when one equation already has an unknown as the subject; otherwise elimination is usually faster for two linear equations.
是兩條圖像相交的點。若兩條互相平行，便沒有解。
It is the point where the two graphs cross. If they are parallel there is no solution.
大部分認可計數機都有二元或三元方程的模式。可用來檢查答案，但卷面仍要寫出步驟。
Most approved calculators have an equation mode for 2 or 3 unknowns. Use it to check your working, but still show the steps in the paper.