📚 Solving Two Simultaneous Equations – One Linear and One Non-Linear | 解两个联立方程——一个线性，一个非线性
In IGCSE Mathematics, solving two simultaneous equations where one is linear and the other is non-linear (usually quadratic) is a core algebraic skill. You will often be asked to find the intersection points of a straight line and a curve. The method always involves substitution, leading to a quadratic equation that you can solve by factorising or using the quadratic formula.
在IGCSE数学中，解一个线性方程和一个非线性方程（通常是二次方程）的联立方程组是一项核心代数技能。题目常要求你求出直线与曲线的交点。解题方法始终围绕代入法，最终会得到一个二次方程，你可以通过因式分解或使用二次公式来求解。
1. What Are Linear and Non-Linear Equations? | 什么是线性方程和非线性方程？
s a straight-line graph and has variables only to the power of 1, such as y = 2x + 3 or 3x – CSE non-linear equations include y = ax² + bx + c (a parabola) or x² + y² = r² (a circle).
线性方程的图像是一条直线，变量只出现一次幂，例如 y = 2x + 3 或 3x – 4y = 12。方程中不含 x²、y²、xy 或 1/x 这样的项。非线性方程则含有 x²、y²、xy 或其他函数项，图像为曲线。典型的IGCSE非线性方程包括 y = ax² + bx + c（抛物线）或 x² + y² = r²（圆）。
2. Understanding the Problem Type | 理解题型
You are given a pair of equations, for instance y = 2x + 1 and y = x² + 3x – d the curve. Because one equation is non-linear, there could be two solutions, one solution (if the line is tangent), or no real solutions (if the line does not intersect the curve).
你会得到一组方程，例如 y = 2x + 1 和 y = x² + 3x – 5。你的任务是找出同时满足这两个方程的 (x, y) 坐标组。在几何意义上，这些就是直线与曲线的交点。由于其中一个是非线性方程，解的可能情况有三种：两个解、一个解（直线与曲线相切）或无实数解（直线与曲线不相交）。
3. The Substitution Method: General Steps | 代入法：一般步骤
Solve the quadratic by factorising or using the quadratic formula. Step 5: Substitute each x-value back into the linear equation to find the corresponding y-values. Step 6: Write the solutions as coordinate pairs (x, y).
第一步：如果线性方程尚未写成 y=... 或 x=... 的形式，先把它整理成主项明确的形式。第二步：将这个表达式代入非线性方程。第三步：化简并整理得到一个二次方程（通常是 ax² + bx + c = 0）。第四步：通过因式分解或二次公式解这个二次方程，求出 x。第五步：将每个 x 值代回线性方程，求出对应的 y 值。第六步：将解写成坐标 (x, y) 的形式。
4. Worked Example 1: Line and Parabola | 例题1：直线与抛物线
Solve the simultaneous equations: y = 3x – 2 and y = x² – 4x + 5.
Substitute y from the linear into the quadratic: 3x – 2 = x² – 4x + 5. Rearrange: 0 = x² – 7x + 7. This does not factorise nicely, so use the quadratic formula: x = [7 ± √49 – 28] / 2 = [7 ± √21] / 2.
So x = (7 + √21)/2 and x = (7 – √21)/2.
Substitute back into y = 3x – 2: for x₁, y₁ = 3((7+√21)/2) – 2 = (21 + 3√21)/2 – 2 = (17 + 3√21)/2. For x₂, y₂ = (17 – 3√21)/2. The solutions are these two coordinate pairs.
解联立方程：y = 3x – 2 和 y = x² – 4x + 5。
将一次方程中的 y 代入二次方程：3x – 2 = x² – 4x + 5。移项整理：0 = x² – 7x + 7。该方程不易因式分解，故使用二次公式：x = [7 ± √49 – 28] / 2 = [7 ± √21] / 2。
因此 x = (7 + √21)/2 和 x = (7 – √21)/2。
代回 y = 3x – 2：将 x₁ 代入，y₁ = 3((7+√21)/2) – 2 = (21 + 3√21)/2 – 2 = (17 + 3√21)/2；同理，x₂ 对应的 y₂ = (17 – 3√21)/2。最终解为这两个坐标点。
5. Worked Example 2: Line and Circle | 例题2：直线与圆
Solve the simultaneous equations: x² + y² = 25 and y = x + 1.
into the circle equation: x² + (x + 1)² = 25. Expand: x² + x² + 2x + 1 = 25 → 2x² + 2x + 1 – 25 = 0 → 2x² + 2x – 24 = 0. Divide by 2: x² + x – 12 = 0. Factorise: (x + 4)(x – 3) = 0, so x = –4 or x = 3.
For x = –4, y = –4 + 1 = –3. For x = 3, y = 3 + 1 = 4. The intersection points are (–4, –3) and (3, 4). Always check these satisfy the original circle equation.
解联立方程：x² + y² = 25 和 y = x + 1。
将 y = x + 1 代入圆的方程：x² + (x + 1)² = 25。展开：x² + x² + 2x + 1 = 25 → 2x² + 2x + 1 – 25 = 0 → 2x² + 2x – 24 = 0。两边除以 2：x² + x – 12 = 0。因式分解：(x + 4)(x – 3) = 0，得 x = –4 或 x = 3。
当 x = –4 时，y = –3；当 x = 3 时，y = 4。交点坐标为 (–4, –3) 和 (3, 4)。最后务必验证这两组解是否满足原来的圆的方程。
6. Using the Discriminant to Determine Number of Solutions | 利用判别式判断解的个数
Once you rearrange into a quadratic ax² + bx + c = 0, the discriminant Δ = b² – is exactly one solution (the line is tangent to the curve). If Δ < 0, there are no real solutions (the line misses the curve completely). This can save time in exam questions that ask about the nature of intersection.
当你将方程整理成二次方程 ax² + bx + c = 0 后，判别式 Δ = b² – 4ac 能告诉你交点的个数。若 Δ > 0，则有两个不同的实数解（直线与曲线相交于两点）。若 Δ = 0，则恰好有一个解（直线与曲线相切）。若 Δ < 0，则无实数解（直线与曲线不相交，完全分离）。在考试中遇到询问交点性质的题目时，利用判别式可以节省时间。
7. Graphical Interpretation | 图形解释
rsecting a parabola may cross it twice, touch it once, or not meet it at all. With a circle, a line can be a secant (two intersections), a tangent (one intersection) or an external line (no intersection). Sketching a rough graph can clarify whether your algebraic solutions make sense. For the examples above, imagine the line and curve; the coordinates should lie realistically on both graphs.
理解你在计算的几何意义有助于减少错误。一条直线 y = mx + c 与抛物线相交时，可能穿过两次、相切一次或完全不相交。直线与圆的关系可能是割线（两个交点）、切线（一个交点）或外离（无交点）。简单地画一个草图，就能帮助你判断代数解是否合理。对上述例题，可以想象直线和曲线的位置，所得坐标在图上应落在两者的交点上。
8. Checking Your Solutions | 验证你的解
nd y back into the original equations, especially the non-linear one, to verify your answers. In a circle equation, both (–4)² + (– s catches algebraic slips such as sign errors when expanding brackets or misapplying the quadratic formula. Even in the exam, a quick mental check can save marks.
将求出的 x 和 y 一起代回原始方程（尤其是非线性方程）来验证答案是必不可少的步骤。在圆的例子中，(–4)² + (–3)² = 16 + 9 = 25，确认了坐标的正确性。这样做能发现代数运算中的小错误，例如展开括号时的正负号错误或二次公式使用不当。即使在考试中，快速的心算验证也能为你保住分数。
9. Common Mistakes and How to Avoid Them | 常见错误及其避免方法
d y after solving the quadratic, but incorrectly substituting into the non-linear one. Mistake 4: Not simplifying the quadratic before solving, making factorisation harder. Mistake 5: When the non-linear equation is x² + y² = r², forgetting to square the entire substituted expression properly. Write (x+1)² = x² + 2x + 1, not x²+1. Take your time with expansions.
错误一：未能正确移项整理，例如等号两边都保留了变量 y。一定要将其中一个变量单独表示出来。错误二：求出 x 后忘记代回求 y。每个 x 只能从线性方程求得一个对应的 y，而不是两个。错误三：解完二次方程后，用非线性方程求 y，而不是用线性方程，导致错误。错误四：解二次方程前没有先化简，使因式分解更困难。错误五：当非线性方程为 x² + y² = r² 时，忘记把代入的整个表达式正确平方。务必记住 (x+1)² = x² + 2x + 1，而不是 x² + 1。展开时请细心。
10. Dealing with Non-Standard Non-Linear Equations | 处理非典型的非线性方程
y = 6 and y = 2x + 1. The method is still substitution. Replace y: x(2x + 1) = 6 → 2x² + x – 6 = 0, which you can factorise (2x – lying through by x. The key is always to eliminate one variable and obtain an equation in one unknown.
有时非线性方程会含有 xy 项，例如 xy = 6 和 y = 2x + 1。解题方法依然是代入。将 y 换掉：x(2x + 1) = 6 → 2x² + x – 6 = 0，然后可因式分解为 (2x – 3)(x + 2) = 0。只要保持冷静，按步骤处理即可。其他形式如 y = 1/x 与直线联立，也可通过代入并在方程两边同乘 x 来得到一个二次方程。核心思路总是消去一个变量，得到只含一个未知数的方程。
11. Exam Tip: Showing Sufficient Working | 考试技巧：展示充分步骤
Edexcel IGCSE examiners expect clear logical steps. Even if you can do some mental arithmetic, write down the substitution, the simplification, the quadratic in standard form, factorising (or formula) steps, and the back-substitution. Marks are awarded for method. If you make a slip but your method is correct, you can still gain most of the marks. Always state your final answer as coordinate pairs clearly, e.g., (3, 4) and (–4, –3).
Edexcel IGCSE考官希望看到清晰、有逻辑的解题步骤。即使有些心算可以跳过，也请写出代入过程、化简过程、标准二次方程、因式分解（或公式）的步骤以及回代过程。步骤是有步骤分的。如果你犯了小错误但方法正确，仍能拿到大部分分数。务必以坐标对的形式明确写出最终答案，例如 (3, 4) 和 (–4, –3)。
12. Summary and Practice Advice | 总结与练习建议
tforward process: isolate, substitute, solve the quadratic, back-substitute. Practice with a variety of non-linear equations— Once confident, apply the discriminant to questions that ask for the number of intersection points or values of k for which a line is tangent to a curve. Mastery of this topic will also strengthen your skills in algebra and coordinate geometry.
解一个线性和一个非线性的联立方程组是一个很直接的过程：分离变量、代入、解二次方程、回代求值。通过针对不同非线性方程的练习来巩固，包括抛物线、圆、双曲线（如 xy = k）以及倒数函数。熟练之后，可以将判别式用于解答求交点个数或求直线与曲线相切时参数 k 的取值等问题。掌握本专题将同时增强你的代数与坐标几何的能力。
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