Simultaneous equations and feasible regions · Foundation
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| elimination/ɪˌlɪmɪˈneɪʃn/ | 消元法 | xiāo yuán fǎ |
Can one equation determine two counts?
- Two ticket types raise a total amount. A single equation cannot identify both unknown counts.
- This lesson studies elimination 消元法: Combining equations to remove one variable while retaining the same solutions.
一个方程能否确定两个数量？
- 两种票种共筹集一定金额。单个方程无法同时确定两个未知数量。
- 本课学习消元法：联立方程以消除一个变量，同时保持解集不变。
Choose the mathematical structure
- For two linear equations, use elimination or substitution and check both equations. For a line and a quadratic, substitute the linear relation first; then solve the resulting quadratic. For inequalities, shade the region satisfying every condition.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
选择数学结构
- 对于两个线性方程，使用消元法或代入法，并检验两个方程。对于直线与二次曲线，先代入线性关系；再求解所得二次方程。对于不等式，涂色表示满足所有条件的区域。
- 计算前请先明确允许的输入项和单位。方程应表达关系本身，而不仅仅是记录计算器按键过程。
Which description correctly defines elimination?
Combining equations to remove one variable while retaining the same solutions.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
If x+y=12 and 3x+2y=31, subtract twice the first equation from the second to obtain x=7, then y=5. For y=x+1 and y=x²-1, x²-x-2=0 gives x=2 or -1, with y=3 or 0.
通过验证案例进行推导
- 将结果与初始量进行比对。代入原始关系式，或在适当情况下对比图表与数值答案。
若 x+y=12 且 3x+2y=31，用第二个方程减去第一个方程的两倍可得 x=7，进而 y=5。对于 y=x+1 和 y=x²-1，由 x²-x-2=0 可得 x=2 或 -1，对应的 y=3 或 0。
Simultaneous equations and feasible regions
For two linear equations, use elimination or substitution and check both equations
Compare the model with the worked case and explain one change.
Solve x+y=12 and 3x+2y=31. Find x.
Subtract twice x+y=12 from 3x+2y=31: x=31-24=7.
Test a tempting shortcut
- One equation checked is not enough. A line can meet a quadratic twice, so retain both solutions unless the context removes one. Inequality boundaries may be included or excluded according to the sign.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
A pair of simultaneous equations is solved by checking just one of them. This claim is false. Explain which definition or assumption it violates.
检验一个诱人的捷径
- 仅检验一个方程是不够的。直线可能与二次曲线有两个交点，因此除非题目背景排除了其中一个解，否则应保留两个解。不等式的边界是否包含取决于符号。
- 当捷径失效时，找出其违背的假设。保留精确值直到题目要求的最终舍入步骤。
仅通过检验其中一个方程即可解出一组联立方程。该说法是错误的。请说明它违反了哪一定义或假设。
For those equations, find y.
Use x+y=12, so y=12-7=5.
A pair of simultaneous equations is solved by checking just one of them.
One equation checked is not enough. A line can meet a quadratic twice, so retain both solutions unless the context removes one. Inequality boundaries may be included or excluded according to the sign.
Interpret a new situation
- Define the variables and their units. If they count objects, both must be nonnegative integers. For a feasible region, test one point on the required side of each boundary and then take the intersection.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
解读新情境
- 定义变量及其单位。若变量代表物体数量，则两者必须为非负整数。对于可行域，需在每条边界的指定侧测试一点，然后取交集。
- 完整解答需给出数学结果并解释其含义。检查其在所述背景下是否可行。
Find the larger x-coordinate where y=x+1 meets y=x²-1.
Equate the graphs: x²-x-2=(x-2)(x+1)=0. The larger root is 2.
Match each part of a complete solution to its purpose.
An assumption justifies the model; a check tests the result; interpretation connects it to the question.
Use this in your course
- 4MA1 · Foundation · 2. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
Combining equations to remove one variable while retaining the same solutions. Choose the relationship, show the method, check its assumptions and interpret the result.
将其应用于你的课程
- 4MA1 · 基础级 · 2. 在分配拓展内容前，请匹配目标层级和教学大纲要求。
- 在给出最终答案前先展示解题方法，并遵循试卷的计算器及公式使用规则。通过定位第一个无效步骤来复盘错误答案。
联立方程以消除一个变量，同时保持解集不变。选择关系式，展示解题方法，检验假设条件并解释结果。