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CIE 9709 2026 June Paper 15 Q1

A Level / CIE / P1

CIE 9709 2026 June Paper 15 Paper · Question 1

题目

Problem

Solve the simultaneous equations

4x2+2y2−3y=21and2x−y−5=0.4x^2+2y^2-3y=21\qquad\text{and}\qquad2x-y-5=0.
(5)
题目中文翻译

解联立方程

4x2+2y2−3y=21,2x−y−5=0.4x^2+2y^2-3y=21,\qquad2x-y-5=0.
(5)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

解法一

思路

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用直线方程消去 xx,把联立方程化成关于 yy 的二次方程。求根后回代求 xx。

答题过程

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Since 2x=y+52x=y+5, substitution gives

(y+5)2+2y2−3y= 21⟹3y2+7y+4=(3y+4)(y+1)= 0.\begin{align*} (y+5)^2+2y^2-3y=&\,21\\ \quad\Longrightarrow\quad 3y^2+7y+4=(3y+4)(y+1)=&\,0. \end{align*}

For y=−4/3y=-4/3, we have x=(5−4/3)/2=11/6x=(5-4/3)/2=11/6. For y=−1y=-1, we have x=2x=2. The solutions are (11/6,−4/3)(11/6,-4/3) and (2,−1)(2,-1).