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IAL 2020 Oct Q4

A Level / Edexcel / P1

IAL 2020 Oct Paper · Question 4

题目

Problem

Solve the simultaneous equations

y−3x= 4,x2+y2+6x−4y= 4,\begin{align*} y-3x=&\,4,\\ x^2+y^2+6x-4y=&\,4, \end{align*}

showing all your working.

(7)

解答

解法一

思路

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先由直线方程得到 y=3x+4y=3x+4，再代入第二个方程。这样会得到一个关于 xx 的二次方程，解出 xx 后再代回直线求 yy。

答题过程

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From

y−3x=4,\begin{align*} y-3x=4, \end{align*}

we have

y=3x+4.\begin{align*} y=3x+4. \end{align*}

Substitute this into the second equation:

x2+(3x+4)2+6x−4(3x+4)= 4x2+9x2+24x+16+6x−12x−16= 410x2+18x−4= 05x2+9x−2= 0.\begin{align*} x^2+(3x+4)^2+6x-4(3x+4)=&\,4\\ x^2+9x^2+24x+16+6x-12x-16=&\,4\\ 10x^2+18x-4=&\,0\\ 5x^2+9x-2=&\,0. \end{align*}

Factorise:

5x2+9x−2= 0(5x−1)(x+2)= 0.\begin{align*} 5x^2+9x-2=&\,0\\ (5x-1)(x+2)=&\,0. \end{align*}

Thus

x=15orx=−2.\begin{align*} x=\frac15\quad\text{or}\quad x=-2. \end{align*}

When x=15x=\frac15,

y=3(15)+4=235.\begin{align*} y=3\left(\frac15\right)+4=\frac{23}{5}. \end{align*}

When x=−2x=-2,

y=3(−2)+4=−2.\begin{align*} y=3(-2)+4=-2. \end{align*}

Therefore the solutions are

(15,235)and(−2,−2).\begin{align*} \left(\frac15,\frac{23}{5}\right) \quad\text{and}\quad (-2,-2). \end{align*}