Posted on by

202301170910 Exercise 3.2

Solve these simultaneous equations:

1. \begin{cases} x+y=5 \\ xy=6 \\ \end{cases}

Roughwork. Read More

    Substituting x=5-y for x:

    \begin{aligned} (5-y)y & = 6 \\ y^2-5y+6 & = 0 \\ (y-2)(y-3) & = 0 \\ y & = 2,3 \\ \begin{pmatrix}x \\ y \end{pmatrix} = \bigg\{ \begin{pmatrix} 2 \\ 3 \end{pmatrix},\begin{pmatrix} 3 \\ 2 \end{pmatrix} \bigg\} & \\ \end{aligned}

    Substituting x=1+2y for x:

    \begin{aligned} (1+2y)^2 + y^2 & =29 \\ (1+4y+4y^2)+y^2 & = 29 \\ 5y^2 +4y -28 & = 0 \\ (5y+14)(y-2) & = 0 \\ y & = -\frac{14}{5},2 \\ \begin{pmatrix}x\\y\end{pmatrix} = \bigg\{ \begin{pmatrix}5\\2\end{pmatrix}, \begin{pmatrix}-23/5\\-14/5\end{pmatrix} \bigg\} & \\ \end{aligned}

    Substituting y=5-2x for y:

    \begin{aligned} x^2-x(5-2x) & = 12 \\ 3x^2 -5x -12 & = 0 \\ (3x+4)(x-3) & = 0 \\ x & = -\frac{4}{3},3 \\ \begin{pmatrix}x\\y\end{pmatrix} = \bigg\{ \begin{pmatrix}-4/3\\7/3\end{pmatrix},\begin{pmatrix}3\\-1\end{pmatrix}\bigg\} &\\ \end{aligned}

    Substituting y=3x-7 for y:

    \begin{aligned} x^2-x(3x-7)+(3x-7)^2 & = 7 \\ 7x^2 -35x +42 & = 0 \\ (7x+7)(x-6) & = 0 \\ x & = -1,6 \\ \begin{pmatrix}x\\y\end{pmatrix} = \bigg\{ \begin{pmatrix}-1\\-10\end{pmatrix} , \begin{pmatrix}6\\11\end{pmatrix}\bigg\} & \\ \end{aligned}

    Substituting y=9-5x for y:

    \begin{aligned} 3x(9-5x)+(9-5x)^2 & = -5 \\ 10x^2-63x +86 & = 0 \\ (10x-43)(x-2) & = 0 \\ x & = \frac{43}{10},2 \\ \begin{pmatrix}x\\y\end{pmatrix} = \bigg\{ \begin{pmatrix}43/10\\-25/2\end{pmatrix} & , \begin{pmatrix}2\\-1\end{pmatrix}\bigg\} \\ \end{aligned}

    Substituting \displaystyle{y = \frac{13-3x}{2}} for y:

    \begin{aligned} 3x^2 + \bigg(\frac{13-3x}{2}\bigg)^2 = 31 \\ 7x^2 - 26x+15 & = 0 \\ (7x-5)(x-3) & = 0 \\ x & = \frac{7}{5},3 \\ \begin{pmatrix}x\\y\end{pmatrix} = \bigg\{ \begin{pmatrix}7/5\\22/5\end{pmatrix},\begin{pmatrix}3\\-13\end{pmatrix} \bigg\} & \\ \end{aligned}

    Substituting \displaystyle{y=\frac{2x+11}{3}} for y:

    \begin{aligned} 2x^2-x\bigg( \frac{2x+11}{3}\bigg) & =36 \\ 4x^2-11x-108 &= 0 \\ (4x-27)(x+4) & = 0 \\ x & = -4,\frac{27}{4} \\ \begin{pmatrix}x\\y\end{pmatrix} & = \bigg\{ \begin{pmatrix}-4\\1\end{pmatrix},\begin{pmatrix}27/4\\49/6\end{pmatrix}\bigg\} \\ \end{aligned}

    Substituting \displaystyle{y=\frac{2x-1}{5}} for y:

    \begin{aligned} x^2-x\bigg(\frac{2x-1}{5}\bigg)+3\bigg(\frac{2x-1}{5}\bigg)^2 & = 9 \\ 27x^2 -7x -222 & = 0 \\ (27x+74)(x-3) & = 0 \\ x & = -\frac{74}{27},3 \\ \begin{pmatrix}x\\y\end{pmatrix} = \bigg\{ \begin{pmatrix}-74/27\\-35/27\end{pmatrix} , \begin{pmatrix}3\\1\end{pmatrix} \bigg\} & \\ \end{aligned}

    Substituting \displaystyle{x=\frac{7-3y}{2}} for x:

    \begin{aligned} y^2 & = 26-\bigg(\frac{7-3y}{2}\bigg)^2 \\ 0 & = 13y^2 -42y -55 \\ 0 & = (13y-55)(y+1) \\ y & = -1,\frac{55}{13} \\ \begin{pmatrix}x\\y\end{pmatrix} & = \bigg\{ \begin{pmatrix}5\\-1\end{pmatrix} , \begin{pmatrix}74/26\\55/13\end{pmatrix}\bigg\} \\ \end{aligned}

    Substituting \displaystyle{x=\frac{-7-3y}{5}} for x:

    \begin{aligned} 3y^2 & = \bigg(\frac{-7-3y}{5}\bigg)^2-4y+3 \\ 0 & = 33y^2+29y-62 \\ 0 & = (33y+62)(y-1) \\ y & = 1,-\frac{62}{33}\\ \begin{pmatrix}x\\y\end{pmatrix} & = \bigg\{\begin{pmatrix} -2 \\1\end{pmatrix},\begin{pmatrix} -3/11\\-62/33\end{pmatrix}\bigg\} \\ \end{aligned}

    Extracted from A. Godman & J. F. Talbert. (1975). Additional Mathematics Pure and Applied in SI Units.

    This problem is not to be attempted.