3.3.4 System of Equations, Long Questions (Question 7 & 8)

Question 7:

Solve the following simultaneous equations.

5y – 6x= 2

\frac{4 y}{x} - \frac{3 x}{y} = 4.

$\frac{4 y}{x} - \frac{3 x}{y} = 4.$

Solution:

5y – 6x= 2 —– (1)

\begin{matrix} \frac{4 y}{x} - \frac{3 x}{y} = 4 ------- (2) From (1), y = \frac{2 + 6 x}{5} \end{matrix}

$\begin{array}{l} \frac{4 y}{x} - \frac{3 x}{y} = 4 ------- (2) \\ From (1), \\ y = \frac{2 + 6 x}{5} \end{array}$

Substitute (3) into (2),

\begin{matrix} \frac{4 (\frac{2 + 6 x}{5})}{x} - \frac{3 x}{(\frac{2 + 6 x}{5})} = 4 \frac{8 + 24 x}{5 x} - \frac{15 x}{2 + 6 x} = 4 \frac{(8 + 24 x) (2 + 6 x) - (15 x) (5 x)}{5 x (2 + 6 x)} = 4 \end{matrix}

$\begin{array}{l} \frac{4 (\frac{2 + 6 x}{5})}{x} - \frac{3 x}{(\frac{2 + 6 x}{5})} = 4 \\ \frac{8 + 24 x}{5 x} - \frac{15 x}{2 + 6 x} = 4 \\ \frac{(8 + 24 x) (2 + 6 x) - (15 x) (5 x)}{5 x (2 + 6 x)} = 4 \end{array}$

16 + 48x + 48x + 144x² – 75x² = 20x (2 + 6x)

69x² + 96x + 16 = 40x + 120x²

51x² – 56x – 16 = 0

(3x – 4)(17x + 4) = 0

3x – 4 = 0 or 17x + 4 = 0

x = \frac{4}{3} or x = - \frac{4}{17}

$x = \frac{4}{3} or x = - \frac{4}{17}$

\begin{matrix} Substitute x = \frac{4}{3} into (3), y = \frac{2 + 6 (\frac{4}{3})}{5} = 2 \end{matrix}

$\begin{array}{l} Substitute x = \frac{4}{3} into (3), \\ y = \frac{2 + 6 (\frac{4}{3})}{5} = 2 \end{array}$

\begin{matrix} Substitute x = - \frac{4}{17} into (3), y = \frac{2 + 6 (- \frac{4}{17})}{5} = \frac{2}{17} \end{matrix}

$\begin{array}{l} Substitute x = - \frac{4}{17} into (3), \\ y = \frac{2 + 6 (- \frac{4}{17})}{5} = \frac{2}{17} \end{array}$

∴ The solutions are (\frac{4}{3}, 2) and (- \frac{4}{17}, \frac{2}{17}) .

$∴ The solutions are (\frac{4}{3}, 2) and (- \frac{4}{17}, \frac{2}{17}) .$

Question 8:
Solve the following simultaneous equations.

\begin{matrix} x + 2 y = 1 2 x^{2} + y^{2} + x y = 5 \end{matrix}

$\begin{array}{l} x + 2 y = 1 \\ 2 x^{2} + y^{2} + x y = 5 \end{array}$
Give your answer correct to three decimal places.

Solution:

\begin{matrix} x + 2 y = 1.......... (1) 2 x^{2} + y^{2} + x y = 5.......... (2) x = 1 - 2 y .......... (3) Substitute (3) into (2), 2 {(1 - 2 y)}^{2} + y^{2} + (1 - 2 y) y = 5 2 (1 - 4 y + 4 y^{2}) + y^{2} + y - 2 y^{2} = 5 2 - 8 y + 8 y^{2} - y^{2} + y - 5 = 0 7 y^{2} - 7 y - 3 = 0 a = 7, b = - 7, c = - 3 x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} y = \frac{- (- 7) \pm \sqrt{{(- 7)}^{2} - 4 (7) (- 3)}}{2 (7)} y = \frac{7 \pm \sqrt{133}}{14} y = 1.324 or - 0.324 From x = 1 - 2 y When y = 1.324, x = 1 - 2 (1.324) = 1.648 When y = - 0.324, x = 1 - 2 (- 0.324) = 1.648 ∴ The solutions are (1.648, 1.324) and (1.648, - 0.324) . \end{matrix}

$\begin{array}{l} x + 2 y = 1.......... (1) \\ 2 x^{2} + y^{2} + x y = 5.......... (2) \\ x = 1 - 2 y .......... (3) \\ Substitute (3) into (2), \\ 2 {(1 - 2 y)}^{2} + y^{2} + (1 - 2 y) y = 5 \\ 2 (1 - 4 y + 4 y^{2}) + y^{2} + y - 2 y^{2} = 5 \\ 2 - 8 y + 8 y^{2} - y^{2} + y - 5 = 0 \\ 7 y^{2} - 7 y - 3 = 0 \\ a = 7, b = - 7, c = - 3 \\ x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ y = \frac{- (- 7) \pm \sqrt{{(- 7)}^{2} - 4 (7) (- 3)}}{2 (7)} \\ y = \frac{7 \pm \sqrt{133}}{14} \\ y = 1.324 or - 0.324 \\ From x = 1 - 2 y \\ When y = 1.324, \\ x = 1 - 2 (1.324) = 1.648 \\ When y = - 0.324, \\ x = 1 - 2 (- 0.324) = 1.648 \\ ∴ The solutions are (1.648, 1.324) and (1.648, - 0.324) . \end{array}$

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