Solve: 2x+3y=xy and 3x-4y=2xy by Elemination and Substitution Method
Elemination Method
2x+3y=xy —-(i)
3x-4y=2xy —-(ii)
From Equation (i) we get,
$2x + 3y = xy$
$\Rightarrow \frac{2x + 3y}{xy} = 1$
$\Rightarrow \frac{2x}{xy} + \frac{3y}{xy} = 1$
$\Rightarrow \frac{2}{y} + \frac{3}{x} = 1$ —(iii)
From Equation (ii) we get,
$3{\mathrm{x}} – 4{\mathrm{y}} = 2{\mathrm{xy}}$
$\Rightarrow \frac{3{\mathrm{x}} – 4{\mathrm{y}}}{2{\mathrm{xy}}} = 1$
$\Rightarrow \frac{3{\mathrm{x}}}{2{\mathrm{xy}}} – \frac{4{\mathrm{y}}}{2{\mathrm{xy}}} = 1$
$\Rightarrow \frac{3}{2{\mathrm{y}}} – \frac{2}{{\mathrm{x}}} = 1$ —(iv)
Now multiplying equation (iii) by 2 and equation (iv) by 3 we get,
$\frac{4}{{\mathrm{y}}} + \frac{6}{{\mathrm{x}}} = 2$ —-(v)
$\frac{9}{2{\mathrm{y}}} – \frac{6}{{\mathrm{x}}} = 3$ —(vi)
Adding equations (v) and (vi) we get,
$\frac{4}{{\mathrm{y}}} + \frac{9}{2{\mathrm{y}}} = 5$
$\Rightarrow \frac{8 + 9}{2{\mathrm{y}}} = 5$
$\Rightarrow \frac{17}{2{\mathrm{y}}} = 5$
$\Rightarrow {\mathrm{y}} = \frac{17}{2 \times 5}$
$\Rightarrow {\mathrm{y}} = \frac{17}{10}$
Substituting the value of y in equation (i) we get,
$2{\mathrm{x}} + 3\left(\frac{17}{10}\right) = {\mathrm{x}}\left(\frac{17}{10}\right)$
$\Rightarrow 2{\mathrm{x}} + \frac{51}{10} = \frac{17{\mathrm{x}}}{10}$
$\Rightarrow 2{\mathrm{x}} – \frac{17{\mathrm{x}}}{10} = – \frac{51}{10}$
$\Rightarrow \frac{20{\mathrm{x}} – 17{\mathrm{x}}}{10} = – \frac{51}{10}$
$\Rightarrow \frac{3{\mathrm{x}}}{10} = – \frac{51}{10}$
$\Rightarrow 3{\mathrm{x}} = – 51$
$\Rightarrow {\mathrm{x}} = \frac{ – 51}{3}$
$\Rightarrow {\mathrm{x}} = – 17$
∴ Required Solutions are x = – 17 and y=$\frac{17}{10}$
Substitution Method
2x+3y=xy —-(i)
3x-4y=2xy —-(ii)
From equation (i) we get,
$2{\mathrm{x}} = {\mathrm{xy}} – 3{\mathrm{y}}$
$\Rightarrow 2{\mathrm{x}} = {\mathrm{y}}({\mathrm{x}} – 3)$
$\Rightarrow {\mathrm{y}} = \frac{2{\mathrm{x}}}{{\mathrm{x}} – 3}$ —(iii)
Substituting the value of y in equation (ii) we get,
$3{\mathrm{x}} – 4\left(\frac{2{\mathrm{x}}}{{\mathrm{x}} – 3}\right) = 2{\mathrm{x}}\left(\frac{2{\mathrm{x}}}{{\mathrm{x}} – 3}\right)$
$\Rightarrow 3{\mathrm{x}} – \frac{8{\mathrm{x}}}{{\mathrm{x}} – 3} = 2{\mathrm{x}}\left(\frac{2{\mathrm{x}}}{{\mathrm{x}} – 3}\right)$
$\Rightarrow \frac{3{\mathrm{x}}({\mathrm{x}} – 3) – 8{\mathrm{x}}}{{\mathrm{x}} – 3} = \frac{4{\mathrm{x}}^2}{{\mathrm{x}} – 3}$
$\Rightarrow 3{\mathrm{x}}^2 – 9{\mathrm{x}} – 8{\mathrm{x}} = 4{\mathrm{x}}^2$
$\Rightarrow 3{\mathrm{x}}^2 – 4x^2 – 17x = 0$
$\Rightarrow – {\mathrm{x}}^2 – 17{\mathrm{x}} = 0$
$\Rightarrow {\mathrm{x}} + 17 = 0$
$\Rightarrow {\mathrm{x}} = – 17$
From equation (iii) we get,
$\therefore {\mathrm{y}} = \frac{2{\mathrm{x}}}{{\mathrm{x}} – 3}$
$\Rightarrow {\mathrm{y}} = \frac{2\left( – 17\right)}{ – 17 – 3}$
$\Rightarrow {\mathrm{y}} = \frac{ – 34}{ – 20}$
$\Rightarrow {\mathrm{y}} = \frac{17}{10}$
∴ Required Solutions are x = – 17 and y=$\frac{17}{10}$