Solve: 3(2x+y)=7xy and 3(x+3y)=11xy

Solve: 3(2x+y)=7xy and 3(x+3y)=11xy

3(2x+y)=7xy —(i)

3(x+3y)=11xy —(ii)

From Equation (i) we get,

$3(2x + y) = 7xy$

$\Rightarrow 6x + 3y = 7xy$

$\Rightarrow \frac{6x + 3y}{xy} = 7$

$\Rightarrow \frac{6{\mathrm{x}}}{{\mathrm{xy}}} + \frac{3{\mathrm{y}}}{{\mathrm{xy}}} = 7$

$\Rightarrow \frac{6}{{\mathrm{y}}} + \frac{3}{{\mathrm{x}}} = 7$ —-(iii)

From Equation (ii) we get,

$3(x + 3y) = 11xy$

$\Rightarrow 3x + 9y = 11xy$

$\Rightarrow \frac{3x + 9y}{xy} = 11$

$\Rightarrow \frac{3x}{xy} + \frac{9y}{xy} = 11$

$\Rightarrow \frac{3}{y} + \frac{9}{x} = 11$ —-(iv)

Multiplying Equation (iv) by 2 we get,

$\Rightarrow \frac{6}{y} + \frac{18}{x} = 22$ —-(v)

Now Subtracting Equation (iii) from Equation (v) we get,

$\Rightarrow \frac{6}{y} + \frac{18}{x} – \frac{6}{{\mathrm{y}}} – \frac{3}{{\mathrm{x}}}$=22-7

$\Rightarrow \frac{18}{x} – \frac{3}{x} = 15$

$\Rightarrow \frac{15}{x} = 15$

$\Rightarrow x = \frac{15}{15}$

$\Rightarrow {\mathrm{x}} = 1$

Substituting the value of x in equation (i) we get,

3(2x+y)=7xy

⟹ 6x+3y=7xy

⟹ 6(1)+3y=7(1)y

⟹ 6+3y=7y

⟹ 6=7y-3y

⟹ 4y=6

⟹ y = $\frac{6}{4}$

⟹ y=$\frac{3}{2}$

Therefore Required Solutions are x = 1 and y =$\frac{3}{2}$