Question

$$\begin{gathered} 4x+6y=3xy, \\ 8x+9y=5xy\left(x\ne 0,y\ne 0\right). \end{gathered}$$

Answer 1

The given equations are:
4x + 6y = 3xy .......(i)
8x + 9y = 5xy .........(ii)

From equation (i), we have:

$\frac{4x+6y}{xy}=3$
$\Rightarrow \frac{4}{y}+\frac{6}{x}=3$ .............(iii)

For equation (ii), we have:

$\frac{8x+9y}{xy}=5$
$\Rightarrow \frac{8}{y}+\frac{9}{x}=5$ .............(iv)
On substituting $\frac{1}{y}=v\mathrm{and}\frac{1}{\mathrm{x}}=\mathrm{u}$, we get:
4v + 6u = 3 ...........(v)
8v + 9u = 5 ...........(vi)

On multiplying (v) by 9 and (vi) by 6, we get:
36v + 54u = 27 ..............(vii)
48v + 54u = 30 ...............(viii)

On subtracting (vii) from (viii), we get:
12v = 3 $\Rightarrow v=\frac{3}{12}=\frac{1}{4}$
$\Rightarrow \frac{1}{y}=\frac{1}{4}\Rightarrow y=4$

On substituting y = 4 in (iii), we get:
$\frac{4}{4}+\frac{6}{x}=3$
$\Rightarrow 1+\frac{6}{x}=3\Rightarrow \frac{6}{x}=\left(3-1\right)=2$
$\Rightarrow 2x=6\Rightarrow x=\frac{6}{2}=3$
Hence, the required solution is x = 3 and y = 4.