Question

Question 1 (v)
Solve the following pairs of equations by reducing them to a pair of linear equations:
$\frac{7x-2y}{xy}=5$
$\frac{8x+7y}{xy}=15$

Answer 1

$\frac{7x-2y}{xy}=5$
$\Rightarrow \frac{7x}{xy}-\frac{2y}{xy}=5$
$\Rightarrow \frac{7}{y}-\frac{2}{x}=\mathrm{5....}\left(i\right)$
$\frac{8x+7y}{xy}=15$
$\Rightarrow \frac{8x}{xy}+\frac{7y}{xy}=15$
$\Rightarrow \frac{8}{y}+\frac{7}{x}=\mathrm{15...}\left(ii\right)$
$Putting\frac{1}{x}=pand\frac{1}{y}=q\text{in (i) and (ii) we get,}$
7q - 2p = 5 ... (iii)
8q + 7p = 15 ... (iv)
Multiplying equation (iii) by 7 and multiplying equation (iv) by 2 we get,
49q - 14p = 35 ... (v)
16q + 14p = 30 ... (vi)
Now, adding equation (v) and (vi) we get,
49q - 14p + 16q + 14p = 35 + 30
$\Rightarrow$ 65q = 65
$\Rightarrow$ q = 1
Putting the value of q in equation (iv),
8 + 7p = 15
$\Rightarrow$ 7p = 7
$\Rightarrow$ p = 1
Now, $p=\frac{1}{x}=1$
$\Rightarrow \frac{1}{x}=1$
$\Rightarrow x=1$ also,
$q=1=\frac{1}{y}$
$\Rightarrow \frac{1}{y}=1$
$\Rightarrow y=1$
Hence, x =1 and y = 1 is the solution.