Question

Solve the simultaneous equation:
$\frac{148}{x}+\frac{231}{y}=\frac{527}{xy}$; $\frac{231}{x}+\frac{148}{y}=\frac{610}{xy}$

Answer 1

Given

$\frac{148}{x}+\frac{231}{y}=\frac{527}{xy}$

$\frac{231}{x}+\frac{148}{y}=\frac{610}{xy}$

Multiply both equation with xy, we get

$148y+231x=\mathrm{527.....}\left(1\right)$

$231y+148x=\mathrm{610......}\left(2\right)$

Multiply (1) by 148 and (2) by 231, we get

$34188x+21904y=\mathrm{77996....}\left(3\right)$

$34188x+53361y=\mathrm{140910....}\left(4\right)$

Subtracting (3) from (4), we get

$53361y-21904y=140910-77996$

$31457y=62914$

$\therefore y=2$

Put y=2 in (1)

$231x+148\left(2\right)=527$

$231x=527-296$

$231x=231$

$\therefore x=1$