If $29 x+37 y=103$ and $37 x+29 y=95$ then (a) x=1, y=2 (b) x=2, y=1 (c) x=3, y=2 (d) x=2, y=3

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If $29 x+37 y=103$ and $37 x+29 y=95$ then
(a) x=1, y=2
(b) x=2, y=1
(c) x=3, y=2
(d) x=2, y=3

If $29 x+37 y=103$ and $37 x+29 y=95$ then
(a) x=1, y=2
(b) x=2, y=1
(c) x=3, y=2
(d) x=2, y=3

Answer: $(a) x=1, y=2$

Solution:
The given system is
$29 x+37 y=103\dots \dots(i)$
$37 x+29 y=95\dots \dots(ii)$
Adding equation(i) and equation(ii), we get
$66 x+66 y=198$
$\Rightarrow x+y=3\dots \dots(iii)$
Subtracting equation(i) from equation(ii), we obtain
$\begin{array}{l} 8 x-8 y=-8 \\ \Rightarrow x-y=-1\dots \dots(iv) \end{array}$
Adding equation(iii) and equation(iv), we get
$2 x=2 \Rightarrow x=1$
Substituting $\mathrm{x}=1$ in equation(iii), we have
$1+y=3 \Rightarrow y=2$
As a result, $\mathrm{x}=1$ and $\mathrm{y}=2$ .