Question

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

Answer 1

The given equations are
$$\begin{gathered} 29x+37y=103.....\left(i\right) \\ 37x+29y=95.....\left(ii\right) \end{gathered}$$
Adding (i) and (ii), we get
$$\begin{gathered} 66x+66y=198 \\ \Rightarrow x+y=3.....\left(iii\right) \end{gathered}$$
Subtracting (i) from (ii), we get
$$\begin{gathered} 8x-8y=-8 \\ \Rightarrow x-y=-1.....\left(iv\right) \end{gathered}$$
Adding (iii) and (iv), we get
$2x=2\Rightarrow x=1$
Substituting x = 1 in (iii), we have
$1+y=3\Rightarrow y=2$
Thus, x = 1 and y = 2.
Hence, the correct answer is option (a).