Question

If $\frac{2x}{3}-\frac{y}{2}+\frac{1}{6}=0\mathrm{and}\frac{x}{2}+\frac{2y}{3}=3$ then
(a) x = 2, y = 3 (b) x = − 2, y = 3 (c) x = 2, y = − 3 (d) x = − 2, y = − 3

Answer 1

The given system of equations is
$$\begin{gathered} \frac{2x}{3}-\frac{y}{2}=-\frac{1}{6}.....\left(i\right) \\ \frac{x}{2}+\frac{2y}{3}=3.....\left(ii\right) \end{gathered}$$
Multiplying (i) and (ii) by 6, we get
$$\begin{gathered} 4x-3y=-1.....\left(iii\right) \\ 3x+4y=18.....\left(iv\right) \end{gathered}$$
Multiplying (iii) by 4 and (iv) by 3 and adding, we get
$$\begin{gathered} 16x+9x=-4+54 \\ \Rightarrow x=\frac{50}{25}=2 \end{gathered}$$
Now, putting x = 2 in (iv), we have
$3\times 2+4y=18\Rightarrow y=\frac{18-6}{4}=3$
Thus, x = 2 and y = 3.
Hence, the correct answer is option (a).