Question

$$\begin{gathered} 6\left(ax+by\right)=3a+2b, \\ 6\left(bx-ay\right)=3b-2a. \end{gathered}$$

Answer 1

The given equations are:
6(ax + by) = 3a + 2b
⇒ 6ax + 6by = 3a + 2b ...............(i)

and 6(bx − ay) = 3b − 2a
⇒ 6bx − 6ay = 3b − 2a ...................(ii)

On multiplying (i) by a and (ii) by b, we get:
6a²x + 6aby = 3a² + 2ab ................(iii)
6b²x − 6aby = 3b² − 2ab ....................(iv)

On adding (iii) and (iv), we get:
6(a² + b²)x = 3(a² + b²)
$\Rightarrow x=\frac{3\left(a^2+b^2\right)}{6\left(a^2+b^2\right)}=\frac{1}{2}$

On substituting $x=\frac{1}{2}$ in (i), we get:
$6a\times \frac{1}{2}+6by=3a+2b$
⇒ 3a + 6by = 3a + 2b
⇒ 6by = 2b
⇒ y = $\frac{2b}{6b}=\frac{1}{3}$
Hence, the required solution is $x=\frac{1}{2}$ and $y=\frac{1}{3}$.