Question

$$\begin{gathered} 2\left(ax-by\right)+\left(a+4b\right)=0, \\ 2\left(bx+ay\right)+\left(b-4a\right)=0. \end{gathered}$$

Answer 1

The given equations are:
2(ax − by) + (a + 4b) = 0
⇒ 2ax − 2by + a + 4b = 0 ...............(i)

and 2(bx + ay) + b − 4a = 0
⇒ 2bx + 2ay + b − 4a = 0 ...................(ii)

On multiplying (i) by a and (ii) by b, we get:
2a²x − 2aby + a² + 4ab = 0 ................(iii)
2b²x + 2aby + b² − 4ab = 0 ....................(iv)

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

On substituting $x=\frac{-1}{2}$ in (i), we get:
$2a\times \frac{-1}{2}-2by+a+4b=0$
⇒ −2by + 4b = 0
⇒ 2by = 4b
⇒ y = $\frac{4b}{2b}=2$

Hence, the required solution is $x=\frac{-1}{2}$ and y = 2.