Solve: (a+2b)x + (2a-b)y=2 (a-2b)x + …

The given system of equations are :
(a + 2b)x + (2a - b)y = 2
So, (a + 2b)x + (2a - b)y -  2 = 0 ............(i)
And (a - 2b) x + (2a + b) y = 3
So, (a - 2b)x + (2a + b)y - 3 = 0 .........(ii)
The given equations is in the form of
a₁x + b_1y + c₁ = 0
and  a₂x + b_2y + c₂ = 0
So, we get
a₁ = a + 2b, b₁ = 2a - b, c₁ = -(2)
a₂ = a - 2b, b₂ = 2a + b, c₂ = (-3)
By cross-multiplication method:
{tex} \frac{x}{{ - 2a + 5b}} = \frac{y}{{a + 10b}} = \frac{1}{{10ab}}{/tex}
Now, {tex}\frac{x}{{ - 2a + 5b}} = \frac{1}{{10ab}} {/tex}
{tex} ⇒ x = \frac{{5b - 2a}}{{10ab}}{/tex}
And {tex}\frac{y}{{a + 10b}} = \frac{1}{{10ab}} {/tex}
{tex} ⇒ y = \frac{{a + 10b}}{{10ab}}{/tex}
The solution of the system of equations are {tex}x = \frac{{5b - 2a}}{{10ab}}{/tex} and {tex}y = \frac{{a + 10b}}{{10ab}}{/tex} respectively.