Question

Solve the system of equations by using the method of cross multiplication:
2ax + 3by – (a + 2b) = 0,
3ax + 2by – (2a + b) = 0

Answer 1

The given equations may be written as:
2ax + 3by – (a + 2b) = 0 ……(i)
3ax + 2by – (2a + b) = 0 ……(ii)
Here, `a_1 =2a, b_1 =3b, c_1 = -(a + 2b), a_2 = 3a, b_2 = 2b and c_2 = -(2a + b)`
By cross multiplication, we have

`∴ x/([3b ×(−(2a+ b) −2b) ×(−(a+2b))]) = y/([−(a+2b) × 3a −2a ×(−(2a+b))]) = 1/([2a ×2b −3a ×3b])`
`⇒x/((−6ab−3b^2+2ab+4b^2)) = y/((−3a^2−6ab+4a^2+2ab)) = 1/(4ab−9ab)`

`⇒x/(b^2−4ab) = y/(a^2 −4ab) = 1/(−5ab)`
`⇒ x/(−b(4a−b)) = y/(−a(4b−a)) = 1/(−5ab)`
`⇒x = (−b(4a−b))/(−5ab )= ((4a−b))/(5a,) y = (−a(4b−a))/(−5ab) = ((4b−a))/(5b)`
Hence, x = `((4a−b))/(5a) and y = ((4b−a))/(5b)` is the required solution.