Question

Solve the system of equations by using the method of cross multiplication:

2ax + 3by – (a + 2b) = 0, 3ax + 2by – (2a + b) = 0

Solve the system of equations by using the method of cross multiplication:

2ax + 3by = (a + 2b), 3ax + 2by = (2a + b)

Answer 1

The given equations may be written as:

2ax + 3by – (a + 2b) = 0   ...(i)

3ax + 2by – (2a + b) = 0   ...(ii)

Here, a₁ = 2a, b₁ = 3b, c₁ = –(a + 2b), a₂ = 3a, b₂ = 2b and c₂ = –(2a + b)

By cross multiplication, we have

∴ `x/([3b xx (-(2a + b)) - 2b xx (-(a + 2b))]) = y/([-(a + 2b) xx 3a - 2a xx (-(2a + b))]) = 1/([2a xx 2b - 3a xx 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.