Solve each of the following systems of equations by using the method of cross multiplication:
2ax + 3by = (a + 2b).
3ax + 2by = (2a + b).
We have,
2ax + 3by – (a + 2b) = 0 …(i)
3ax + 2by – (2a + b) = 0 …(ii)
From equation (i), we get a1 = 2a, b1 = 3b and c1 = - (a + 2b)
And from equation (ii), we get a2 = 3a, b2 = 2b and c2 = - (2a + b)
Using cross multiplication,
⇒
⇒
⇒
⇒ and
⇒ and
⇒ and
Thus, ,