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, ,


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