Solve each of the following systems of equations by using the method of cross multiplication:
x + 2y + 1 = 0, 2x – 3y – 12 = 0.
We have,
x + 2y + 1 = 0 …(i)
2x – 3y – 12 = 0 …(ii)
From equation (i), we get a1 = 1, b1 = 2 and c1 = 1
And from equation (ii), we get a2 = 2, b2 = - 3 and c2 = - 12
Using cross multiplication,
⇒
⇒
⇒
⇒ and
⇒ and
⇒ x = 3 and y = - 2
Thus, x = 3, y = - 2