Q12 of 244 Page 113

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

Thus, ,

Solve each of the following systems of equations by using the method of cross multiplication:

ax – by = 2ab.

Solve each of the following systems of equations by using the method of cross multiplication:

Where x ≠ 0 and y ≠ 0

Show that each of the following systems of equations has a unique solution and solve it:

3x + 5y = 12, 5x + 3y = 4.

Solve each of the following systems of equations by using the method of cross multiplication:2ax + 3by = (a + 2b).3ax + 2by = (2a + b).