Q. 33

# Solve the following systems of equations:

Answer :

Multiplying eq1 by 5 and eq2 by 2 and subtracting

⇒ 13/(3x – 2y) = 13

⇒ 3x – 2y = 1------- (3)

Multiplying eq2 by 3 and subtracting eq1 from it

⇒ 13/(3x +2y) = 13/5

⇒ 3x + 2y = 5 ------ (4)

(3) + (4)

⇒ 6x = 6

⇒ x = 1

Thus, 3 + 2 = 5

⇒ y = 1

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