Q. 1 D5.0( 4 Votes )

# Solve the followi

The given equations are  = – 2 … (1) + = 13 … (2)

Let a = and b = .

5a – 4b + 2 = 0 … (3)

2a + 3b – 13 = 0 … (4)

For cross multiplication method, we write the coefficients as Hence, we get = =  = =  = = a = = 2

b = = 3

When a = 2, = 2. Thus, x = 1/2.

When b = 3, = 3. Thus, y = .

( , ) is the solution to the given system.

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