Q. 1 C4.2( 25 Votes )

# <span lang="EN-US and Let be p and be q

2p + 3q = 2 …III

4p – 9q = -1 …IV

Multiplying eq. III by 3 and subtracting with eq. IV, we get

6p + 9q = 6

4p – 9q = -1

10p = 5

p = substitute value of p in Eq. III + 3q = 2

3q = 2 – 1

q = But,   x = 4   y = 9

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