Q. 4 B3.7( 10 Votes )

# Attempt any two s

Given equations are 27/(x-2) + 31/(y+3) = 85; 31/(x-2) + 27/(y+3) = 89.

Let 1/(x-2) = a and 1/(y+3) = b

Given equations becomes

27a + 31b = 85 ………………..(1)

31a + 27b = 89 ………………..(2)

27a + 31b + 31a + 27b = 85 + 89

58a + 58b = 174

58 (a + b) = 174

a + b = 3 …………………(3)

Subtracting equation (1) and (2)

27a + 31b – 31a – 27b = 89

-4a + 4b = -4

a – b = 1 …………………(4)

a + b + a – b = 4

2a = 4

a = 4/2

a = 2

Putting value of a in (4)

2 – b = 1

b = 2 – 1

b = 1

Resubstituting

a = 1/(x – 2)

1/(x – 2) = 2

1 = 2x – 4

2x = 1 + 4

x = 5/2

b = 1/(y + 3)

1 = 1/(y + 3)

y + 3 = 1

y = 1 – 3

y = -2

x = 5/2 and y = -2

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