Question

Rd sharma PAGE no. 3.45

Answer 1

Hey there !!
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22 / (x+y) + 15 / (x–y) = 5 ...▶
55 / (x+y) + 45 / (x–y) = 14 ...◀

Let 1 / (x+y) = a and 1 / (x–y) = b

Now, 22a + 15b = 5 ...(1)
55a + 45b = 14 ...(2)

From (1)
22a = 5–15b
a = (5–15b) / 22

Substituting a's value in (2)
55 [ (5–15b) / 22 ] + 45b = 14
275 – 825b + 990b = 308
165b = 308 – 275
165b = 33
b = 33/165
b = 1/5

Putting b = 1/5 in (1)
22a + 15(1/5) = 5
110a + 15 = 25
110a = 25 – 15
110a = 10
a = 10/110
a = 1/11

So, 1/(x+y) = a = 1/11
=> x + y = 11 ...(i)

1/(x–y) = b = 1/5
=> x – y = 5 ...(ii)

Subtracting (i) and (ii)

x + y = 11
x – y = 5
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2y = 6
y = 6/3
y = 2

Putting y = 2 in (ii)
x – 2 = 5
x = 5 + 2
x = 7

Thus, x and y's values are 7 and 2 respectively.
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Hope my ans.'s helpful. (^-^)