Question

solve by applying method of equations reducible to linear equation......plz be fast.......
that person will be followed who will give the right ans​

Answer 1

Solution :-

Given pair of linear equations :

→ 20/(x + y) + 3/(x - y) = 7

→ 8/(x - y) - 15/(x + y) = 5

Sustituting 1/(x + y) = a and 1/(x - y) = b in the equations we get,

→ 20a + 3b = 7 --- eq(1)

→ - 15a + 8b = 5 ---- eq(2)

LCM of 20, 15 is 60

Multiplying eq(1) by 3 and eq(2) by 4 we get,

(20a + 3b = 7) * 3

( - 15a + 8b = 5) * 4

→ 60a + 9b = 21 -- eq(3)

→ - 60a + 32b = 20 --- eq(4)

Adding eq(3) and eq(4)

⇒ 60a + 9b + ( - 60a + 32b ) = 21 + 20

⇒ 60a + 9b - 60a + 32b = 41

⇒ 41b = 41

⇒ b = 41/41 = 1

But, b = 1/(x - y)

⇒ 1/(x - y) = 1

⇒ x - y = 1 --- eq(5)

Substituting b = 1 in eq(1)

⇒ 20a + 3b = 7

⇒ 20a + 3(1) = 7

⇒ 20a + 3 = 7

⇒ 20a = 7 - 3

⇒ 20a = 4

⇒ a = 4/20 = 1/5

But, a = 1/(x + y)

⇒ 1/(x + y) = 1/5

⇒ x + y = 5 -- eq(6)

Adding eq(5) and eq(6)

⇒ x - y + ( x + y ) = 1 + 5

⇒ x - y + x + y = 6

⇒ 2x = 6

⇒ x = 6/2 = 3

Substituting x = 3 in x + y = 5

⇒ 3 + y = 5

⇒ y = 5 - 3 = 2

Hence, the required solution is x = 3 and y = 2.