Question

Pls solve Question 15 second part

Answer 1

Answer:

3,2

Step-by-step explanation:

(ii)

The given equations are,

(i) (10/x + y) + (2/x - y) = 4

(ii) (15/x + y) - (5/x - y) = -2

Let (1/x + y) = u and (1/x - y) = v.

Then, the equations can be written as,

(iii) 10u + 2v = 4

(iv) 15u - 5v = -2

On solving (iii) * 6 & (iv) * 4, we get

⇒ 60u + 12v = 24

⇒ 60u - 20v = -8

   ------------------------

           32v = 32

          v = 1

Substitute v = 1 in (iii), we get

⇒ 10u + 2v = 4

⇒ 10u + 2(1) = 4

⇒ 10u + 2 = 4

⇒ 10u = 4 - 2

⇒ 10u = 2

⇒ u = 1/5

So, 1(x + y) = 1/5 and (1/x - y) = 1.

∴ x + y = 5 and x - y = 1

On solving both, we get

⇒ x + y = 5

⇒ x - y = 1

   ------------

  2x = 6

    x = 3.

Substitute x = 3 in above equations, we get

⇒ x + y = 5

⇒ 3 + y = 5

⇒ y = 2.

∴ The values of x and y are 3 & 2.

Hope it helps!

Answer 2

10/x+y +2/x-y=4________(1)

15/x+y-5/x-y=-2________(2)

let 1/x+y=A and 1/x-y=Bua

therefore eqn 1 and 2 will become

10A+2B=4

15A-5B=-2

Now multiplying eqn 1 by -5 and eqn 2 by 2 we get

-50A-10B=-20

30A-10B=-4

Now by elimination method we have A=16/80=1/5

and B=1

but 1/x+y=A and 1/x-y =B

putting the values of A and B we get x+y=5____(3) and x-y=1______(4)

Now from eqn 3 we get x=5-y Now putting the value of x in eqn 4 we get y=2 and x=5-y i.e x=5-2=3. Therefore x=3 and y=2