Question

question:-
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Answer 1

Step-by-step explanation:

Given :-

[10/(x+y)] + [2/(x-y)] = 4

[15/(x+y)] - [5/(x-y)] = -2

To find :-

Solve for x and y ?

Solution :-

Given equations are :

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

[15/(x+y)] - [5/(x-y)] = -2--------(2)

Put 1/(x+y) = a and 1/(x-y) = b then

(1) becomes

10a + 2b = 4

=> 2(5a+b) = 4

=> 5a + b = 4/2

=> 5a + b = 2 --------------(3)

=> b = 2-5a -----------------(4)

and

(2) becomes

15a - 5b = -2 --------------(5)

On Substituting the value of b in (5)

=> 15a - 5(2-5a) = -2

=> 15a -10+25a = -2

=> 40a -10 = -2

=> 40a = -2+10

=> 40a = 8

=> a = 8/40

=> a = 1/5

On Substituting the value of a in (4)

=> b = 2-5(1/5)

=> b = 2-(5/5)

=> b = 2-1

=> b = 1

We have,

a = 1/5

=> 1/(x+y) = 1/5

=> x+y = 5---------------(6)

and

b = 1

=> 1/(x-y) = 1

=> x-y = 1 ---------------(7)

On adding (6)&(7)

x+y = 5

x-y = 1

(+)

________

2x+0 = 6

________

=> 2x = 6

=> x = 6/2

=> x = 3

On Substituting the value of x in (6)

=> 3+y = 5

=> y = 5-3

=> y = 2

Therefore, x = 3 and y = 2

Answer:-

The value of x = 3 and y = 2

The solution for the given problem is (3,2)

Check:-

If x = 3 and y = 2 then LHS of (1)

10/(3+2) + 2/(3-2)

=> 10/5 + 2/1

=> 2+2

=> 4

=> LHS = RHS is true for x= 3 and y = 2

If x = 3 and y = 2 then LHS of (2)

15/(3+2) - 5/(3-2)

=> 15/5 - 5/1

=> 3-5

=> -2

=> LHS = RHS is true for x= 3 and y = 2

Verified the given relations in the given problem.

Used Methods :-

• Reducible to the linear equations in two variables
• Substitution method.