Question

Solve the pair of linear equations by reducing them to a pair of linear equations :-

$\frac{10}{x + y} + \frac{2}{x - y} = 4$
And,

$\frac{15}{ x + y} + \frac{5}{x - y} = - 2$
( Detailed Solution needed )

Answer 1

HELLO DEAR,

let 1/(X+Y)=P-------------(1)

AND, 1/(x-y)=q-------------(2)

now put the in both Equations
we get,

10p+2q=4---------------(3)

15p+5q=-2 --------------(4)

now ,
{subtract the both Equations
and multiply by(5)in--(3)
and by (2)in----(4)}

we get,

50p+10q=20
30p+10q= -4
- - +
--------------------
20p=24
p=6/5 -----put in Equation--(3)

we get,

10(6/5)+2q=4

2*6+2q=4

2q=4-12

q=-8/2

q= (-4)

now we get the values of
(p)and(q)

put the value of p in (1)

we get,
p=1/(x+y)

6/5=1/(x+y)

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

and put the value of (q) in --(2)

we get,

1/(x-y)= -4

=> -4x+4y=1

=> 4x-4y= -1 -------------(6)

from (5) and (2)

multiply by (4)in---(5)
and in(6) by 6

now
we get,

24x+24y=20
24x-24y= -6
----------------------
48x=14
x=7/24. ------put in (6)

we get,

4(7/24) - 4y = -1

7/6 - 4y= -1
(7-24y)/6= -1

7-24y=-6

24y=13

y=13/24

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Heya !!!

This is Your answer...

Given :- 

$\frac{10}{x+y} + \frac{2}{x-y} = 4, \: and \\ \\ \frac{15}{x+y} + \frac{5}{x-y} = -2$.

$\frac{10}{x+y} = 10 X \frac{1}{x+y} \: and \: so \: on.$.

$Let \: \frac{1}{x+y} \: be \: a \: \frac{1}{x-y} \: be \: b. \\So, \frac{10}{x+y} = 10a \: \& \: \frac{2}{x-y} = 2b, \: and \\ \\ \frac{15}{x+y} = 15 a \: \& \: \frac{5}{x-y} = 5b.$

So, according to the Question,
10a + 2b = 4      ............(i)
15a + 5b = -2     ...........(ii)

Multiply Eq. (i) by 5 and second equation by 2, we get
50a + 10b = 20               ....(iii)
30a + 10b = -4                ....(iv)

Subtract  Eq. (iv) from eq. (iii),  we get..
$. \:\:\:\:\: 50a + 10b = 20 \\ . \:\:\:\:\:30a + 10b = -4 \\ . \:\: (-) \:\:\: (-) \:\:\:\:\:\:\: (+) \\ . ---------- \\ . \:\:\:\:\: 20a \:\:\:\:\:\:\:\:\:\:\:\: = 24$
a = 24/20 = 6/5.

Putting a = 6/5 in eq. (i), 
10 X 6/5 + 2b = 4
12 + 2b = 4
2b = -8
b = -4

Now, 
$a = \frac{1}{x+y} = \frac{6}{5} \\=\ \textgreater \ 6 (x+y) = 5 \\ =\ \textgreater \ 6x + 6y = 5 .... (v) \\ \\ b = \frac{1}{x-y} = -4 \\ =\ \textgreater \ -4 (x-y) = 1 \\=\ \textgreater \ -4x + 4y = 1. \\ Multiply \: both \: sides \: by \: (-), \: we \: get \\ 4x - 4y = -1... (vi)$

Now, multiply eq. (v) by 4 and eq. (vi) by 6, we get
24x + 24y = 20, (vii)
24x - 24y = -6.   (viii)
Adding eq. (vii) and (viii), we get
$. \: \: \: \: \: \: 24x + 24y = 20 \\ + \\ .\: \: \: \: \: \: 24x - 24y = -6 \\ ----------- \\ . \: \: \: \: \: \: 48x \: \: \: \: \: \: \: \: \: \: \: \: \: = 14$
48x = 14
x= 14/48 = 7/24.
Putting x = 7/24 in eq. (v), we get
6 X 7/24 + 6y = 5
7/4 + 6y = 5
6y = 5 - 7/4
6y = 20/4 - 7/4
6y = 13/4
y = 13/4(6)
y = 13/24.

Hence, x= 7/24 and y = 13/24.

Hope This Helps