Question

Find the value of x and y.
$\frac{5}{x+y}+\frac{3}{x-y}=4\frac{2}{x+y}+\frac{5}{x-y}=5\frac{2}{5}$

• A. 1, 3
• B. 2, 5
• C. 3, 2
• D. 4, 6
• E. 5, 8

Answer 1

The correct option is C

3, 2

$Let\frac{1}{x+y}=a...\left(i\right),\frac{1}{x+y}=b$...(ii)

5a+3b=4 ...........(iii)

$2a+5b=\frac{27}{5}.............\left(iv\right)$

Multiplying equation (iii) by 2 and equation (iv) by 5 and subtracting (iv) from (iii) we get

10a+6b - (10a+25b) = 8 - 27

$⟹$ -19b=-19

$⟹$b=$\frac{-19}{-19}$=1

Substituting value of b in equation (iii), we get

5a+3 $\times$1=4

$⟹$ 5a+3=4

$⟹$ 5a=1

$a=\frac{1}{5}Now,\frac{1}{x+y}=aand\frac{1}{x-y}=b\therefore \frac{1}{x+y}=\frac{1}{5}and\frac{1}{x-y}=1$

$\therefore$ x+y=5..............(v) and x-y=1 .................(vi)

Subtracting (vi) from (v), we get

x + y - x + y = 5 - 1

$⟹$ 2y=4

$⟹$ y=2

Substituting value of y in equation (vi) we get

x-2=1

$⟹$x=3

$\therefore$ x=3, y=2