Question

(22/x+y) + (15/x-y) = 5 and (55/x+y) (45/x-y) = 14 solve by simultaneus linear equations

Answer 1

Answer:

$x = 8 \: and \: y = - 3$

Step-by-step explanation:

$\frac{22}{x + y} + \frac{15}{x - y} = 5 \: \: \: \: \: \: - ( i)$

$\frac{55}{x + y} + \frac{45}{ x - y} = 14 \: \: \: \: \: - (ii)$

$let \: \frac{1}{x + y} = a \: \: and \: \frac{1}{x - y} = b$

$(i) = > 22a + 15b = 5 \: \: \: \: - (iii)$

$(ii) = > 55a + 45b = 14 \: \: \: \: - (iv)$

Subtract equation (iii) from equation (iv)...

$= > 33a + 30b = 9 \\ divide \: by \: 3 \\ = > 11a + 10b = 3 \: \: \: \: - (v)$

From (v) derive value of 'a'...

$a = \frac{3 - 10b}{11}$

Insert value of 'a' in equation (iii)

$22( \frac{3 - 10b}{11} ) + 15b = 5 \\ = > 2(3 - 10b) + 15b = 5 \\ = > 6 - 20b + 15b = 5 \\ = > - 5b = - 1 \\ = > b = \frac{1}{5}$

Insert value of 'b' in value of 'a'

$a = \frac{3 - 10 \times ( \frac{1}{5}) }{11} \\ = > \frac{3 - 2}{11} \\ = > \frac{1}{11}$

$therefore \: \frac{1}{x + y} = a = \frac{1}{11} \: \\ and \: \frac{1}{x - y} = b = \frac{1}{5}$

Now, By reciprocal,

$x + y = 5 \: \: \: \: \: - (vi)$

$x - y = 11 \: \: \: \: \: - (vii)$

Add equation (vi) and equation (vii)...

$2x = 16 \\ = > x = 8$

Insert value of 'x' in equation (vi)

$8 + y = 5 \\ = > y = - 3$