Question

if 6/x=1/a+1/b. show that (x+3a)/(x-3a)+ (x+3b)/(x-3b)=2, a≠b

Answer 1

Answer:

Step-by-step explanation:

6/x = 1/a + 1/b

=> 6/x = a + b / ab

=> x = 6ab/a+b

=> x/3a = 2b/a+b - ----- (1)

Apply compendo and divdendo using (1),

x + 3a/ x-3a  = 2b + a + b / 2b - (a+b)  = 3b + a / b - a

Again,

x = 6ab/a + b =>  x/3b = 2a/a+b  ------------- (2)

Apply Compendo and dividendo on (2)

x + 3b/ x - 3b      = 2a + a + b / 2a -(a+b) = 3a + b / a - b

L.H.S:

(x+3a)/(x-3a) + (x+3b)/(x-3b)

= 3b + a / b - a   + 3a + b / a- b

= 1/a-b[ - 3b - a + 3a + b]

= 2a - 2b / a - b

= 2(a - b) / a - b

= 2.

= R.H.S

Hence proved.