`sqrt(1-2x)-sqrt(16-x) = 5`

I will rearrange the above equation as,

`sqrt(1-2x) = 5 -sqrt(16-x)`

Taking the squared value of both sides,

`(sqrt(1-2x))^2 = (5 - sqrt(16-x))^2`

`(1-2x)= 25 - 10sqrt(16-x)+(16-x)`

Separating x terms and sqrt terms on to either side.

`10sqrt(16-x) = 25+16-1+2x-x`

`10sqrt(16-x) = x+40`

Again taking the squared value of bothe sides,

`(10sqrt(16-x))^2 = (x+40)^2`

`100(16-x) = x^2+80x+1600`

This gives a quadratic equation as below,

`x^2+180 = 0`

`x(x+180) = 0`

Therefore `x = 0 ` or `x =-180` .

Now try to substitute these answers in the original equation and check whether these satisfy it.

`x = 0,`

`LHS = sqrt(1)-sqrt(16) = 1 -4 = -3`

Therefore , `LHS != RHS` ,

`x = 0` is not a solution

`x = -180`

`LHS = sqrt(1+360) - sqrt(16+180)`

`= sqrt(361) -sqrt(196)`

`= 19 - 14`

`= 5`

`= RHS`

Therefore x =-180 is the correct answer.

**The answer is x = -180.**