2022/058) Given $\frac{1}{x} + \frac{1}{y} = \frac{1}{x+y}$ find $\frac{x}{y} + \frac{y}{x} $

Multiply both sides by x+ y to get

$\frac{x+y}{x} + \frac{x+y}{y} = 1$

or $ (1+ \frac{y}{x}) + (\frac{x}{y} + 1)  = 1$

or   $ \frac{y}{x} +\frac{x}{y} + 2  = 1$

or $\frac{y}{x} + \frac{x}{y}   = -1$

2022/057) Given $(a-5)^2 + (b-7)^2 = 4$ find the minimum of $(a+7)^2 + (b+2)^2$

$(a-5)^2 + (b-7)^2 = 4$ is a circle whose centre is $(5,7)$ and radius is 2

We need to find minimum of $(a+7)^2 + (b+2)^2$ that is distance of the point A with coordinates a,b  fom (-7,-2).

This is minimum when the point lies on the straight line from (-7,2) , (5,7) . 

this distane from (-7,2) to (5,7) = $\sqrt{(5+7)^2 + (7-2)^2} = 13$

the distance of a from (5,7) is 2

so distance of point from (-7,2)is 13 - 2 = 11

so minmum of  $(a+7)^2 + (b+2)^2$ is 121

  

2022/056) What is the value of a, b, c If $2^a+4^b+8^c=328$ and a, b, c are natural numbers?

 Let us express 328 as sum of power of 2 (as 2,4,8 all are power of 2)

$328 = 256 + 64 + 8 = 2^8 + 2^ 6 + 2^3$

Only $2^6$ or $2^3$ are power of 8

So let us consider these 2 cases

1) $2^6 = 8^2$ and 8 is not power of 4 so $8 = 2^3$ and $256=4^4$ giving a = 3, b = 4, c= 1

2) $2^3 = 8^1$ and this gives 2 cases 

        case 1 $4^4 = 256, 2^6 = 64$ giving a = 6, b = 4, c = 1

        case 2  $2^8 = 256, 4^3 = 64$ giving a = 8, b = 3, c = 1

The above 3 are solutions