(a) Solve the simultaneous equation : \(\frac{1}{x} + \frac{1}{y} = 5 ; \frac{1}{y} - \frac{1}{x}

Explanation

(a) \(\frac{1}{x} + \frac{1}{y} = 5 ..... (1)\)

\(\frac{1}{y} - \frac{1}{x} = 1 ........ (2)\)

Substitute p and q for \(\frac{1}{x}\) and \(\frac{1}{y}\) respectively.

\(p + q = 5 ......... (1a)\)

\(q - p = 1 ....... (2a)\)

From (2a), q = 1 + p.

Putting into (1a), we have:

\(p + (1 + p) = 5\)

\(2p + 1 = 5 \implies 2p = 5 - 1 = 4\)

\(p = 2\)

\(q = 1 + p = 1 + 2 = 3\)

\(p = \frac{1}{x} = 2\)

\(\implies x = \frac{1}{2}\)

\(q = \frac{1}{y} = 3\)

\(\implies y = \frac{1}{3}\)

Hence, \(x = \frac{1}{2} ; y = \frac{1}{3}\).

(b) Total distance travelled = 48 km

Total time taken = 45 minutes = \(\frac{45}{60} = 0.75 hr\)

Let the man travel a distance of x km on a good surface. Then he travel a distance of (48 - x) km on bad surface.

Let the time taken to travel on good surface be t hours, then the time taken to travel on bad surface = (0.75 - t) hrs.

Using \(speed = \frac{distance}{time}\) in each case

On good surface : \(72 = \frac{x}{t}\)

\(\implies x = 72t ... (1)\)

On bad surface : \(48 = \frac{48 - x}{0.75 - t}\)

\(48(0.75 - t) = 48 - x\)

\(\implies 36 - 48t = 48 - x ...... (2)\)

Substitute 72t = x in (2) :

\(36 - 48t = 48 - 72t \implies 72t - 48t = 48 - 36\)

\(24t = 12 \implies t = \frac{12}{24} = 0.5 hours\)

From (1), x = 72t

\(x = 72 \times 0.5 = 36 km\)

Hence, the man travels a distance of 36 km on good road surface.