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.