(a) Solve the simultaneous equation : \(\frac{1}{x} + \frac{1}{y} = 5 ; \frac{1}{y} – \frac{1}{x} = 1\).
(b) A man drives from Ibadan to Oyo, a distance of 48km in 45 minutes. If he drives at 72 km/h where the surface is good and 48 km/h where it is bad, find the number of kilometers of good surface.
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.