(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.