(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.
(a) Solve \(\dfrac{1}{x}+\dfrac{1}{y}=5\) and \(\dfrac{1}{y}-\dfrac{1}{x}=1\)
Let \(u=\dfrac{1}{x}\) and \(v=\dfrac{1}{y}\). Then
\[u+v = 5 \quad\text{and}\quad v-u = 1\]
Add the two equations:
\[2v = 6 \;\Rightarrow\; v = 3 \;\Rightarrow\; \frac{1}{y}=3 \;\Rightarrow\; y = \frac{1}{3}\]
Then \(u = 5 - v = 2\), so \(\dfrac{1}{x}=2 \Rightarrow x = \dfrac{1}{2}\).
\(x = \tfrac{1}{2},\; y = \tfrac{1}{3}\).
(b) Length of good surface
Total distance \(48\) km in \(45\) minutes \(= 0.75\) h. Let good surface \(= g\) km at \(72\) km/h and bad surface \(= b\) km at \(48\) km/h.
Distance: \(g + b = 48\).
Time: \(\dfrac{g}{72} + \dfrac{b}{48} = 0.75\). Multiply through by \(144\):
\[2g + 3b = 108\]
Substitute \(b = 48 - g\):
\[2g + 3(48 - g) = 108 \;\Rightarrow\; 2g + 144 - 3g = 108 \;\Rightarrow\; -g = -36\]
\[g = 36\]
The good surface is 36 km.