(a) If p varies directly as \(r^{2}\) and p = 3.2 when r = 4, find the value of p when r = 6.5.
(b) Solve the simultaneous equations :
\(\frac{x}{2} + \frac{y}{4} = 1 ; \frac{x}{3} – \frac{y}{4} = \frac{-1}{6}\)
Explanation
(a) \(p \propto r^{2}\)
\(\implies p = kr^{2}\)
\(3.2 = k(4^{2}) \implies 3.2 = 16k\)
\(k = \frac{3.2}{16} = 0.2\)
\(\therefore p = 0.2r^{2}\)
When r = 6.5,
\(p = 0.2 (6.5^{2}) \)
= \(0.2(42.25)\)
\(p = 8.45\)
(b) \(\frac{x}{2} + \frac{y}{4} = 1 ... (1)\)
\(\frac{x}{3} - \frac{y}{4} = -\frac{1}{6} .... (2)\)
Adding equation (1) and (2), we have
\(\frac{x}{2} + \frac{x}{3} = 1 - \frac{1}{6}\)
\(\frac{5x}{6} = \frac{5}{6} \implies x = 1\)
Put x = 1 in (1),
\(\frac{1}{2} + \frac{y}{4} = 1 \implies \frac{y}{4} = 1 - \frac{1}{2}\)
\(\frac{y}{4} = \frac{1}{2} \implies y = 2\)
\((x, y) = (1, 2)\).