(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), \(\frac{1}{2} + \frac{y}{4} = 1 \implies \frac{y}{4} = 1 - \frac{1}{2}\) \(\frac{y}{4} = \frac{1}{2} \implies y = 2\) \((x, y) = (1, 2)\).