(a) Make q the subject of the relation \(t = \sqrt{\frac{pq}{r} - r^{2}q}\).
(b) If \(9^{(1 - x)} = 27^{y}\) and \(x - y = -1\frac{1}{2}\), find the value of x and y.
Show Answer Show Explanation Explanation (a) \(t = \sqrt{\frac{pq}{r} - r^{2}q}\) Squaring both sides, \(t^{2} = \frac{pq}{r} - r^{2}q\) \(t^{2} = \frac{pq - r^{3}q}{r}\) \(t^{2}r = pq - r^{3}q \implies t^{2}r = q(p - r^{3})\) \(q = \frac{t^{2}r}{p - r^{3}}\). (b) \(9^{(1 - x)} = 27^{y}\) \(\implies 3^{2(1 - x)} = 3^{3y}\) \(2 - 2x = 3y ..... (1)\) \(x - y = -1\frac{1}{2} ....... (2)\) From (2), \(x = -1\frac{1}{2} + y\) \(\therefore 2 - 2(-1\frac{1}{2} + y) = 3y\) \(2 + 3 - 2y = 3y \implies 5 = 3y + 2y = 5y\) \(\implies y = 1\) \(x = -1\frac{1}{2} + 1 = -\frac{1}{2}\) \(\therefore x + y = -\frac{1}{2} + 1 = \frac{1}{2}\)