(a) Make q the subject of the relation \(t = \sqrt{\frac{pq}{r} - r^{2}q}\). (b) If...

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

\(\therefore 2 - 2(-1\frac{1}{2} + y) = 3y\)

\(2 + 3 - 2y = 3y \implies 5 = 3y + 2y = 5y\)

\(\implies y = 1\)

\(x = -1\frac{1}{2} + 1 = -\frac{1}{2}\)

\(\therefore x + y = -\frac{1}{2} + 1 = \frac{1}{2}\)