3.2.2 Algebraic Formulae, PT3 Practice - Mathematics

Question 5:

Given

\sqrt{\frac{8 s - 3 h}{4}} = 2

$\sqrt{\frac{8 s - 3 h}{4}} = 2$ , express s in terms of h.

Solution:

\begin{array}{l} \sqrt{\frac{8 s - 3 h}{4}} = 2 \\ \frac{8 s - 3 h}{4} = 4 \leftarrow Square both sides \\ 8 s - 3 h = 16 \\ 8 s = 16 + 3 h \\ s = \frac{16 + 3 h}{8} \\ s = 2 + \frac{3 h}{8} \end{array}

$\begin{array}{l} \sqrt{\frac{8 s - 3 h}{4}} = 2 \\ \frac{8 s - 3 h}{4} = 4 \leftarrow Square both sides \\ 8 s - 3 h = 16 \\ 8 s = 16 + 3 h \\ s = \frac{16 + 3 h}{8} \\ s = 2 + \frac{3 h}{8} \end{array}$

Question 6:

Given 2 a b = 3 a - \frac{b^{2}}{2} . Express a in terms of b .

$Given 2 a b = 3 a - \frac{b^{2}}{2} . Express a in terms of b .$

Solution:

\begin{array}{l} 2 a b = 3 a - \frac{b^{2}}{2} \\ 3 a - 2 a b = \frac{b^{2}}{2} \\ a (3 - 2 b) = \frac{b^{2}}{2} \\ a = \frac{b^{2}}{2 (3 - 2 b)} \end{array}

$\begin{array}{l} 2 a b = 3 a - \frac{b^{2}}{2} \\ 3 a - 2 a b = \frac{b^{2}}{2} \\ a (3 - 2 b) = \frac{b^{2}}{2} \\ a = \frac{b^{2}}{2 (3 - 2 b)} \end{array}$

Question 7:

Given a = \frac{\sqrt{b + 1}}{a} . Express b in terms of a .

$Given a = \frac{\sqrt{b + 1}}{a} . Express b in terms of a .$

Solution:

\begin{array}{l} a = \frac{\sqrt{b + 1}}{a} \\ \sqrt{b + 1} = a^{2} \\ {[{(b + 1)}^{\frac{1}{2}}]}^{2} = {(a^{2})}^{2} \\ b + 1 = a^{4} \\ b = a^{4} - 1 \end{array}

$\begin{array}{l} a = \frac{\sqrt{b + 1}}{a} \\ \sqrt{b + 1} = a^{2} \\ {[{(b + 1)}^{\frac{1}{2}}]}^{2} = {(a^{2})}^{2} \\ b + 1 = a^{4} \\ b = a^{4} - 1 \end{array}$

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