Algebra
(a) $\frac {4x}{5y} = \frac xy \times \frac 45 = \frac 54 \times \frac 45 = 1$
(b) $\frac {4x}{5y}=1$ so that $x=\frac {5y}{4}$ and substitute in $\frac {x-y}{x+y}$,
$\displaystyle \frac {x-y}{x+y} = \frac {\frac{5y}{4}-y}{\frac{5y}{4}+y}=\frac{\frac {5y-4y}{4}}{\frac {5y+4y}{4}}=\frac{\frac{y}{4}}{\frac{9y}{4}}= \frac 19$
$\frac 1a = \frac 1b - \frac ca$ , [placed the required value (c) in left side of equation]
$\frac ca = \frac 1b -\frac 1a$
$c=a(\frac 1b - \frac 1a)$
$c=\frac ab - 1$
$c^2=9^3+6^4$
$c^2=729+1296$
$c^2=2025$
$c = \sqrt {2025} $
$c=45$
$(x-a)(x+b)=3bx$
$(x-a)=\frac {3bx}{x+b}$ [the required expression 'a' lies in left side of equation]
$-a=\frac{3bx}{x+b}-x$
$a=x-\frac{3bx}{x+b}$
$a=\frac{x^2+bx-3bx}{x+b}$
$a=\frac{x^2-2bx}{x+b}$
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