Question

Using the properties of proportion, solve the following equation for $x$:
$16{\left(\frac{a-x}{a+x}\right)}^2=\frac{a+x}{a-x}$

Answer 1

$16(\frac{a-n}{a+n}{)}^2=\frac{a+n}{a-n}$

$\Rightarrow 16\left({\frac{a-n}{a+n}}^2\right).\left(\frac{a-n}{a+n}\right)=1$

$\Rightarrow (\frac{a-n}{a+n}{)}^3=\frac{1}{\left(16\right)}$

$\frac{a-n}{a+n}=\frac{1}{(10{)}^{1/3}}$

$\left(16{)}^{1/3}\right[a-n]=a+n$

$(16{)}^{1/3}a-a=n+(16{)}^{1/3}n$

$a\left[\right(16{)}^{1/3}-1]=[1+\left(16{)}^{1/3}\right]n$

$n=\frac{a\left[\right(16{)}^{1/3}-1]}{\left[\right(16{)}^{1/3}+1]}$

$n=\frac{a\left[\right(16{)}^{1/3}-1]}{\left[\right(16{)}^{1/3}+1]}$