Question

Using properties of proportion, solve each of the following for $x$:

(iii) $16\frac{{(a-x)}^3}{{(a+x)}^3}=\frac{a+x}{a-x}$.

Answer 1

Step1: Simplify the given equation

If $\frac{a}{b}=\frac{c}{d}$, then the property of componendo and dividendo states that $\frac{a+b}{a-b}=\frac{c+d}{c-d}$.

If $\frac{a}{b}=\frac{c}{d}$, then the property of invertendo states that $\frac{b}{a}=\frac{d}{c}$.

The given proportion can be simplified as,

$$\begin{gathered} 16=\frac{{(a+x)}^4}{{(a-x)}^4}(Multiplyingbothsidesby\frac{{(a+x)}^3}{{(a-x)}^3}) \\ \Rightarrow \frac{{(a+x)}^4}{{(a-x)}^4}=2^4 \\ \Rightarrow \frac{(a+x)}{(a-x)}=\pm 2(Takingthefourthrootonbothsides) \end{gathered}$$

Step2: Considering the positive value.

$$\begin{gathered} \frac{a+x}{a-x}=2 \\ \Rightarrow \frac{a+x+a-x}{a+x-a+x}=\frac{2+1}{2-1}(Bycomponendoanddividendo) \\ \Rightarrow \frac{2a}{2x}=\frac{3}{1} \\ \Rightarrow \frac{a}{x}=3 \\ \Rightarrow \frac{x}{a}=\frac{1}{3}(Byinvertendo) \\ \therefore x=\frac{a}{3} \end{gathered}$$

Step3: Considering the negative value.

$$\begin{gathered} \frac{a+x}{a-x}=-2 \\ \Rightarrow \frac{a+x+a-x}{a+x-a+x}=\frac{-2+1}{-2-1}(Bycomponendoanddividendo) \\ \Rightarrow \frac{2a}{2x}=\frac{-1}{-3} \\ \Rightarrow \frac{a}{x}=\frac{1}{3} \\ \Rightarrow \frac{x}{a}=3(Byinvertendo) \\ \therefore x=3a \end{gathered}$$

Final Answer: The values of $x$ are $3a,\frac{a}{3}$.