# Evaluating Algebraic Expressions using Ratios

Given $\displaystyle \frac{a+b}{3} = \frac{b+c}{4} = \frac{c+a}{5}$, evaluate the following.

(a)   $a : b : c$

\displaystyle \begin{align} \require{AMSsymbols} \text{Let }\frac{a+b}{3} &= \frac{b+c}{4} = \frac{c+a}{5} = k \\ a+b &= 3k \cdots (1) \\ b+c &= 4k \cdots (2) \\ c+a &= 5k \cdots (3) \\ (a+b)-(c+a) &= 3k-5k &\color{green}{(1)-(3)} \\ b-c &= -2k \cdots (4) \\ (b+c)+(b-c) &=4k-2k &\color{green}{(2)+(4)} \\ 2b &= 2k \\ b &= k \\ a+k &= 3k &\color{green}{\text{Substitute } b=k \text{ into } a+b=3k\cdots(1)} \\ a &= 2k \\ k+c &= 4k &\color{green}{\text{Substitute } b=k \text{ into } b+c=4k\cdots(2)} \\ c &= 3k \\ a : b : c &= 2k : k : 3k \\ \therefore a : b : c &= 2 : 1 : 3 \end{align}

(b)   $\displaystyle \frac{1}{ab} : \frac{1}{bc} : \frac{1}{ca}$

\displaystyle \begin{align} \frac{1}{ab} : \frac{1}{bc} : \frac{1}{ca} &= \frac{1}{2k \times k} : \frac{1}{k \times 3k} : \frac{1}{3k \times 2k} \\ &= \frac{1}{2k^2} : \frac{1}{3k^2} : \frac{1}{6k^2} \\ &= \frac{1}{2} : \frac{1}{3} : \frac{1}{6} \\ &= \frac{1}{2} \times 6 : \frac{1}{3} \times 6 : \frac{1}{6} \times 6 \\ \therefore \frac{1}{ab} : \frac{1}{bc} : \frac{1}{ca} &= 3 : 2 : 1 \end{align}

(c)   $\displaystyle \frac{ab+bc+ca}{a^2+b^2+c^2}$

\displaystyle \begin{align} \frac{ab+bc+ca}{a^2+b^2+c^2} &= \frac{2k \times k + k \times 3k + 3k \times 2k}{(2k)^2 + k^2 + (3k)^2} \\ &= \frac{2k^2+3k^2+6k^2}{4k^2+k^2+9k^2} \\ &= \frac{11k^2}{14k^2} \\ \therefore \frac{ab+bc+ca}{a^2+b^2+c^2} &= \frac{11}{14} \end{align}