Question

Fill in the blanks of the following

(i) $\frac{x}{7}=\frac{y}{3}=\frac{3x+5y}{..........}=\frac{7x-9y}{.........}$

(ii) $\frac{a}{3}=\frac{b}{4}=\frac{c}{7}=\frac{a-2b+3c}{.......}=\frac{..........}{6-8+14}$

Answer 1

(i)
$$\begin{gathered} \frac{x}{7}=\frac{y}{3}=\frac{3\times x}{3\times 7}=\frac{5\times y}{5\times 3} \\ \Rightarrow \frac{{\displaystyle x}}{{\displaystyle 7}}=\frac{{\displaystyle y}}{{\displaystyle 3}}=\frac{{\displaystyle 3x}}{{\displaystyle 21}}=\frac{{\displaystyle 5y}}{{\displaystyle 15}}=\frac{3x+5y}{21+15}\left(\mathrm{Theorem}\mathrm{of}\mathrm{equal}\mathrm{ratios}\right) \\ \Rightarrow \frac{{\displaystyle x}}{{\displaystyle 7}}=\frac{{\displaystyle y}}{{\displaystyle 3}}=\frac{3x+5y}{36} \end{gathered}$$
Also,

$$\begin{gathered} \frac{x}{7}=\frac{y}{3}=\frac{7\times x}{7\times 7}=\frac{-9\times y}{-9\times 3} \\ \Rightarrow \frac{{\displaystyle x}}{{\displaystyle 7}}=\frac{{\displaystyle y}}{{\displaystyle 3}}=\frac{{\displaystyle 7x}}{{\displaystyle 49}}=\frac{-9y}{-27}=\frac{7x-9y}{49+\left(-27\right)}\left(\mathrm{Theorem}\mathrm{of}\mathrm{equal}\mathrm{ratios}\right) \\ \Rightarrow \frac{{\displaystyle x}}{{\displaystyle 7}}=\frac{{\displaystyle y}}{{\displaystyle 3}}=\frac{7x-9y}{22} \end{gathered}$$
$\therefore \frac{x}{7}=\frac{y}{3}=\frac{3x+5y}{\underline{\mathbf{36}}}=\frac{7x-9y}{\underline{\mathbf{22}}}$

(ii)
$$\begin{gathered} \frac{a}{3}=\frac{b}{4}=\frac{c}{7}=\frac{-2\times b}{-2\times 4}=\frac{3\times c}{3\times 7} \\ \Rightarrow \frac{{\displaystyle a}}{{\displaystyle 3}}=\frac{{\displaystyle b}}{{\displaystyle 4}}=\frac{{\displaystyle c}}{{\displaystyle 7}}=\frac{{\displaystyle -2b}}{{\displaystyle -8}}=\frac{{\displaystyle 3c}}{{\displaystyle 21}}=\frac{a+\left(-2b\right)+3c}{3+\left(-8\right)+21}\left(\mathrm{Theorem}\mathrm{of}\mathrm{equal}\mathrm{ratios}\right) \\ \Rightarrow \frac{{\displaystyle a}}{{\displaystyle 3}}=\frac{{\displaystyle b}}{{\displaystyle 4}}=\frac{{\displaystyle c}}{{\displaystyle 7}}=\frac{a-2b+3c}{16} \end{gathered}$$
Also,

$$\begin{gathered} \frac{a}{3}=\frac{b}{4}=\frac{c}{7}=\frac{2\times a}{2\times 3}=\frac{-2\times b}{-2\times 4}=\frac{2\times c}{2\times 7} \\ \Rightarrow \frac{{\displaystyle a}}{{\displaystyle 3}}=\frac{{\displaystyle b}}{{\displaystyle 4}}=\frac{{\displaystyle c}}{{\displaystyle 7}}=\frac{{\displaystyle 2a}}{{\displaystyle 6}}=\frac{{\displaystyle -2b}}{{\displaystyle -8}}=\frac{2c}{14}=\frac{2a+\left(-2b\right)+2c}{6+\left(-8\right)+14}\left(\mathrm{Theorem}\mathrm{of}\mathrm{equal}\mathrm{ratios}\right) \\ \Rightarrow \frac{{\displaystyle a}}{{\displaystyle 3}}=\frac{{\displaystyle b}}{{\displaystyle 4}}=\frac{{\displaystyle c}}{{\displaystyle 7}}=\frac{2a-2b+2c}{6-8+14} \end{gathered}$$
$\therefore \frac{{\displaystyle a}}{{\displaystyle 3}}=\frac{{\displaystyle b}}{{\displaystyle 4}}=\frac{{\displaystyle c}}{{\displaystyle 7}}=\frac{a-2b+3c}{\underline{\mathbf{16}}}=\frac{\underline{\mathbf{2}\mathbf{a}\mathbf{-}\mathbf{2}\mathbf{b}\mathbf{+}\mathbf{2}\mathbf{c}}}{6-8+14}$