Question

If $\frac{a}{b}=\frac{7}{3}$ then find the values of the following ratios.

(i) $\frac{5a+3b}{5a-3b}$ (ii) $\frac{2a^2+3b^2}{2a^2-3b^2}$ (iii) $\frac{a^3-b^3}{b^3}$ (iv) $\frac{7a+9b}{7a-9b}$

Answer 1

$$\begin{gathered} \frac{a}{b}=\frac{7}{3} \\ \Rightarrow \frac{a}{7}=\frac{b}{3}=k \\ \Rightarrow a=7k,b=3k \end{gathered}$$
(i)
$$\begin{gathered} \frac{5a+3b}{5a-3b} \\ =\frac{5\times 7k+3\times 3k}{5\times 7k-3\times 3k} \\ =\frac{35k+9k}{35k-9k} \\ =\frac{44k}{26k} \\ =\frac{22}{13} \end{gathered}$$
(ii)
$$\begin{gathered} \frac{2a^2+3b^2}{2a^2-3b^2} \\ =\frac{2\times {\left(7k\right)}^2+3\times {\left(3k\right)}^2}{2\times {\left(7k\right)}^2-3\times {\left(3k\right)}^2} \\ =\frac{98k^2+27k^2}{98k^2-27k^2} \\ =\frac{125k^2}{71k^2} \\ =\frac{125}{71} \end{gathered}$$
(iii)
$$\begin{gathered} \frac{a^3-b^3}{b^3} \\ =\frac{{\left(7k\right)}^3-{\left(3k\right)}^3}{{\left(3k\right)}^3} \\ =\frac{343k^3-27k^3}{27k^3} \\ =\frac{316k^3}{27k^3} \\ =\frac{316}{27} \end{gathered}$$
(iv)
$$\begin{gathered} \frac{7a+9b}{7a-9b} \\ =\frac{7\times 7k+9\times 3k}{7\times 7k-9\times 3k} \\ =\frac{49k+27k}{49k-27k} \\ =\frac{76k}{22k} \\ =\frac{38}{11} \end{gathered}$$