Question

If $\frac{x}{y}=\frac{13}{6}$ then find the value of $\left(\mathrm{i}\right)\frac{3x+5y}{3x-5y}\left(\mathrm{ii}\right)\frac{2x^2+5y^2}{2x^2-5y^2}$.

Answer 1

$$\begin{gathered} \frac{x}{y}=\frac{13}{6} \\ \left(\mathrm{i}\right)\frac{3x+5y}{3x-5y} \\ \mathrm{Dividing}\mathrm{the}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}y,\mathrm{we}\mathrm{get}: \\ \frac{3\times {\displaystyle \frac{x}{y}}+5}{3\times {\displaystyle \frac{x}{y}}-5} \\ =\frac{3\times {\displaystyle \frac{13}{6}}+5}{3\times {\displaystyle \frac{13}{6}}-5} \\ =\frac{{\displaystyle \frac{13}{2}}+5}{{\displaystyle \frac{13}{2}}-5} \\ =\frac{23}{3} \end{gathered}$$


$\left(\mathrm{ii}\right)\frac{2x^2+5y^2}{2x^2-5y^2}$
$$\begin{gathered} \mathrm{Dividing}\mathrm{the}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}{y}^2,\mathrm{we}\mathrm{get}: \\ \frac{2{\left({\displaystyle \frac{x}{y}}\right)}^2+5}{2{\left(\frac{x}{y}\right)}^2-5} \\ =\frac{2\times {\left({\displaystyle \frac{13}{6}}\right)}^2+5}{2\times {\left(\frac{13}{6}\right)}^2-5} \\ =\frac{{\displaystyle \frac{169}{18}}+5}{{\displaystyle \frac{169}{18}}-5} \\ =\frac{{\displaystyle \frac{169+90}{18}}}{\frac{169-90}{18}} \\ =\frac{259}{79} \end{gathered}$$