Given: \(\frac{x + y}{x - y} = \frac{a}{b}\) Prove that: \(\frac{y^2 + xy}{x^2 - xy} = \frac{a^2 - ab}{b^2 + ab}\)

Q.Given: \(\frac{x + y}{x - y} = \frac{a}{b}\) Prove that: \(\frac{y^2 + xy}{x^2 - xy} = \frac{a^2 - ab}{b^2 + ab}\)

\(\frac{x + y}{x - y} = \frac{a}{b} \) — (i) Or, \(\frac{x + y + x - y}{x + y - x + y} = \frac{a + b}{a - b}\) [By applying the process of addition and subtraction] Or, \(\frac{2x}{2y} = \frac{a + b}{a - b}\) Or, \(\frac{x}{y} = \frac{a + b}{a - b}\) ∴ \(\frac{y}{x} = \frac{a - b}{a + b}\) — (ii) ∴ Left-hand side = \(\frac{y^2 + xy}{x^2 - xy}\) = \(\frac{y(x + x)}{x(x - y)}\) = \(\frac{y}{x} \times \frac{x + y}{x - y}\) = \(\frac{a - b}{a + b} \times \frac{a}{b}\) [Substituting values from equations (i) and (ii)] = \(\frac{a^2 - ab}{ab + b^2}\) = \(\frac{a^2 - ab}{b^2 + ab}\) = Right-hand side (Proved)