InterviewSolution
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InterviewSolution
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1). 125/1212). 2153). 131/1254). 315 |
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Answer» Given 7x - 15y = 4x + y ⇒ 3X = 16y ⇒ $(\frac{{\RM{x}}}{{\rm{y}}} = \frac{{16}}{3})$ Squaring both sides, we get $(\begin{array}{l} \Rightarrow {\rm{\;}}\frac{{{{\rm{x}}^2}}}{{{{\rm{y}}^2}}} = \frac{{256}}{9}\\ \Rightarrow {\rm{\;}}\frac{{3{{\rm{x}}^2}}}{{2{{\rm{y}}^2}}} = \frac{3}{2} \TIMES \frac{{256}}{9}\\ \Rightarrow {\rm{\;}}\frac{{3{{\rm{x}}^2}}}{{2{{\rm{y}}^2}}} = \frac{{128}}{3} \END{array})$ $(\Rightarrow {\rm{\;}}\frac{{3{{\rm{x}}^2}{\rm{\;}} + {\rm{\;}}2{{\rm{y}}^2}}}{{3{{\rm{x}}^2}\; -\;2{{\rm{y}}^2}}} = \frac{{128{\rm{\;}} + {\rm{\;}}3}}{{128\; -\; 3}})$ (By componendo and dividendo) $(\Rightarrow {\rm{\;}}\frac{{3{{\rm{x}}^2}{\rm{\;}} + {\rm{\;}}2{{\rm{y}}^2}}}{{3{{\rm{x}}^2}\; - \;2{{\rm{y}}^2}}} = \frac{{131}}{{125}})$ |
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