Value matters.

Algebra Level 2

If x y = 1 3 \dfrac{x}{y} = \dfrac{1}{3} , then what is x 2 + y 2 x 2 − y 2 \dfrac{x^2+y^2}{x^2-y^2} ?

5 3 \dfrac{5}{3} 1 2 \dfrac{1}{2} − 10 9 -\dfrac{10}{9} − 5 4 -\dfrac{5}{4} − 5 3 -\dfrac{5}{3} 5 5 \dfrac{5}{5} 2 4 \dfrac{2}{4}

Munem Shahriar by Munem Shahriar , 18, Bangladesh

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2 solutions

Munem Shahriar
Aug 19, 2017

x y = 1 3 \dfrac{x}{y} = \dfrac{1}{3}

Cross multiply,

⇒ 3 x = y . \Rightarrow 3x = y.

Square both sides,

⇒ 9 x 2 = y 2 \Rightarrow 9x^2 = y^2

Therefore, ⇒ x 2 + y 2 x 2 − y 2 = x 2 + 9 x 2 x 2 − 9 x 2 = 10 x 2 − 8 x 2 ⟹ − 10 8 = − 5 4 \Rightarrow \dfrac{x^2 + y^2}{x^2 - y^2} = \dfrac{x^2 + 9x^2}{x^2- 9x^2} = \dfrac{10x^2}{-8x^2} \implies -\dfrac{10}{8} = \boxed{-\dfrac{5}{4}}

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Chew-Seong Cheong
Aug 20, 2017

X = x 2 + y 2 x 2 − y 2 Divide up and down with x y = x y + y x x y − y x = 1 3 + 3 1 3 − 3 = 10 − 8 = − 5 4 \begin{aligned} X & = \frac {x^2+y^2}{x^2-y^2} & \small \color{#3D99F6} \text{Divide up and down with }xy \\ & = \frac {\frac xy+\frac yx}{\frac xy-\frac yx} = \frac {\frac 13+3}{\frac 13-3} \\ & = \frac {10}{-8} = \boxed{ - \dfrac 54} \end{aligned}

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