Fractional Misstep

Algebra Level 1

Find the number of ordered pairs of positive real numbers ( x , y ) (x,y) that satisy the equation

1 5 + x y = 1 + x 5 + y . \dfrac15 + \dfrac xy = \dfrac{1+x}{5+y}.

0 1 2 3

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Prince Raj by Prince Raj , 19, India

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2 solutions

Andy Hayes
Nov 11, 2016

Combine the fractions on the left side of the equation:

y + 5 x 5 y = 1 + x 5 + y \frac{y+5x}{5y}=\frac{1+x}{5+y}

Cross multiply, then simplify:

( y + 5 x ) ( 5 + y ) = 5 y ( 1 + x ) y 2 + 5 x y + 5 y + 25 x = 5 y + 5 x y y 2 = − 25 x \begin{aligned} (y+5x)(5+y) &= 5y(1+x) \\ y^2+5xy+5y+25x &= 5y+5xy \\ y^2 &= -25x \end{aligned}

y 2 y^2 is always a positive number, but − 25 x -25x is positive only when x x is negative. Therefore, there are no solutions with both x x and y y positive.

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there is a word "satisy" in the problem... I think it has to be satisfy....

shithil Islam - 4 years, 6 months ago

@Andy Hayes ± i 2 = − 1 \pm i ^ { 2 } = -1 , then ± 5 i 2 = − 25 \pm 5i ^ { 2 } = -25 .

. . - 2 months, 4 weeks ago
Rico Lee
Nov 11, 2016

After some quick maths, you will get to the form 25x + y^2 = 0 Obviously, one number has to be negative to satisfy this equation. So, there are no ordered pairs of positive x and y.

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