An algebra problem

Algebra Level pending

Positive integers x x and y y are such that:

{ E ( x , y ) = x + 1 2 ( y + 2 ) E ( x , y ) + E ( y , x ) = 3 x + y \begin{cases} E(x,y) = \dfrac {x+1}{2(y+2)} \\ E(x,y)+E(y,x) = \dfrac 3{x+y} \end{cases}

Find x x .


The answer is 2.

Irina Stanciu by Irina Stanciu , 21, United Kingdom

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1 solution

Irina Stanciu
Dec 2, 2016

We've got: x + 1 2 ( y + 2 ) + y + 1 2 ( x + 2 ) = 3 x + y \frac{x+1}{2(y+2)}+ \frac{y+1}{2(x+2)}=\frac{3}{x+y} Multiplying by 2, we have x + 1 y + 2 + y + 1 x + 2 = 6 x + y ⁣ o r ⁣ x + 1 y + 2 + 1 + y + 1 x + 2 + 1 = 6 x + y + 2 \frac{x+1}{y+2} +\frac {y+1}{x+2}=\frac{6}{x+y} \! or \! \frac{x+1}{y+2} + 1 + \frac{y+1}{x+2} + 1=\frac{6}{x+y} + 2 => x + y + 3 y + 2 + x + y + 3 x + 2 = 2 ∗ x + y + 3 x + y \frac{x+y+3}{y+2} + \frac{x+y+3}{x+2}= 2 *\frac{x+y+3}{x+y} Dividing by (x+y+3), we get: 1 y + 2 + 1 x + 2 = 2 x + y = > x 2 + y 2 = 8 = > x = y = 2 \frac{1}{y+2} +\frac{1}{x+2}=\frac{2}{x+y} => x^{2} + y^{2}=8 => x=y=2

3 Helpful 0 Interesting 0 Brilliant 0 Confused

You may use LATEX,so that your answer looks interesting to read:

https://brilliant.org/discussions/thread/beginner-latex-guide/

Try the link.

Anandmay Patel - 4 years, 6 months ago

Thank you!

Irina Stanciu - 4 years, 6 months ago

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