One equation with two unknowns

Algebra Level 3

Find the value of x y xy , where x x and y y are real numbers that satisfy the equation below. x − 30 y + y − 40 x + 625 x y = 0 \dfrac{x-30}y+\dfrac{y-40}x+\dfrac{625}{xy} =0


The answer is 300.

Choi Chakfung by choi chakfung , 28, Hong Kong

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1 solution

x − 30 y + y − 40 x + 625 x y = 0 x 2 − 30 x + y 2 − 40 y + 625 x y = 0 x 2 − 30 x + y 2 − 40 y + 625 = 0 x 2 − 30 x + 225 + y 2 − 40 y + 400 = 0 ( x − 15 ) 2 + ( y − 20 ) 2 = 0 \begin{aligned} \frac{x-30}y+\frac{y-40}x+\frac{625}{xy} & = 0 \\ \frac{x^2 - 30 x + y^2 - 40y + 625}{xy} & = 0 \\ x^2 - 30 x + y^2 - 40y + 625 & = 0 \\ x^2 - 30 x + 225 + y^2 - 40y + 400 & = 0 \\ (x-15)^2 + (y-20)^2 & = 0 \end{aligned}

We note that the L H S = ( x − 15 ) 2 + ( y − 20 ) 2 ≥ 0 LHS = (x-15)^2 + (y-20)^2 \ge 0 has a unique minimum value 0 0 . Therefore, there is only a unique solution for the equation when the L H S LHS is minimum or when x = 15 x=15 and y = 20 y=20 , and then x y = 300 xy = \boxed{300}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice explanation sir.+1 Btw it was a nice question!!!

Rishabh Tiwari - 5 years ago

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