Solve $x, y \in \mathbb{R}$ given $x(12-x)+y(16-y)=100$.

Question

Solve the equation in $\mathbb{R}$ $$x(12-x)+y(16-y)=100$$

MY IDEAS:

\begin{aligned} & x(12-x)+y(16-y)=100 \\ & 12 x-x^2+y^{16}-y^2=100 \\ & 12 x+16 y-x^2-y^2=100 \\ & 12 x+16 y-x^2-y^2-100=0 \\ & -12 x-16 y+x^2+y^2+100=0 \\ & x^2+y^2+100=+12 x+16 y=12 x+16 y \\ & 2 x^2+2 y^2+100-x^2-y^2=12 x+16 y \end{aligned}

This is all I could thought of. I tried in the last part to use the formula

$a^{2}-b^{2}=(a+b)(a-b)$

I don't know how to continue! Any ideas are welcome! Thank you!

Answer 1

This is a completing the square exercise. So first you open the brackets, then you complete the square, and then the answer is obvious. $$x(12-x) + y(16-y) = -(x^2 - 12x)-(y^2-16y) = -((x-6)^2-36)-((y-8)^2 - 64)=\ -(x-6)^2 + 36 -(y-8)^2 + 64 = 100 \\ \Rightarrow -(x-6)^2 -(y-8)^2 + 100 = 100\\ \Rightarrow (x-6)^2 + (y-8)^2 = 0 $$ Note that the sum of squares is positive unless each of the squares is zero. Hence the unique answer ix $x=6, y=8$