MHB Find $m$ in Real Numbers: $x,y\in R$

[Albert1]

$x,y\in R$

$if: \sqrt {3x+5y-2-m}+\sqrt {2x+3y-m}=\sqrt {x-200+y}\,\times\sqrt {200-x-y}$

$find:\,m=?$

 

[anemone]

$x,y\in R$

$if: \sqrt {3x+5y-2-m}+\sqrt {2x+3y-m}=\sqrt {x-200+y}\,\times\sqrt {200-x-y}$

$find:\,m=?$

My solution:

Spoiler

If we let $k=x+y$, we see that the RHS of the given equation, i.e.$\sqrt{x-200+y}\times \sqrt{200-x-y}=\sqrt {(k-200)(200-k)}=\sqrt{-(k-200)^2}$. Since $-(k-200)^2 \le 0$ and we cannot take a square root of a negative number, thus we must have $k=200$, this gives the RHS of the equation as zero.

Thus, each of the term of the LHS must equal to zero and with the relation $k=x+y=200$, we get

$3x+5y-2-m=0$ yields $m=3x+5y-2$

and

$2x+3y-m=0$ yields $m=2x+3y$

Equating these two equations we have $2=x+2y$ or $2=200+y$ which then gives $y=-198$ thus $x=398$.

$\therefore m=2(398)+3(-198)=202$