Review : Systems of Equations — Question 2

Question 2 – Solution

Strategy. Adding and subtracting separates the sum of the unknowns from their difference, exposing both exceptional parameter values.

Step 1: Reduce reversibly. The sum and difference equations are $$\lambda(x+y)=2,\qquad (\lambda-2)(x-y)=0.$$ They are equivalent to the originals: taking half their sum and half their difference recovers the two original equations. If $\lambda\ne 0,2$, then $$\boxed{x=y=1/\lambda.}$$ At $\lambda=0$, the sum is $0=2$, giving no solution. At $\lambda=2$, only $x+y=1$ remains, giving infinitely many solutions.

Step 2: Retain the exceptional equations. Directly, $\lambda=0$ gives $-x+y=1$ and $x-y=1$, incompatible parallel lines. At $\lambda=2$, both equations become $x+y=1$. Dividing by a factor that can be zero discards these cases rather than solving them. The figure displays the two exceptional geometries.

Step 3: Compare limits with exact parameter values. As $\lambda\to 2$ through the unique-solution cases, $(x,y)\to(1/2,1/2)$. This is one point of the entire line present at $\lambda=2$, not the whole limiting parameter’s solution set. As $\lambda\to 0^+$ both coordinates increase without bound; as $\lambda\to 0^-$ both decrease without bound. There is no finite limiting solution, consistent with the contradiction at zero.

Step 4: Enforce nonnegativity. For $\lambda\ne 0,2$, the solution is nonnegative exactly when $\lambda>0$. At $\lambda=2$, the nonnegative solutions are $(x,y)=(u,1-u)$, $0\le u\le 1$. Therefore a nonnegative solution exists exactly for $\boxed{\lambda>0}$, with the entire stated segment at $2$.