Review : Systems of Equations — Question 1

Question 1 – Solution

Strategy. Inspect relationships between whole equations before committing to a long elimination.

Step 1: Test consistency. The third left-hand side is the sum of the first two. Any common solution must therefore satisfy $k=2+5=7$. If $k\ne 7$, subtracting the first two equations from the third gives the contradiction $0=k-7$.

Step 2: Find and verify the full family. When $k=7$, the third equation contributes no new restriction. Subtract twice the first equation from the second to get $y-z=1$. Setting $z=t$ gives $$\boxed{(x,y,z)=(1-2t,\,1+t,\,t),\qquad t\in\mathbb R.}$$ The first left side is $2$, the second is $5$, and the third is $7$ for every $t$. Conversely, every solution must have $y=1+z$ and $x=1-2z$, so no solutions have been omitted. Thus $k=7$ gives infinitely many solutions; no value of $k$ gives exactly one solution in the original system.

Step 3: Use the additional measurement. On this family, $x-y=-3t$. The new measurement forces $t=0$, giving $$\boxed{(x,y,z)=(1,1,0)\quad\text{if }k=7.}$$ If $k\ne 7$, the contradiction from the original equations remains, so adding a measurement cannot create a solution.

Step 4: Distinguish equation count from information. For $k=7$, the third equation merely repeats a consequence already known. For $k\ne 7$, it contradicts that consequence. It never removes just the one remaining freedom. By contrast, $x-y$ varies as $-3t$ along the family, so its measured value selects exactly one member. Counting equations and unknowns alone cannot distinguish these situations; their relationships matter.