Review : Systems of Equations — Question 9

Question 9 – Solution

Strategy. Passing through a target is necessary, but the measurement must also distinguish points along the first equation’s line.

Step 1: Require both passage and uniqueness. The target lies on the second line exactly when $c=(a+2b)/3$. Substituting $y=1-x$ gives $(a-b)x=c-b$. Hence unique selection requires $$\boxed{c=(a+2b)/3,\qquad a\ne b.}$$ If $a=b$, the target condition forces $c=a$ and the second equation is a multiple of $x+y=1$ (possibly $0=0$). The whole first line remains possible.

Step 2: List all normalized designs. Put $b=3-a$, with $0\le a\le 3$. Then $$\boxed{b=3-a,\quad c=2-a/3,\quad a\in[0,3]\setminus\{3/2\}.}$$ The midpoint is excluded because both coefficients would equal $3/2$.

Step 3: Derive the exact measurement sensitivity. Subtract the target equations from the perturbed equations. With $\Delta x=x-1/3$ and $\Delta y=y-2/3$, one obtains $\Delta x+\Delta y=0$ and $a\Delta x+b\Delta y=\delta$. Thus $$\boxed{\Delta x=\frac\delta{a-b},\qquad \Delta y=-\frac\delta{a-b}.}$$ The worst coordinate error over $|\delta|\le\eta$ is exactly $\eta/|a-b|$, attained at either endpoint of the allowed error interval.

Step 4: Optimize within the allowed normalization. Since $a+b=3$ and both are nonnegative, $|a-b|\le 3$, with equality exactly at $(a,b)=(3,0)$ or $(0,3)$. Therefore the minimum worst error is $\boxed{\eta/3}$, attained by exactly two designs: $$\boxed{3x=1\quad\text{or}\quad 3y=2.}$$ Perturbing these readings directly changes the measured coordinate by $\delta/3$ and the other by its negative. Both attain the claimed bound. Without the normalization, arbitrary rescaling would change the meaning of an absolute right-hand-side error, making this comparison ill-defined.