Review : Systems of Equations — Question 6

Question 6 – Solution

Strategy. A homogeneous system can fix ratios while leaving the overall scale free. A normalization must actually measure that remaining freedom.

Step 1: Solve the homogeneous equations. From the second equation, $y=2z-x$. Substitution into the first gives $3x-z=0$, hence $z=3x$ and $y=5x$. Thus $$\boxed{(x,y,z)=t(1,5,3),\qquad t\in\mathbb R.}$$ Substitution gives $2t-5t+3t=0$ and $t+5t-6t=0$. Every solution has this form, including the zero solution at $t=0$. Dividing by an unknown coordinate would risk losing that solution.

Step 2: Apply the linear normalization. The coordinate sum is $9t$. The condition $9t=9$ fixes $t=1$, giving $\boxed{(1,5,3)}$ uniquely.

Step 3: Compare the quadratic normalization. The squared length is $(1+25+9)t^2=35t^2$. Therefore the quadratic condition gives $t^2=1$ and exactly two solutions: $$\boxed{(1,5,3)\quad\text{and}\quad(-1,-5,-3).}$$ Squaring loses the overall sign. If nonnegative coordinates are also required, only the positive triple remains. The zero triple does not satisfy either stated normalization.

Step 4: Classify an arbitrary linear measurement. On the homogeneous family, the extra left side is $(a+5b+3c)t$. Set $K=a+5b+3c$. Then $$\boxed{\begin{array}{ll} K\ne 0:&t=m/K\text{ is the unique solution scale},\\ K=0,\ m\ne 0:&\text{no solution},\\ K=0,\ m=0:&\text{every real }t\text{ remains possible}. \end{array}}$$ A nonzero-looking added equation can still vanish on the entire family. What matters is its value on the remaining freedom, not the visual presence of three coefficients.