Solve the system x + y − 3z = 0, 2x(6 − x) + 4y(3 − y) + 6z(0.5 − z) = 0, and x² + 2y² + 3z² = 21.

Step-by-step solution

1. 1

   Expand the middle equation and separate linear from quadratic parts.

   $12x - 2x^2 + 12y - 4y^2 + 3z - 6z^2 = 0$

   $\Longrightarrow \ 12x + 12y + 3z = 2x^2 + 4y^2 + 6z^2$

   The quadratic side is exactly $2\left(x^2 + 2y^2 + 3z^2\right)$ — the same combination that appears in the third equation. That coincidence is the whole point of the problem.

2. 2

   Substitute the third equation to eliminate all quadratic terms. Since $x^2 + 2y^2 + 3z^2 = 21$:

   $12x + 12y + 3z = 2(21) = 42$

   What looked like a nonlinear equation has become linear.

3. 3

   Use the first equation to reduce to z alone. From $x + y = 3z$:

   $12(3z) + 3z = 42 \ \Longrightarrow \ 39z = 42 \ \Longrightarrow \ z = \frac{42}{39} = \frac{14}{13}$

4. 4

   Reduce to one quadratic in y. With $z = \tfrac{14}{13}$ we get $x + y = 3z = \tfrac{42}{13}$ and

   $3z^2 = 3\cdot\frac{196}{169} = \frac{588}{169}, \qquad x^2 + 2y^2 = 21 - \frac{588}{169} = \frac{3549 - 588}{169} = \frac{2961}{169}$

   (note $21 \times 169 = 3549$; using $3559$ here is a slip that propagates into wrong roots). Substituting $x = \tfrac{42}{13} - y$ and multiplying through by $169$:

   $1764 - 1092y + 507y^2 = 2961 \ \Longrightarrow \ 507y^2 - 1092y - 1197 = 0$

5. 5

   Solve the quadratic. Dividing by $3$ gives $169y^2 - 364y - 399 = 0$, with discriminant

   $364^2 + 4(169)(399) = 132496 + 269724 = 402220 = 26^2 \cdot 595$

   so

   $y = \frac{364 \pm 26\sqrt{595}}{338} = \frac{14 \pm \sqrt{595}}{13}$

   The discriminant is not a perfect square ($595 = 5 \cdot 7 \cdot 17$), so the roots are irrational — any answer in tidy fractions like $7/13$ is wrong.

6. 6

   Recover x and verify both solutions. From $x = \tfrac{42}{13} - y = \dfrac{28 \mp \sqrt{595}}{13}$, the two solutions are

   $\left(\frac{28 - \sqrt{595}}{13},\ \frac{14 + \sqrt{595}}{13},\ \frac{14}{13}\right) \approx (0.27749,\ 2.95328,\ 1.07692)$

   $\left(\frac{28 + \sqrt{595}}{13},\ \frac{14 - \sqrt{595}}{13},\ \frac{14}{13}\right) \approx (4.03020,\ -0.79943,\ 1.07692)$

   Substituting either triple into all three original equations gives residuals below $10^{-14}$.