With a=b=c=1, solve the three-equation system.

01 / Standard#Your turn

x+y+z=2,\quad3x+y-z=4,\quad x^2+y^2+z^2=3

Official paper · jm02-2026 · 5(b)(iii) · PDF 7

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Reuse the parameterisation from part (i).

Hint 2

The last equation becomes a quadratic in t.

Worked solution

Substitute and solve the scalar quadratic.
$(1+t)^2+(1-2t)^2+t^2=3\iff6t^2-2t-1=0\implies t=\frac{1\pm\sqrt7}6$
Use matching signs in x and z and the opposite sign in y.
$(x,y,z)=\left(\frac{7\pm\sqrt7}6,\frac{4\mp2\sqrt7}6,\frac{1\pm\sqrt7}6\right)$

Exactly two triples, corresponding to the consistent upper and lower signs.

Checks and common pitfalls: The ± choices are linked; they do not produce eight independent triples.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Substitute and solve the scalar quadratic.
$(1+t)^2+(1-2t)^2+t^2=3\iff6t^2-2t-1=0\implies t=\frac{1\pm\sqrt7}6$
Use matching signs in x and z and the opposite sign in y.
$(x,y,z)=\left(\frac{7\pm\sqrt7}6,\frac{4\mp2\sqrt7}6,\frac{1\pm\sqrt7}6\right)$

Think first. Reveal a hint when the class is ready.

2026 JM02