Find every p,q giving more than one solution, and give each full solution family.

01 / Standard#Your turn

\begin{cases}x+y+pz=1\\px+y+z=p\\x+py+z=q\end{cases}

Official paper · jm02-2022 · 5(b)(ii) · PDF 7

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Only p=1 and p=−2 can fail uniqueness.

Hint 2

For each, test consistency before parameterising.

Worked solution

At p=1 all left sides equal x+y+z; consistency requires q=1.
$p=1,q=1:\quad(x,y,z)=(1-s-t,s,t),\quad s,t\in\mathbb R$
At p=−2 the sum of all equations is 0=q−1, again requiring q=1.
$x+y-2z=1,\quad-2x+y+z=-2$
Solve these independent equations and verify the third.
$p=-2,q=1:\quad(x,y,z)=(1+t,t,t),\quad t\in\mathbb R$

For (p,q)=(1,1): (1−s−t,s,t). For (−2,1): (1+t,t,t). No other multiple-solution cases.

Checks and common pitfalls: A singular coefficient matrix may still give no solution when q≠1.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
At p=1 all left sides equal x+y+z; consistency requires q=1.
$p=1,q=1:\quad(x,y,z)=(1-s-t,s,t),\quad s,t\in\mathbb R$
At p=−2 the sum of all equations is 0=q−1, again requiring q=1.
$x+y-2z=1,\quad-2x+y+z=-2$
Solve these independent equations and verify the third.
$p=-2,q=1:\quad(x,y,z)=(1+t,t,t),\quad t\in\mathbb R$

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

2022 JM02