Classify all cases with more than one solution, including conditions on p,q and the general solutions.

01 / Standard#Your turn

kx+2y-z=p,\quad ky+z=q,\quad kx+3y=6

Official paper · jm02-2025 · 5(b)(ii) · PDF 7

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Only the singular values k=0 and k=1 need consideration.

Hint 2

Check the remaining equation after eliminating variables.

Worked solution

When k=0, the last two equations fix y and z, leaving x free if the first equation agrees.
$k=0:\quad p+q=4,\quad(x,y,z)=(t,2,q),\ t\in\mathbb R$
When k=1, add the first two equations and compare with the third.
$k=1:\quad p+q=6,\quad(x,y,z)=(6-3t,t,q-t),\ t\in\mathbb R$
If the corresponding consistency condition fails, there is no solution; no other k can give multiple solutions.

k=0 requires p+q=4; k=1 requires p+q=6, with the parameterisations above.

Checks and common pitfalls: Both singular cases must be included.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
When k=0, the last two equations fix y and z, leaving x free if the first equation agrees.
$k=0:\quad p+q=4,\quad(x,y,z)=(t,2,q),\ t\in\mathbb R$
When k=1, add the first two equations and compare with the third.
$k=1:\quad p+q=6,\quad(x,y,z)=(6-3t,t,q-t),\ t\in\mathbb R$
If the corresponding consistency condition fails, there is no solution; no other k can give multiple solutions.

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

2025 JM02