Systems in Three Variables: Additional Practice (Set 1)

Subtract the first equation from the second: $-2y + z = 3$. Subtract twice the first from the third: $-y - 3z = -2$, that is $y + 3z = 2$. Solving this pair gives $y = -1$, $z = 1$.

$x = 2 - y - z = 2 - (-1) - 1 = 2$

The solution is $(2, -1, 1)$.

The row $0 = 4$ can never be satisfied, so no triple solves the system.

$0 = 4 \text{ is false} \ \Rightarrow \ \text{empty solution set}$

The system is inconsistent.

The second left side is twice the first, so the equations coincide only when the right side doubles too.

$m = 2(3) = 6$

Any other $m$ makes them parallel and inconsistent.

No point lies on all three planes at once, since the pairwise intersection lines never coincide.

$\text{no common point} \ \Rightarrow \ \text{no solution}$

Elimination exposes it as a false row.

Subtracting the equations gives $2y = 0$, so $y = 0$; adding them gives $2x + 2z = 0$, so $x = -z$.

$(x, y, z) = (1, 0, -1)$

Check: $1 + 0 - 1 = 0$ and $1 - 0 - 1 = 0$, both hold.

A whole plane of solutions means two free variables, so only one real constraint survived.

$3 - r = 2 \ \Rightarrow \ r = 1$

Exactly one independent equation remains; the other two are multiples of it.

Two equations with the same left side but different right sides can never both hold.

$x + y + z = 1 \text{ and } x + y + z = 2 \ \Rightarrow \ 1 = 2, \text{ impossible}$

The other pairs describe planes that do intersect.

Test each triple in all three equations. Only one satisfies every one.

$2 + 1 + 3 = 6, \quad 2(2) + 1 - 3 = 2, \quad 2 - 1 + 2(3) = 7$

So $(2, 1, 3)$ works; each other triple fails the second equation.

One equation in three unknowns leaves two degrees of freedom.

$x = 8 + y - 4z, \quad y \text{ and } z \text{ free}$

Two free variables give a whole plane of solutions.

Let the numbers be $x$, $y$, $z$. Then $x + y + z = 12$ and $x + z = 8$ give $y = 4$. Since $y = z - 3$, the third number is $z = y + 3 = 7$.

$z = 4 + 3 = 7$

Then $x = 8 - z = 1$, so the numbers are $1$, $4$, $7$.

The sum $x + y + z = 12$ together with $x + z = 8$ gives $y = 4$, and $y = z - 3$ gives $z = 7$. Now use $x + z = 8$.

$x = 8 - 7 = 1$

The numbers are $1$, $4$, $7$.

The classification allows a point, a line, a plane, or the empty set, nothing else.

$\text{possible} : \text{point},\ \text{line},\ \text{plane},\ \text{empty}$

A single free variable already gives infinitely many points, so exactly two points can never occur.