Systems in Three Variables: Additional Practice (Set 2)

The second left side is exactly twice the first, so the equations match only if the right side is also doubled.

$2(x + 2y + z) = 2(3) = 6 \ \Rightarrow \ k = 6$

Any other $k$ makes the pair inconsistent instead of identical.

Eliminating $x$ gives $-2y + z = 3$ and $y + 3z = 2$. From the first, $z = 3 + 2y$; substitute into the second.

$y + 3(3 + 2y) = 2 \ \Rightarrow \ 7y = -7 \ \Rightarrow \ y = -1$

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

Clearing $x$ gives $(2) - 2(1): y - 3z = -3$ and $(3) - 3(1): y - 3z = 0$. Subtract these.

$0 = 3$

The false row means the system is inconsistent, so there is no solution.

From the bottom, $z = \frac{2}{3}$, then $y = 1 + \frac{2}{3} = \frac{5}{3}$. Substitute both into the top row.

$x = -y - z = -\frac{5}{3} - \frac{2}{3} = -\frac{7}{3}$

Back-substitution carries the fractions upward without trouble.

Add the first and third to clear $z$: $2x + 3y = -4$. Add the second and third: $3x + y = 1$, so $y = 1 - 3x$. Substitute.

$2x + 3(1 - 3x) = -4 \ \Rightarrow \ -7x = -7 \ \Rightarrow \ x = 1, \ y = -2$

The solution is $(1, -2, 3)$.

Subtracting the first equation from the second gives $y + 2z = 3$, and from the third gives $2y + 4z = 6$, the same relation, so one row vanishes. Setting $z = t$ gives $y = 3 - 2t$ and $x = 3 - y - z = t$.

$(x, y, z) = (t,\ 3 - 2t,\ t)$

One free variable makes the solution set a line.

The three copies impose a single condition, so $r = 1$ and two variables are free.

$3 - 1 = 2 \text{ free variables}$

The solutions fill the entire plane $x + y + z = 6$.

Three pivots leave no free variable, so back-substitution forces one value per unknown.

$3 - 3 = 0 \text{ free variables} \ \Rightarrow \ \text{one point}$

The planes cross at a single shared point.

Adding the first two equations gives $2x + 3y + 4z = 0$. The third matches this exactly when $a = 4$, making it redundant.

$a = 4 \ \Rightarrow \ \text{third equation is redundant}$

With a redundant equation the all-zero-right-side system gains a free variable and nonzero solutions.

Substituting all zeros makes every left side $0$, matching every right side.

$a(0) + b(0) + c(0) = 0$

So $(0, 0, 0)$ always works; a homogeneous system is never inconsistent.

The second equation is twice the first, so it vanishes and two constraints remain: $x + y - z = 1$ and $x - y + z = 3$. Adding them gives $2x = 4$, so $x = 2$; subtracting gives $y - z = -1$. Let $z = t$.

$x = 2, \quad y = t - 1 \ \Rightarrow \ (2,\ t - 1,\ t)$

One free variable makes the solution set a line.

A false numerical row means no triple satisfies all three equations.

$0 = -7 \text{ is impossible} \ \Rightarrow \ \text{no common point}$

The planes are arranged so that no single point lies on all three.