Systems in Three Variables: Practice

Adding the first equation to each of the others gives $2y + z = 2$ both times, so one row vanishes and a variable is free. Setting $z = 2t$ gives $y = 1 - t$ and $x = t$.

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

The alternative $(t, 1 + t, 2t)$ fails the first equation, since $t + 3(1 + t) + 2t = 6t + 3 \neq 3$.

The second equation is $-2$ times the first and the third is $3$ times the first, so both reduce to $0 = 0$ and only one constraint survives. Solving $2x - y + z = 3$ for $y$ with $x = s$, $z = t$ free gives $y = 2s + t - 3$.

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

Two free variables fill a plane.

The second left side is twice the first, so it is forced to equal $2(2) = 4$. Any other value contradicts it.

$k = 5 \neq 4 \ \Rightarrow \ 0 = 1, \text{ impossible}$

At $k = 4$ the equations coincide (infinitely many solutions); at $k = 5$ they are parallel and inconsistent.

Let $t$, $w$, $f$ be the counts. Then $t + w + f = 24$, $t + 2w + 5f = 56$ (value in tens of dollars), and $t - w = 2$. Subtracting the first from the second gives $w + 4f = 32$; using $t = w + 2$ in the first gives $2w + f = 22$.

$w + 4f = 32, \quad 2w + f = 22 \ \Rightarrow \ f = 6$

Then $w = 8$ and $t = 10$.

Each pair of planes meets in a line, but the three lines never cross at one shared point.

$\text{no point on all three} \ \Rightarrow \ \text{no solution}$

Elimination reveals it as a false row such as $0 = 3$.

Two vanishing rows drop out, leaving one constraint, so $r = 1$.

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

Two free variables fill a plane.

A point is on the line only if all three coordinates match one value of $t$. The first coordinate $2$ forces $t = 2$.

$(t,\ 1 - t,\ 2t) = (2,\ 1 - 2,\ 4) = (2, -1, 4)$

So $(2, -1, 4)$ works; the others fail at least one coordinate.

On the line, $x = t$ and $z = 2t$, so compare $z$ with $x$.

$z = 2t = 2x$

This holds for all $t$, while $z = x$, $y = 2x$, and $x = 2z$ each fail except at isolated values.

Subtract twice the first from the second: $-3y - z = -1$. Subtract the first from the third: $2y - 3z = 8$. Solving this pair gives $y = 1$ and $z = -2$.

$-3(1) - z = -1 \ \Rightarrow \ z = -2$

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

The facts are $x + y + z = 100$, $10x + 20y + 50z = 2200$ (acid in mL times $100$), and $x = y$. With $x = y$, the first gives $2x + z = 100$ and the second gives $3x + 5z = 220$.

$2x + z = 100, \quad 3x + 5z = 220 \ \Rightarrow \ z = 20$

Then $x = y = 40$.

The two coincident planes give one constraint, and the crossing plane gives a second independent one, so $r = 2$.

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

Two distinct planes meet in a line, so the solution set is that line.

A plane of solutions comes from one surviving constraint, so $r = 1$.

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

A line needs $1$ parameter and a point needs $0$; a plane needs $2$.