Systems in Three Variables: Practice

Eliminating $x$ gives $y + 2z = 1$ and $y - z = 4$. Subtract the second from the first to strip out $y$.

$(y + 2z) - (y - z) = 1 - 4 \ \Rightarrow \ 3z = -3 \ \Rightarrow \ z = -1$

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

"Twice as much house as decaf" means the house amount equals two times the decaf amount.

$h = 2k$

The phrase "$2h = k$" would instead say the decaf is double the house, which is the reverse.

Set $z = t$ and solve each surviving equation for its leading variable.

$x = 5 - t, \qquad y = 1 + t$

So the solution line is $(5 - t,\ 1 + t,\ t)$.

Since $h = d + k$, the first equation gives $2(d + k) = 12$, so $d + k = 6$ and $h = 6$. Dividing the cost equation by $3$ and substituting $h = 6$ gives $5d + 6k = 34$.

$(5d + 6k) - 5(d + k) = 34 - 30 \ \Rightarrow \ k = 4$

So $4$ pounds of decaf, with $d = 2$ and $h = 6$.

First solve the bottom row: $2z = 5$ gives $z = \frac{5}{2}$. Substitute into $y + z = 4$.

$y = 4 - \frac{5}{2} = \frac{8}{2} - \frac{5}{2} = \frac{3}{2}$

Back-substitution handles fractions the same way it handles whole numbers.

Subtract $2(x + 2y + z) = 2 x + 4y + 2z$ from $2x - y + 3z$, matching the right sides.

$(2x - y + 3z) - (2x + 4y + 2z) = 4 - 6 \ \Rightarrow \ -5y + z = -2$

The $x$ terms cancel, leaving an equation in $y$ and $z$.

With $y = s$ and $z = t$ free, solve the one equation for $x$.

$x = 1 - y - z = 1 - s - t$

So every solution is $(1 - s - t,\ s,\ t)$. Two free letters sweep out a plane.

Subtract the first equation from the second to isolate $x$.

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

Then $(3) - (1)$ gives $y = 3$, and the first equation gives $z = -2$.

From the differences of equations, $x = 1$ and $y = 3$. Substitute into the first equation.

$1 + 3 + z = 2 \ \Rightarrow \ z = -2$

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

With three pivots there are no free variables, so back-substitution forces one value per unknown.

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

The three planes meet at a single point.

Two pivots means two constraints and one free variable.

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

One free variable traces a line.

One pivot means one constraint and two free variables.

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

Two free variables fill a plane.