Systems with More Variables: Practice

A solution assigns one value to each of the three variables, recorded in a fixed order.

$(x,\ y,\ z)$

The order matters, exactly as it did for ordered pairs.

The $z$ terms are $+z$ and $-z$, opposites that cancel when you add.

$(x + y + z) + (x + y - z) = 7 + 3 \;\Rightarrow\; 2x + 2y = 10$

Only $z$ disappears.

Add the left sides and the right sides; the $z$ terms cancel.

$2x + 2y = 10$

Dividing by $2$ would give $x + y = 5$, but the raw sum is $2x + 2y = 10$.

Subtracting flips every sign in the second equation, so the $x$ terms cancel and $2y - (-y) = 3y$.

$(x + 2y + z) - (x - y + 2z) = 8 - 3 \;\Rightarrow\; 3y - z = 5$

A linear equation has each variable only to the first power, never multiplied together, squared, or in a denominator.

$x + y + z = 6$

The others multiply the variables, square one, and divide by one.

Each pair gives one equation without $z$, and two pairs give two of them.

$\text{pair 1} \rightarrow (A), \qquad \text{pair 2} \rightarrow (B)$

Both $(A)$ and $(B)$ involve only $x$ and $y$.

Substitute the coordinates and compute the left side.

$3(2) + 0 - 1 = 6 - 1 = 5$

The result is $5$, which matches the right side, so the triple satisfies the equation.

The $+y$ and $-y$ cancel, and the $x$ and $z$ terms combine.

$(x + y + z) + (2x - y + z) = 6 + 5 \;\Rightarrow\; 3x + 2z = 11$

In general position the three planes cross at a single shared point.

$\text{three planes} \rightarrow \text{one common point}$

No solution and infinitely many are the special exceptions.

The $+z$ and $-z$ cancel when you add.

$(x + z) + (x - z) = 5 + 1 \;\Rightarrow\; 2x = 6$

So $x = 3$.

A solution must satisfy all three equations at once, so every one of them must be tested.

$\text{check } (x, y, z) \text{ in (1), (2), and (3)}$

Passing only one or two of them is not enough.

Every equation reducing to $0 = 0$ means nothing is left to narrow the triples down, so every triple works.

$\text{no surviving constraint} \;\Rightarrow\; \text{infinitely many}$

The two conditions in the prompt are both doing work. One equation reducing to $0 = 0$ only says that equation was redundant, and the equations still standing decide: $x + y + z = 1$ with $2x + 2y + 2z = 2$ collapses to $0 = 0$, but adding $x + y + z = 5$ gives $0 = 4$ and there is no solution at all. A single $0 = k$ with $k \neq 0$ anywhere overrules every vanished equation.