Systems with More Variables: Additional Practice (Set 2)

Add the first two equations to cancel $z$: $4x + 3y = -2$. Subtract three times the second from the third to cancel $z$: $x + 7y = -13$.

$4x + 3y = -2, \quad x + 7y = -13 \;\Rightarrow\; y = -2$

Eliminate $z$. Add the second and third equations: $3x + y = 8$. Add the first and third: $2x + 3y = 10$.

$3x + y = 8, \quad 2x + 3y = 10 \;\Rightarrow\; x = 2,\ y = 2$

Then $z = 2$, so the solution is $(2, 2, 2)$.

A vanished equation is a redundant one, and with nothing left to pin the third coordinate the survivors describe a whole line or a whole plane.

$\text{redundant equations, no contradiction} \;\Rightarrow\; \text{infinitely many solutions}$

The planes share a whole line, or coincide as one plane, so every point on it solves the system. The second condition in the prompt is what makes this safe to say: a $0 = 0$ on its own only tells you one equation was redundant. Take $x + y + z = 1$ and $2x + 2y + 2z = 2$, which collapse to $0 = 0$, and add $x + y + z = 5$: now $0 = 4$ and the planes never meet at all.

Add the first two equations; the $y$ and $z$ terms cancel.

$(x + y + z) + (x - y - z) = 6 + 4 \;\Rightarrow\; 2x = 10 \;\Rightarrow\; x = 5$

Add the first two equations to cancel $y$ and $z$: $2x = 8$, so $x = 4$. Then the first gives $y + z = 1$ and the third gives $y - z = -3$; add them.

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

Eliminate $z$. Add the first and third equations: $2x + 3y = 11$. Subtract the second from the first: $-x + 2y = 5$.

$2x + 3y = 11, \quad -x + 2y = 5 \;\Rightarrow\; y = 3$

Let Carl be $c$. Then Ana is $2c$ and Ben is $c + 3$, and the three ages add to $27$.

$2c + (c + 3) + c = 27 \;\Rightarrow\; 4c = 24 \;\Rightarrow\; c = 6$

Ana is $2c = 12$.

Eliminate $x$. Three times the first minus twice the second gives $7y - z = 5$; the first minus twice the third gives $-y - 5z = -11$, that is $y + 5z = 11$.

$7y - z = 5, \quad y + 5z = 11 \;\Rightarrow\; y = 1,\ z = 2$

Three planes tilted in general directions cross at a single shared point.

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

That point's coordinates are the unique triple that solves the system.

Subtract the second equation from the first to get $2z = 10$, so $z = 5$; subtract the third from the first to get $2y = 6$, so $y = 3$.

$2z = 10 \;\Rightarrow\; z = 5, \qquad 2y = 6 \;\Rightarrow\; y = 3$

Then $x = 2$, so the solution is $(2, 3, 5)$.

One round removes a single variable and a single equation.

$5 \times 5 \;\rightarrow\; 4 \times 4$

Four more rounds would bring the system down to one equation in one unknown.

Subtract the first equation from the second to get $x - y = 2$, and from the third to get $-y + z = 1$. So $x = y + 2$ and $z = y + 1$; substitute into the first equation.

$(y + 2) + 2y + (y + 1) = 7 \;\Rightarrow\; 4y = 4 \;\Rightarrow\; y = 1$

Then $x = 3$.