Systems of Three Equations in Three Variables

$(2, 3, 4)$

Subtracting the second from the first: $2y = 6$, so $y = 3$. Then $x + z = 6$ and $2x - z = 0$. Adding: $3x = 6$, so $x = 2$ and $z = 4$.

$(2, 3, 2)$

Eliminate $z$: the first plus the third gives $5x + 2y = 16$; three times the first plus the second gives $7x + y = 17$. Then $y = 17 - 7x$ and $5x + 34 - 14x = 16$, so $x = 2$, $y = 3$, and $z = 2x + y - 5 = 2$.

$(1, 2, 3)$

Eliminate $y$: the first plus twice the second gives $5x + 5z = 20$, so $x + z = 4$; the second plus the third gives $5x - z = 2$. Adding: $6x = 6$, so $x = 1$, $z = 3$, and $y = 2x + z - 3 = 2$.

$(2, 3, 4)$

Adding all three: $2(x + y + z) = 18$, so $x + y + z = 9$. Subtract each equation from this: $z = 4$, $x = 2$, $y = 3$.

No solution

Subtracting the first two gives $0 = 2$. The first two planes are parallel.

Infinitely many

The second equation is twice the first: the same plane. The two remaining planes meet in a line, and every point on it is a solution.

$10$ nickels, $5$ dimes, $15$ quarters

$n + d + q = 30$, $5n + 10d + 25q = 475$, $n = 2d$. Then $3d + q = 30$ and $20d + 25q = 475$. With $q = 30 - 3d$: $20d + 750 - 75d = 475$, so $d = 5$, $n = 10$, $q = 15$.

$y = 2x^2 - 3x + 1$

$(0, 1)$ gives $c = 1$. Then $a - b = 5$ and $4a + 2b = 2$, so $2a + b = 1$. Adding: $3a = 6$, $a = 2$, $b = -3$.

$8$, $4$ and $12$

With $x = 2y$ and $z = x + 4 = 2y + 4$: $2y + y + 2y + 4 = 24$, so $y = 4$.

Which variables are left in each new equation?

The student eliminated different variables. Both eliminations must remove the same one.

The first new equation has $x$ and $y$; the second has $x$ and $z$. Together they have three unknowns. Eliminating $z$ from a second pair, such as $(2)$ and $(3)$, would give two equations in $x$ and $y$ alone.