Systems with More Variables: Practice

Eliminate $z$. Add the first and third equations to get $5x + 4y = 13$; subtract twice the third from the second to get $5x + 3y = 11$.

$5x + 4y = 13, \quad 5x + 3y = 11 \;\Rightarrow\; y = 2,\ x = 1$

The third equation gives $z = 3$, so the solution is $(1, 2, 3)$.

Add the first two equations to cancel $y$: $2x + 3z = 3$. Add the second and third to cancel $y$ again: $3x + z = 8$.

$2x + 3z = 3, \quad 3x + z = 8 \;\Rightarrow\; x = 3,\ z = -1$

Let $s$ be the smallest. The middle number is $s + 3$ and the largest is $4s$, and the three add to $15$.

$s + (s + 3) + 4s = 15 \;\Rightarrow\; 6s = 12 \;\Rightarrow\; s = 2$

The largest is $4s = 8$.

Subtract twice the first equation from the second; the whole left side vanishes.

$(2x + 2y + 2z) - 2(x + y + z) = 7 - 6 \;\Rightarrow\; 0 = 1$

That is false, so the system has no solution.

A solution is a point common to all three planes, so without one no triple satisfies all three equations.

$\text{no common point} \;\Rightarrow\; \text{no solution}$

Such a system is called inconsistent.

Each round of elimination removes one variable and one equation.

$4 \times 4 \;\rightarrow\; 3 \times 3$

Repeating the round drives the system down toward a single equation in one unknown.

Add the first two equations to cancel $y$: $3x + 2z = 4$. Add twice the second to the third to cancel $y$: $5x + z = 9$.

$3x + 2z = 4, \quad 5x + z = 9 \;\Rightarrow\; x = 2$

Let $n$, $d$, $q$ be the counts. Since $n = 2q$, the count $n + d + q = 20$ becomes $3q + d = 20$, and the value $5n + 10d + 25q = 215$ becomes $35q + 10d = 215$. Substitute $d = 20 - 3q$.

$35q + 10(20 - 3q) = 215 \;\Rightarrow\; 5q = 15 \;\Rightarrow\; q = 3$

Then $n = 6$ and $d = 20 - 9 = 11$.

Add all three equations: $6x + 6y + 6z = 18$, so $x + y + z = 3$. Subtracting this from the first gives $x + 2y = 3$, and subtracting twice it from the third gives $x - y = 0$.

$x + 2y = 3, \quad x = y \;\Rightarrow\; x = y = 1,\ z = 1$

The solution is $(1, 1, 1)$.

Subtract the first equation from the third to get $x = 2$. Subtract the second from the first to get $3y = 0$, so $y = 0$. Then use the first equation.

$2 + 0 + z = -1 \;\Rightarrow\; z = -3$

The first two planes already meet along a whole line, and the third equation follows from them, so it adds no new restriction.

$(1) + (2) = (3) \;\Rightarrow\; \text{only two independent equations}$

Every point on that shared line satisfies all three equations, so the system has infinitely many solutions.

Subtract the third equation from the first to cancel $x$ and $y$: $3z = 3$, so $z = 1$. Add the first two to cancel $y$: $3x + 2z = 11$.

$3x + 2(1) = 11 \;\Rightarrow\; 3x = 9 \;\Rightarrow\; x = 3$