Systems with More Variables: Additional Practice (Set 1)

Eliminate $z$. Add the second and third equations: $5x + 2y = 9$. Subtract three times the second from the first: $-5x - y = -7$, that is $5x + y = 7$.

$5x + 2y = 9, \quad 5x + y = 7 \;\Rightarrow\; y = 2,\ x = 1$

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

Subtract the third equation from the first; the $x$ and $y$ terms cancel.

$(x + y + z) - (x + y - z) = 7 - (-1) \;\Rightarrow\; 2z = 8 \;\Rightarrow\; z = 4$

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

$(2x + y - z) + (x - y + z) = 7 + 2 \;\Rightarrow\; 3x = 9 \;\Rightarrow\; x = 3$

Let $a$, $r$, $m$ be the per-kg prices. Subtract the second purchase from the first to get $-a + r = 1$, and subtract the second from the third to get $-a + 2m = 0$. So $r = a + 1$ and $m = \tfrac{a}{2}$. Substitute into the second purchase $2a + r + m = 8$.

$2a + (a + 1) + \tfrac{a}{2} = 8 \;\Rightarrow\; \tfrac{7}{2}a = 7 \;\Rightarrow\; a = 2$

Then $m = \tfrac{a}{2} = 1$, so plums cost $1$ dollar per kg.

The single expression $x + y + z$ cannot equal both $2$ and $5$ at once.

$x + y + z = 2 \text{ and } x + y + z = 5 \;\Rightarrow\; 0 = 3$

The contradiction means no triple works, no matter what the third equation says.

Let $s$ be the smallest. The middle is $s + 4$ and the largest is $3s$, and the three add to $24$.

$s + (s + 4) + 3s = 24 \;\Rightarrow\; 5s = 20 \;\Rightarrow\; s = 4$

The second equation is exactly twice the first, so it repeats the same condition.

$2x + 2y + 2z = 8 \text{ is } 2 \times (x + y + z = 4)$

That leaves only two independent equations for three unknowns, so infinitely many triples solve the system.

Let $a$, $s$, $c$ be the counts, with $s = 3a$. The count becomes $4a + c = 300$, and the revenue $12a + 8s + 5c = 2300$ becomes $36a + 5c = 2300$. Substitute $c = 300 - 4a$.

$36a + 5(300 - 4a) = 2300 \;\Rightarrow\; 16a = 800 \;\Rightarrow\; a = 50$

Then $s = 150$ and $c = 300 - 200 = 100$ child tickets.

Subtract the second equation from the first to get $2y = 2$, so $y = 1$. Add the first and third to cancel $z$: $3x + 2y = 14$, so $x = 4$. Then the first equation gives $z$.

$z = 3 - 4 - 1 = -2$

If two planes never meet, no point can lie on both of them, let alone on all three.

$\text{two parallel distinct planes} \;\Rightarrow\; \text{no common point}$

So the system has no solution.

Eliminate $z$. Add the first and second equations: $3x + 2y = 1$. Subtract twice the first from the third: $-x - 3y = 2$, that is $x + 3y = -2$.

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

A genuine solution satisfies every equation, and the equation you never revisited during elimination is the one most likely to expose a slip.

$\text{check } (1, 2, 3) \text{ in equations (1), (2), and (3)}$

Verify all three before calling the system solved.