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Systems with More Variables: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 The expanded record

Write $2(x+z)+3(y-z)=x+8$ in the form $ax+by+cz=d$ , with integer coefficients having no common factor greater than $1$ and a positive coefficient of $x$ .
1. Hint 1
  Expand both groups and collect matching variables.
2. Hint 2
  Move the $x$ on the right to the left.
Answer

$x+3y-z=8$ .

Full solution

Expanding gives
$2x+2z+3y-3z=x+8$
Collect $z$ terms and subtract $x$ from both sides.
$x+3y-z=8$

The three variables all appear to the first power, and reversing the rearrangement gives the original equation.
The coefficients $1$ , $3$ and $-1$ have no common factor greater than $1$ , and the coefficient of $x$ is positive, so this is the requested form.

Answer

$x+3y-z=8$ .

Key idea

Collecting terms reveals the standard form of a linear equation with three variables.
Problem 1: Mark complete
Problem 2 A full reduction

Solve the system below, and check your triple in all three equations.
$x-2y-z=1$ $x+3y+2z=5$ $2x+y+3z=12$
1. Hint 1
  Choose one variable to remove, and remove that same variable from two different pairs of equations.
2. Hint 2
  The first equation has $-z$ : add twice the first equation to the second, and three times it to the third.
3. Hint 3
  Solve the two new equations for $x$ and $y$ , then put both values into the first equation to find $z$ .
Answer

$(x,y,z)=(2,-1,3)$ .

Full solution

Eliminate $z$ .
Twice the first equation plus the second gives
$3x-y=7$
Three times the first equation plus the third gives
$5x-5y=15$
Dividing by $5$ gives
$x-y=3$

These two equations in $x$ and $y$ form an ordinary system.
Subtracting $x-y=3$ from $3x-y=7$ gives
$2x=4$ $x=2$
Then $y=x-3$ , so $y=-1$ .

Back-substitute both values into the first equation.
$2-2(-1)-z=1$ $z=3$

Check all three equations: $2+2-3=1$ , $2-3+6=5$ , and $4-1+9=12$ .
The solution is $(2,-1,3)$ .

Answer

$(x,y,z)=(2,-1,3)$ .

Key idea

Removing the same variable from two different pairs leaves a two-variable system, and back-substitution then recovers the third value.
Problem 2: Mark complete
Problem 3 Three conditions at once

Find every triple $(x,y,z)$ that satisfies all three equations.
$x-y-2z=2$ $3x+y+3z=9$ $5x-y-z=11$
1. Hint 1
  Reduce the system as usual; a leftover statement with no variables in it tells you how many solutions there are.
2. Hint 2
  Adding the second equation to the first, and then to the third, removes $y$ both times.
3. Hint 3
  Compare twice the first new equation with the second new equation, and read what is left when you subtract.
Answer

There is no such triple: the system has no solution.

Full solution

Eliminate $y$ .
Adding the first two equations gives
$4x+z=11$
Adding the second and third equations gives
$8x+2z=20$

Twice the first new equation is
$8x+2z=22$
The same expression $8x+2z$ cannot equal both $22$ and $20$ .
Subtracting the two equations leaves
$0=-2$

This statement is false.
Every step kept the solutions of the original system, so no triple satisfies all three equations, and the system has no solution.
Its three planes share no common point.

Answer

There is no such triple: the system has no solution.

Key idea

A leftover false statement such as $0=-2$ means the system has no solution.
Problem 3: Mark complete
Problem 4 The storage transfer

Three bins A, B, and C initially contain $24$ folders altogether. If two folders move from A to B, B then has twice as many folders as A. If instead four folders move from C to A, A and C then have equal counts. These are separate proposed transfers from the original arrangement. Find the original counts if possible, or explain why no whole-number counts fit.
1. Hint 1
  Give each original bin count its own variable.
2. Hint 2
  For each proposed transfer, change both the sending and receiving counts.
3. Hint 3
  Use the two transfer relations to express B and C in terms of A.
Answer

No original whole-number counts fit all the conditions.

Full solution

Let $a,b,c$ be the original counts.
The total gives
$a+b+c=24$
The first transfer gives $b+2=2(a-2)$ , so
$b=2a-6$
The second gives $a+4=c-4$ , so
$c=a+8$

Substitute both expressions into the total.
$a+(2a-6)+(a+8)=24$ $4a=22$ $a=\frac{11}{2}$
Then $b=5$ and
$c=\frac{27}{2}.$
These are not all whole-number counts, so the proposed records cannot describe folders.

Answer

No original whole-number counts fit all the conditions.

Key idea

A system for indivisible objects must produce whole-number counts to fit the situation.
Problem 4: Mark complete
Problem 5 The paired measurements

Three readings satisfy the following records, but the constant $k$ has been erased. It is also known that $x=y$ . Recover $k$ and the triple $(x,y,z)$ by eliminating $z$ from two different pairs of records, then check all three completed records.
$2x+y+z=13$ $3x+2y-z=7$ $x-y+2z=k$
1. Hint 1
  With $k$ unknown there are four unknowns, so you need four conditions; the fact $x=y$ is the fourth.
2. Hint 2
  Adding the first two records removes $z$ ; twice the first minus the third also removes $z$ .
3. Hint 3
  Use $x=y$ in the reduced equations to find $x$ and $y$ , then recover the erased constant.
Answer

$k=11$ ; $(x,y,z)=\left(\frac52,\frac52,\frac{11}{2}\right)$ , or $(2.5,\,2.5,\,5.5)$ .

Full solution

Adding the first two equations gives
$5x+3y=20.$
Twice the first minus the third gives
$3x+3y=26-k.$
The extra condition says $x=y$ .
The first reduced equation therefore gives $8x=20$ , so
$x=y=\frac52.$

In the other reduced equation, this yields
$15=26-k,$
hence $k=11$ .
Returning to the first original record gives $z=13-5-\frac52$ , so
$z=\frac{11}{2}.$

The completed original left sides are $13$ , $7$ , and $11$ , respectively, and $x=y$ holds, so the triple and recovered constant meet every condition.

Answer

$k=11$ ; $(x,y,z)=\left(\frac52,\frac52,\frac{11}{2}\right)$ , or $(2.5,\,2.5,\,5.5)$ .

Key idea

Eliminating the same variable from different pairs can combine with an extra relationship to recover a missing record.
Problem 5: Mark complete
Problem 6 The unknown constant

In the system below, $m$ is a fixed number. For which values of $m$ does the system have at least one solution? For each such $m$ , find every solution triple.
$x-2y+2z=2$ $2x-3y+2z=m$ $x-3y+4z=1$
1. Hint 1
  Reduce the system as usual; the value of $m$ decides whether the final leftover statement is true or false.
2. Hint 2
  Subtract twice the first equation from the second, and the first equation from the third, so that $x$ is removed from both.
3. Hint 3
  For the allowed $m$ , pick any value of $z$ ; the reduced equation then gives $y$ , and the first equation gives $x$ .
Answer

Only $m=5$ ; the solutions are then $(4+2t,\,1+2t,\,t)$ for every real $t$ , or equivalently: any $z$ , with $y=1+2z$ and $x=4+2z$ .

Full solution

Remove $x$ .
The second equation minus twice the first gives
$y-2z=m-4$
The third equation minus the first gives
$-y+2z=-1$

Adding these two equations leaves
$0=m-5$
If $m\ne5$ , this is false and the system has no solution.
So $m=5$ , and then the leftover is $0=0$ , which adds no new condition.
This matches the left sides: the second is three times the first minus the third, and
$3\cdot 2-1=5.$

With $m=5$ the first reduced equation reads $y-2z=1$ .
Put $z=t$ for any real $t$ , so that $y=1+2t$ , and the first equation gives
$x=2+2(1+2t)-2t$ $x=4+2t$

The first equation gives $(4+2t)-2(1+2t)+2t=2$ , the second gives $2(4+2t)-3(1+2t)+2t=5$ , and the third gives $(4+2t)-3(1+2t)+4t=1$ , whatever the value of $t$ .
So the three planes share a whole line of solutions.

Answer

Only $m=5$ ; the solutions are then $(4+2t,\,1+2t,\,t)$ for every real $t$ , or equivalently: any $z$ , with $y=1+2z$ and $x=4+2z$ .

Key idea

When one left side is a combination of the others, the system has solutions only if its constant is the same combination of their constants.
Problem 6: Mark complete
Problem 7 The shared edge

The figure shows three planes. The two tilted planes have equations $2x+3y+2z=18$ and $2x-y+2z=10$ , and they meet along the highlighted line. The horizontal plane has equation $z=1$ . Find every point that lies on all three planes. Then the horizontal plane is replaced by $z=4$ , with the tilted planes unchanged: find the point where it now crosses the highlighted line.

Three planes of the system, in a schematic view that is not to scale.

Text description of this figure

A schematic three-dimensional view of three translucent planes, drawn as flat patches with no coordinate axes and no numerical scale. The patch labeled 2 x plus 3 y plus 2 z equals 18 leans to the left and the patch labeled 2 x minus y plus 2 z equals 10 leans to the right, so together they form an open V, each with a short flap reaching past the other. The two tilted patches pass through each other, and the line where they meet is drawn as a single solid highlighted line running through both patches from their top edges to their bottom edges. A horizontal patch labeled z equals 1 lies across the picture and cuts through both tilted patches. The highlighted line passes through the horizontal patch, going from above it to below it, and the part of the line below it is drawn fainter where it is seen through the patch. The crossing point is not marked with a dot, a letter or coordinates.
1. Hint 1
  A point common to all three planes must satisfy all three equations at once.
2. Hint 2
  Subtract the second equation from the first to find $y$ ; the third gives $z$ , and then the first equation gives $x$ .
3. Hint 3
  Every point of the highlighted line satisfies both tilted equations, so its $y$ value is the same all along the line.
Answer

Exactly one point, $(5,2,1)$ . The plane $z=4$ crosses the line at $(2,2,4)$ .

Full solution

Subtracting the second equation from the first gives
$4y=8$ $y=2$
The third equation fixes $z=1$ .
The first then gives
$2x+6+2=18$ $x=5$

Each coordinate was forced in turn, so $(5,2,1)$ is the only common point.
It checks: $10+6+2=18$ , $10-2+2=10$ , and $z=1$ .

The subtraction used only the two tilted planes, so every point of the highlighted line has $y=2$ .
With $z=4$ , the first equation gives
$2x+6+8=18$ $x=2$
The new crossing is $(2,2,4)$ , and $4-2+8=10$ checks the second equation.

Answer

Exactly one point, $(5,2,1)$ . The plane $z=4$ crosses the line at $(2,2,4)$ .

Key idea

The line where two planes meet holds every point satisfying both of their equations, and a third plane crossing it picks out the one point that satisfies all three.
Problem 7: Mark complete
Problem 8 The unused record

The records are $x+y+z=8$ , $2x+2y+2z=16$ , and $y-z=2$ . Ren subtracts twice the first equation from the second and gets $0=0$ . Ren claims the third record can be discarded because any triple now works. Is this correct? Give two valid triples and one triple that satisfies the first two records but fails the third.
1. Hint 1
  An identity removes one restriction, not the restrictions in the other records.
2. Hint 2
  Choose a value of $z$ , then use the third and first equations in that order.
Answer

No. Valid examples: $(6,2,0)$ and $(4,3,1)$ . Failing example: $(8,0,0)$ .

Full solution

The identity shows that the second equation repeats the first.
The third still requires $y=z+2$ .
With $z=0$ , this gives $y=2$ and $x=6$ .
With $z=1$ , it gives $y=3$ and $x=4$ .

Both triples have total $8$ and satisfy $y-z=2$ , so they satisfy every record.
But $(8,0,0)$ gives totals $8$ and $16$ in the first two records while
$y-z=0$
This fails the third.
Ren is incorrect.

Answer

No. Valid examples: $(6,2,0)$ and $(4,3,1)$ . Failing example: $(8,0,0)$ .

Key idea

An always-true leftover does not erase the conditions that survive elsewhere in a system.
Problem 8: Mark complete
Problem 9 The reversed progress

A learner starts with three unknowns. After one full elimination round, two equations involve only $x$ and $y$ , while a saved original equation still contains $z$ with a nonzero coefficient. The learner says finding $z$ at the end would restart the whole three-unknown problem. Is the learner right? Explain.
1. Hint 1
  Distinguish a letter still printed in an equation from a value that remains unknown.
2. Hint 2
  Once two coordinates are known, substitute those numbers into the saved equation.
Answer

No; the saved equation then has just the one unknown $z$ .

Full solution

Write the saved equation as $ax+by+cz=d$ , where $c\ne0$ .
After finding values $x=u$ and $y=v$ , it becomes
$au+bv+cz=d$
The terms $au$ and $bv$ are known numbers.

Thus
$cz=d-au-bv$ $z=\frac{d-au-bv}{c}$
Dividing is permitted because $c\ne0$ .
Back-substitution fills in a missing value rather than making known values unknown again, so it does not restart the process.

Answer

No; the saved equation then has just the one unknown $z$ .

Key idea

Back-substitution moves through saved equations with more known values at each step.
Problem 9: Mark complete
Problem 10 The shared equation

Kai solves the system below. He adds the first two equations to get $3x-y=11$ , then adds the second and third equations to get $5x-y=19$ . Mia objects that Kai used the second equation twice. Kai replies that his two new equations, together with the first equation, have exactly the same solutions as the original system. Is Kai right? Solve the system and explain.
$x-3y+z=3$ $2x+2y-z=8$ $3x-3y+z=11$
1. Hint 1
  What does a solution of the original system do to each of Kai's new equations? And can the original equations be recovered from his?
2. Hint 2
  Which combination of the first equation and $3x-y=11$ gives back the second equation? Then use the second to get back the third.
3. Hint 3
  Subtract $3x-y=11$ from $5x-y=19$ to find $x$ , then find $y$ , and put both into the first equation to find $z$ .
Answer

Yes, Kai is right; the only solution is $(4,1,2)$ .

Full solution

Every solution of the original system satisfies $3x-y=11$ and $5x-y=19$ , because each is a sum of two of its equations.
So Kai's new system loses no solution.

The steps can also be undone.
The second equation is $3x-y=11$ minus the first equation, and the third is $5x-y=19$ minus the second.
So any triple that satisfies the first equation and both new equations satisfies all three originals, and no solution is gained either.

Mia's worry would matter only if an original equation had been lost; Kai used all three, and the previous step shows each can be recovered, so reusing the second equation loses nothing.
Kai is right.

Subtracting $3x-y=11$ from $5x-y=19$ gives
$2x=8$ $x=4$
Then $y=3x-11$ , so $y=1$ .
The first equation gives
$4-3+z=3$ $z=2$

Check: $4-3+2=3$ , $8+2-2=8$ , and $12-3+2=11$ .
The one solution is $(4,1,2)$ .

Answer

Yes, Kai is right; the only solution is $(4,1,2)$ .

Key idea

Two different pairs may share an equation: each step can be undone, so the reduction keeps exactly the solutions of the original system.
Problem 10: Mark complete

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