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Systems in Three 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 A full elimination sweep

Solve the system $x+y+2z=9$ , $2x-y+z=3$ , and $x+3y-z=4$ for real $x,y,z$ .
1. Hint 1
  Eliminate the same variable from the other two equations using the first equation.
2. Hint 2
  Combine the two resulting equations to eliminate a second variable, then back-substitute.
Answer

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

Full solution

Eliminate $x$ using the first equation, $x=9-y-2z$ .
Substituting into the other two:
$\begin{aligned} 2(9-y-2z)-y+z&=3 \\ -3y-3z&=-15 \end{aligned}$ $\begin{aligned} (9-y-2z)+3y-z&=4 \\ 2y-3z&=-5 \end{aligned}$

Dividing the first reduced equation by $-3$ gives $y+z=5$ , so $y=5-z$ .
Substituting into the second:
$\begin{aligned} 2(5-z)-3z&=-5 \\ z&=3 \end{aligned}$
Then
$\begin{aligned} y&=5-3 \\ &=2 \end{aligned}$
and back-substituting into $x=9-y-2z$ gives
$\begin{aligned} x&=9-2-6 \\ &=1 \end{aligned}$

Answer

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

Key idea

A forward elimination sweep to triangular form, then back-substitution, solves a system with no equation already isolated to one variable.
Problem 1: Mark complete
Problem 2 Three recorded totals

Determine the solution set of $x+y=4$ , $z=2$ , and $x+y+z=9$ over the real numbers.
1. Hint 1
  Combine the first two facts before trying individual values.
2. Hint 2
  Compare the implied total with the last condition.
Answer

No solution.

Full solution

Adding the first two equations forces $x+y+z=6$ .
Subtract that from the third equation.
$0=3$
The conditions disagree, so no triple works.

Answer

No solution.

Key idea

A single impossible consequence makes the entire system inconsistent.
Problem 2: Mark complete
Problem 3 A family of triples

Parameterize the real solutions of $z=2x-3y+1$ using $x=s$ and $y=t$ .
1. Hint 1
  The equation determines the third coordinate after the first two are chosen.
2. Hint 2
  Keep both freely chosen letters in the third coordinate.
Answer

$(s,t,2s-3t+1)$ for real $s,t$ .

Full solution

The choices $x=s$ and $y=t$ force
$z=2s-3t+1$
Every real pair $s,t$ supplies a solution, and every solution can be obtained by choosing its first two coordinates.
One genuine linear equation in three variables describes a plane.

Answer

$(s,t,2s-3t+1)$ for real $s,t$ .

Key idea

Two freely chosen coordinates parameterize a plane in three-dimensional space.
Problem 3: Mark complete
Problem 4 Three route counters

Three counters measure real flows $a,b,c$ in liters per minute. Their readings are $a+b=9$ , $b+c=7$ , and $a+c=10$ . Find the three flows.
1. Hint 1
  Each reading combines two flows; combine readings to isolate one flow.
2. Hint 2
  Add the first and third readings, then subtract the second.
Answer

$a=6$ , $b=3$ , $c=4$ liters per minute.

Full solution

Adding the first and third readings gives $(a+b)+(a+c)=2a+b+c$ ; subtracting the second reading $b+c=7$ removes $b$ and $c$ , leaving $2a$ alone.
$\begin{aligned} 2a&=9+10-7 \\ a&=6 \\ b&=9-6 \\ c&=10-6 \end{aligned}$

Thus $b=3$ and $c=4$ .
All readings check: $6+3=9$ , $3+4=7$ , and $6+4=10$ .

Answer

$a=6$ , $b=3$ , $c=4$ liters per minute.

Key idea

Pairwise totals can determine three individual quantities.
Problem 4: Mark complete
Problem 5 Three intersecting conditions

Solve $x+2y=1$ , $y+z=4$ , and $2x+3y-z=-2$ for real $x,y,z$ . Describe every solution and its geometric shape.
1. Hint 1
  Test whether the last condition adds a restriction beyond the first two.
2. Hint 2
  Subtract twice the first equation from the third, and compare with the second.
Answer

$(x,y,z)=(1-2t,t,4-t)$ for real $t$ ; a line.

Full solution

Subtracting twice the first equation from the third gives $-y-z=-4$ , the negative of the second.
The last equation is redundant.
Set $y=t$ .
$\begin{aligned} x&=1-2t \\ z&=4-t \end{aligned}$

The third left side becomes $2(1-2t)+3t-(4-t)=-2$ for every $t$ .
One free parameter gives a line.

Answer

$(x,y,z)=(1-2t,t,4-t)$ for real $t$ ; a line.

Key idea

A redundant equation leaves the solutions constrained by the independent equations that remain.
Problem 5: Mark complete
Problem 6 A missing right side

The system is $x-z=2$ , $2y+z=5$ , and $x+2y=k$ . Find the real $k$ for which solutions exist, then describe all of them.
1. Hint 1
  Look for a total implied by the first two conditions.
2. Hint 2
  Add their left sides and compare with the third.
Answer

$k=7$ ; $(x,y,z)=(2+t,(5-t)/2,t)$ for real $t$ .

Full solution

Adding the first two equations gives $x+2y=7$ , so consistency requires $k=7$ .
If $k\ne7$ , subtraction gives a false row.
At $k=7$ , let $z=t$ .
$\begin{aligned} x&=2+t \\ y&=\frac{5-t}{2} \end{aligned}$

Then
$\begin{aligned} x+2y&=(2+t)+(5-t) \\ &=7 \end{aligned}$
confirming the final equation for all $t$ .

Answer

$k=7$ ; $(x,y,z)=(2+t,(5-t)/2,t)$ for real $t$ .

Key idea

A dependent third condition must repeat the total already forced by the others.
Problem 6: Mark complete
Problem 7 Plane descriptions

Find every real solution triple $(x,y,z)$ of $y-2z=3$ , $4y-8z=12$ , and $-y+2z=-3$ , where $x$ is otherwise unconstrained. Explain why one free letter is insufficient to describe all solutions.
1. Hint 1
  Find how many distinct restrictions the equations impose.
2. Hint 2
  Check which coordinate never appears, as well as the freedom within the surviving equation.
Answer

$(x,y,z)=(s,3+2t,t)$ for real $s,t$ ; a plane with two free variables.

Full solution

The second equation is four times the first and the third is its negative.
Only $y-2z=3$ survives.
Set $x=s$ and $z=t$ .
$y=3+2t$
Both $s$ and $t$ vary independently, so this is a plane and a one-parameter linear family would leave out solutions.

Answer

$(x,y,z)=(s,3+2t,t)$ for real $s,t$ ; a plane with two free variables.

Key idea

Count independent restrictions before deciding how many free parameters are needed.
Problem 7: Mark complete
Problem 8 A proposed point

A student says three planes with a common line can still have exactly one common point after their equations are solved more carefully. Is this possible? Explain.
1. Hint 1
  The phrase common line describes points already lying on every plane.
2. Hint 2
  Compare the size of a line with a single point.
Answer

No; their common solution set includes the whole line.

Full solution

Every point of the common line satisfies all three equations by the meaning of common.
A line contains infinitely many points.
Reversible elimination preserves those solutions, so it cannot reduce that set to one point.

Answer

No; their common solution set includes the whole line.

Key idea

Solving equivalent equations cannot remove points known to satisfy the entire system.
Problem 8: Mark complete
Problem 9 Two endings together

A system reduces to $x-y+z=8$ , $0=0$ , and $0=-2$ . A student ignores the second row and concludes there is no solution. Is this conclusion justified? Explain what each numerical row means.
1. Hint 1
  A true numerical row and a false numerical row carry different information.
2. Hint 2
  Ask whether any triple can satisfy the false numerical row.
Answer

Yes; no solution. The zero row is redundant, and the other numerical row is false.

Full solution

The row $0=0$ imposes no condition.
But
$0=-2$
is false independently of $x,y,z$ .
No triple satisfies every row, so the empty-set conclusion is justified even though another row vanished.

Answer

Yes; no solution. The zero row is redundant, and the other numerical row is false.

Key idea

A false row overrides any redundant rows in a system.
Problem 9: Mark complete
Problem 10 Two descriptions of a line

Are $(2+t,1-t,3t)$ for real $t$ and $(2+u/3,1-u/3,u)$ for real $u$ the same solution set? Justify your answer in both directions.
1. Hint 1
  The free letter is a label for a point, not an extra constraint.
2. Hint 2
  Relate the parameters by equating the third coordinates.
Answer

Yes; they describe the same line.

Full solution

For any real $t$ , choose $u=3t$ .
Substitution in the second description produces $(2+t,1-t,3t)$ .
Conversely, given any real $u$ , choose
$t=\frac{u}{3}$
Substitution in the first description produces the second.
Each family therefore covers all points of the other.

Answer

Yes; they describe the same line.

Key idea

Different parameter choices can describe exactly the same solution set.
Problem 10: Mark complete

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