Algebra 3 Quiz: Three Variable Systems
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Three Variable SystemsQuestion 1 of 12

Which ordered triple satisfies the system $$ \begin{cases} 2x+y-z=4 \ x-2y+3z=-1 \ 3x+y+z=6 \end{cases}

(43,53,13)\left(\frac{4}{3}, \frac{5}{3}, \frac{1}{3}\right)
(53,43,13)\left(\frac{5}{3}, \frac{4}{3}, \frac{1}{3}\right)
(43,13,53)\left(\frac{4}{3}, \frac{1}{3}, \frac{5}{3}\right)
(1,2,0)(1,2,0)
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Algebra 3 Quiz

Algebra 3 Quiz: Three Variable Systems

Practice Three Variable Systems in Algebra 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Three Variable Systems, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Which ordered triple satisfies the system $$ \begin{cases} 2x+y-z=4 \ x-2y+3z=-1 \ 3x+y+z=6 \end{cases}

  1. (43,53,13)\left(\frac{4}{3}, \frac{5}{3}, \frac{1}{3}\right) (correct answer)
  2. (53,43,13)\left(\frac{5}{3}, \frac{4}{3}, \frac{1}{3}\right)
  3. (43,13,53)\left(\frac{4}{3}, \frac{1}{3}, \frac{5}{3}\right)
  4. (1,2,0)(1,2,0)
Explanation: When you see a system of equations like this, remember that an ordered triple must satisfy every equation in the system, not just one or two. The quickest check is to substitute each candidate triple into all three equations. For (43,53,13)\left(\frac{4}{3}, \frac{5}{3}, \frac{1}{3}\right), verify:
  • First: 2(43)+53−13=8+5−13=42\left(\frac43\right)+\frac53-\frac13=\frac{8+5-1}{3}=4 ✓
  • Second: 43−2(53)+3(13)=4−10+33=−1\frac43-2\left(\frac53\right)+3\left(\frac13\right)=\frac{4-10+3}{3}=-1 ✓
  • Third: 3(43)+53+13=12+5+13=63\left(\frac43\right)+\frac53+\frac13=\frac{12+5+1}{3}=6 ✓
So this triple works everywhere. The choice (53,43,13)\left(\frac{5}{3}, \frac{4}{3}, \frac{1}{3}\right) is a trap: it swaps the roles of xx and yy. It fails the first equation because 2(53)+43−13=133≠42(\frac53)+\frac43-\frac13=\frac{13}{3}\neq 4. The choice (43,13,53)\left(\frac{4}{3}, \frac{1}{3}, \frac{5}{3}\right) also fails the first equation, giving 43\frac{4}{3} instead of 44; only the third equation happens to work. Finally, (1,2,0)(1,2,0) satisfies the first equation, 2(1)+2−0=42(1)+2-0=4, but it fails the second: 1−2(2)+3(0)=−31-2(2)+3(0)=-3, not −1-1. It also fails the third. Your study tip: when answer choices are given, substitution is fast. But don't stop after one equation — check all three, because a triple can pass one equation while failing the system.

Question 2

The system {x−y+2z=52x+y−z=13x+z=6\begin{cases} x-y+2z=5 \\ 2x+y-z=1 \\ 3x+z=6 \end{cases} has infinitely many solutions. If x=tx=t, which gives the complete solution set?

  1. (t,7−5t,6−3t)(t, 7-5t, 6-3t) (correct answer)
  2. (t,7+5t,6−3t)(t, 7+5t, 6-3t)
  3. (t,7−5t,6+3t)(t, 7-5t, 6+3t)
  4. (t,5t−7,3t−6)(t, 5t-7, 3t-6)
Explanation: When a system has infinitely many solutions, substituting x=tx=t reduces it to finding yy and zz in terms of tt. Start with the third equation, 3x+z=63x+z=6: plugging in x=tx=t gives 3t+z=63t+z=6, so z=6−3tz=6-3t. Then use the first equation: t−y+2(6−3t)=5t-y+2(6-3t)=5, which simplifies to −5t−y+12=5-5t-y+12=5, so −y=−7+5t-y=-7+5t, hence y=7−5ty=7-5t. Thus any solution has the form (t, 7−5t, 6−3t)(t,\,7-5t,\,6-3t). Checking the second equation, 2t+(7−5t)−(6−3t)=12t+(7-5t)-(6-3t)=1, confirms the triple also satisfies the middle equation, so this parameterization is complete. The option (t,7+5t,6−3t)(t,7+5t,6-3t) uses the wrong sign for yy: when solving −y=−7+5t-y=-7+5t, you must multiply both sides by −1-1, not drop the minus sign. The option (t,7−5t,6+3t)(t,7-5t,6+3t) also has a sign error, for zz: 3t+z=63t+z=6 means z=6−3tz=6-3t, not 6+3t6+3t. The option (t,5t−7,3t−6)(t,5t-7,3t-6) reverses both signs, likely from moving terms to the other side without switching signs correctly; it fails the original equations almost everywhere. Strategy: whenever a system has infinitely many solutions, set the given variable, back-substitute starting with the simplest equation, and verify your parameterization in every original equation. Sign errors are the most common trap, so check each substitution carefully.

Question 3

A student solves the first equation in {x−2y+3z=72x+y−z=4−x+3y+2z=5\begin{cases} x-2y+3z=7 \\ 2x+y-z=4 \\ -x+3y+2z=5 \end{cases} for xx and substitutes into the second equation. Which equation results?

  1. 5y+7z=−105y+7z=-10
  2. 5y−7z=−105y-7z=-10 (correct answer)
  3. 5y−7z=105y-7z=10
  4. 3y−4z=−33y-4z=-3
Explanation: Whenever you see a system like this, the substitution method is about isolating one variable and carefully replacing it in another equation. The trap is in the algebra—especially distribution and signed terms. Start by solving the first equation for xx: x=7+2y−3zx = 7 + 2y - 3z Now substitute this expression into the second equation: 2(7+2y−3z)+y−z=42(7 + 2y - 3z) + y - z = 4 Distribute the 2: 14+4y−6z+y−z=414 + 4y - 6z + y - z = 4 Combine like terms: 5y−7z=−105y - 7z = -10 So the resulting equation is indeed 5y−7z=−105y - 7z = -10. Why the others fail:
  • 5y+7z=−105y + 7z = -10 comes from incorrectly adding the z-terms: you must combine −6z-6z and −z-z to get −7z-7z, not +7z+7z.
  • 5y−7z=105y - 7z = 10 has the correct left side but the wrong constant; remember 14+(stuff)=414 + \text{(stuff)} = 4 means the stuff equals 4−14=−104 - 14 = -10, not 1010.
  • 3y−4z=−33y - 4z = -3 likely results from a distribution error, such as not multiplying the 2 through every term, or miscombining 4y4y and yy.
When substituting, always write the full substituted equation before simplifying. Keep a mental checklist: distribute, combine like terms for each variable, then move constants to the other side. That single habit will prevent most sign and arithmetic mistakes on substitution problems.

Question 4

The sum of three numbers is 40. The largest number is 5 less than twice the middle number, and the smallest number is 7 less than the middle number. What is the largest number?

  1. 6
  2. 13
  3. 21 (correct answer)
  4. 26
Explanation: When you see a word problem with several related quantities, your first move should be to pick a variable for the "anchor" number. Here, the middle number is the anchor: both the largest and smallest are described in terms of it. Let the middle number be mm. Then the largest number is 2m−52m - 5, and the smallest number is m−7m - 7. Since the sum is 40: m+(2m−5)+(m−7)=40m + (2m - 5) + (m - 7) = 40 Combine like terms: 4m−12=404m - 12 = 40 4m=524m = 52 m=13m = 13 So the middle number is 13. The largest number is: 2(13)−5=26−5=212(13) - 5 = 26 - 5 = 21 That confirms 21 as the correct answer. Now look at the wrong choices. The choice 13 is simply the middle number, not the largest — a common trap if you stop too early. The choice 6 is the smallest number, since 13−7=613 - 7 = 6. And the choice 26 comes from computing 2×132 \times 13 but forgetting to subtract 5; it represents the "twice the middle" part without the adjustment. Your study tip: always define your variable, translate each relationship into an expression, and then reread the question to confirm what you are actually solving for. Many errors on these problems come from solving an equation correctly but answering the wrong question — here, reporting the middle number instead of the largest.

Question 5

While solving a system of three linear equations, a student adds two equations and obtains 0=00=0. Which conclusion is valid?

  1. The two equations are dependent, so the system may have exactly one solution depending on the third equation.
  2. The two equations are dependent, but the system may still have no solution because of the third equation. (correct answer)
  3. The two equations are inconsistent, so the system has no solution regardless of the third equation.
  4. The third equation must also reduce to 0=00=0 for the system to be consistent.
Explanation: When you see a system of three linear equations, think of each equation as a plane in space. Adding equations is a valid move, but the result tells you about the relationship between the two equations you combined. If adding two equations gives 0=00=0, those two equations are dependent: they represent the same plane and are not contradictory. However, this says nothing about the third equation. The third plane could intersect that shared plane in a line, giving infinitely many solutions, or it could fail to intersect entirely, making the system inconsistent. So the valid conclusion is: the two equations are dependent, but the system may still have no solution because of the third equation. The wrong choices each misread this. Saying the two equations are dependent so the system may have exactly one solution is impossible: with only two independent planes, the solution set cannot be a single point. Saying the two equations are inconsistent and the system has no solution regardless confuses 0=00=0, which is a tautology, with a contradiction like 0=10=1. Finally, saying the third equation must also reduce to 0=00=0 is too strong; the third equation can be different and still be consistent with the other two. Study tip: when elimination produces 0=00=0, ask what has been learned. The answer is: only that those two equations are redundant. Check the remaining equation to decide whether the whole system is consistent.

Question 6

Which ordered triple is NOT a solution of the system $$ \begin{cases} x+y+z=6 \ 2x-y+z=3 \ 3x+2z=9 \end{cases}

  1. (1,2,3)(1,2,3)
  2. (3,3,0)(3,3,0)
  3. (2,52,32)\left(2,\frac{5}{2},\frac{3}{2}\right)
  4. (2,2,2)(2,2,2) (correct answer)
Explanation: When you see a system of equations, recall that a solution must satisfy every equation simultaneously—not just one or two. Here, the quickest path is to substitute each ordered triple into all three equations and see which one breaks the system. Starting with (1,2,3)(1,2,3):
1+2+3=61+2+3=6, 2(1)−2+3=32(1)-2+3=3, and 3(1)+2(3)=93(1)+2(3)=9, so it works. Next, (3,3,0)(3,3,0):
3+3+0=63+3+0=6, 2(3)−3+0=32(3)-3+0=3, and 3(3)+2(0)=93(3)+2(0)=9, so it also works. The fractional triple (2,52,32)\left(2,\frac{5}{2},\frac{3}{2}\right):
2+52+32=62+\frac{5}{2}+\frac{3}{2}=6, 4−52+32=34-\frac{5}{2}+\frac{3}{2}=3, and 6+3=96+3=9, so it satisfies all three equations. But check (2,2,2)(2,2,2): the first equation holds since 2+2+2=62+2+2=6, but the second gives 4−2+2=44-2+2=4, not 33, and the third gives 6+4=106+4=10, not 99. Because it fails two equations, it is not a solution. The trap is stopping after checking one equation. Always test every equation before deciding a triple is valid.

Question 7

What is the solution of the system $$ \begin{cases} x+y+z=6 \ 2x-y+z=3 \ x+2y-z=3 \end{cases}

  1. (97,157,187)\left(\frac{9}{7}, \frac{15}{7}, \frac{18}{7}\right) (correct answer)
  2. (1,2,3)(1,2,3)
  3. (97,187,157)\left(\frac{9}{7}, \frac{18}{7}, \frac{15}{7}\right)
  4. (3,1,2)(3,1,2)
Explanation: When you see a system of three linear equations, you can solve it by elimination or substitution, but on a multiple-choice exam, you can also test each candidate triple in all three equations. A quick check is valuable because every choice here satisfies the first equation, x+y+z=6x+y+z=6, so the first equation alone cannot distinguish them. For the correct triple (97,157,187)\left(\frac{9}{7}, \frac{15}{7}, \frac{18}{7}\right), substitute into each equation. The sum is 9+15+187=427=6\frac{9+15+18}{7}=\frac{42}{7}=6. The second equation gives 2⋅97−157+187=18−15+187=217=32\cdot\frac{9}{7}-\frac{15}{7}+\frac{18}{7}=\frac{18-15+18}{7}=\frac{21}{7}=3. The third gives 97+2⋅157−187=9+30−187=217=3\frac{9}{7}+2\cdot\frac{15}{7}-\frac{18}{7}=\frac{9+30-18}{7}=\frac{21}{7}=3. All three equations hold, so it is the solution. The choice (1,2,3)(1,2,3) works in the first equation and in the second, 2−2+3=32-2+3=3, but in the third it gives 1+4−3=21+4-3=2, not 33. The choice (97,187,157)\left(\frac{9}{7}, \frac{18}{7}, \frac{15}{7}\right) also sums to 66, but its second equation gives 18−18+157=157\frac{18-18+15}{7}=\frac{15}{7}, not 33. Finally, (3,1,2)(3,1,2) fails the second equation immediately: 6−1+2=76-1+2=7, not 33. A strong strategy is to substitute each answer into every equation, not just the first. On this exam, distractors are often designed to satisfy the easiest equation, so always verify the remaining equations before selecting.

Question 8

For what value of kk does the system $$ \begin{cases} x-2y+z=4 \ 2x-4y+2z=8 \ -x+2y-z=k \end{cases}

  1. k=−8k=-8
  2. k=4k=4
  3. k=8k=8
  4. k=−4k=-4 (correct answer)
Explanation: When you see a system of equations and are asked how many solutions it has, think about whether the equations are multiples of one another. Infinitely many solutions means the equations describe the same plane (or line), not just parallel planes. Here, notice the second equation is exactly 22 times the first: 2(x−2y+z)=2(4)⇒2x−4y+2z=8.2(x-2y+z)=2(4)\Rightarrow 2x-4y+2z=8. So those two equations are actually the same plane. Now look at the third equation: −x+2y−z=k.-x+2y-z=k. This is −1-1 times the first equation: −(x−2y+z)=−4⇒−x+2y−z=−4.-(x-2y+z)=-4 \Rightarrow -x+2y-z=-4. Therefore, the third equation will match the first exactly only when k=−4k=-4. Then all three equations represent the same plane, giving infinitely many solutions. If k=4k=4, you are forgetting the negative sign when multiplying the first equation by −1-1; the correct right-hand side would be −4-4, not 44. If k=8k=8, you might be confusing it with the second equation's right-hand side, but the second equation was multiplied by 22, not related to the third. If k=−8k=-8, this likely comes from multiplying −4-4 by 22 or mixing up signs and scaling factors; none of these produce a consistent third equation. Your takeaway: when one equation is a multiple of another, check whether the third equation is also a multiple with the correct constant. Here, the key sign flip was −1-1 times the first equation, so k=−4k=-4.

Question 9

What is the value of zz in the solution of the system $$ \begin{cases} x+y+z=9 \ 2x-y+z=5 \ x+2y-z=4 \end{cases}

  1. 2
  2. 3
  3. 4 (correct answer)
  4. 5
Explanation: When you see a three-variable system, your goal is to reduce it to two variables, solve those, then back-substitute. Here, adding the first and third equations eliminates zz: 2x+3y=13.2x+3y=13. Adding the second and third equations also eliminates zz: 3x+y=9.3x+y=9. Now solve the two-variable system. From 3x+y=93x+y=9, you get y=9−3xy=9-3x. Substitute into 2x+3y=132x+3y=13: 2x+3(9−3x)=13⇒−7x=−14⇒x=2.2x+3(9-3x)=13 \Rightarrow -7x=-14 \Rightarrow x=2. Then y=9−3(2)=3y=9-3(2)=3. Finally, use x+y+z=9x+y+z=9: 2+3+z=9⇒z=4.2+3+z=9 \Rightarrow z=4. So the value of zz is 44. The wrong choices are traps built from the same numbers: 22 is the xx-value, not zz, and 33 is the yy-value. Choosing 55 likely comes from computing x+y=2+3=5x+y=2+3=5 and mistaking that sum for zz. But the equation x+y+z=9x+y+z=9 shows zz must be 9−5=49-5=4. Strategy: in a 3×33\times3 system, look for equation pairs that eliminate the same variable — here, zz cancels by adding first+third and second+third. Then solve the resulting 2×22\times2 system. Finally, always label which variable your answer represents; exam questions often make the other variables distractors.

Question 10

A student is solving the system {x+y+z=6x−y+2z=22x+3y−z=5\begin{cases} x+y+z=6 \\ x-y+2z=2 \\ 2x+3y-z=5 \end{cases} by elimination. She eliminates xx from the first two equations. Which equation results?

  1. 2z−y=42z-y=4
  2. 2y+z=82y+z=8
  3. 2y−z=82y-z=8
  4. 2y−z=42y-z=4 (correct answer)
Explanation: Whenever you solve by elimination, ask: "What can I add or subtract so one variable cancels?" Here, the first two equations both have xx-coefficient 11, so subtracting them eliminates xx. Write it vertically: (x+y+z)−(x−y+2z)=6−2(x+y+z)-(x-y+2z)=6-2 Distribute carefully: x−x+y−(−y)+z−2z=2y−zx-x + y-(-y) + z-2z = 2y - z and the right side is 44. So the resulting equation is 2y−z=42y-z=4. The choice 2y+z=82y+z=8 comes from adding the first two equations instead of subtracting, but adding gives 2x+3z=82x+3z=8, not an xx-free equation. The choice 2y−z=82y-z=8 makes the same correct subtraction but treats 6−26-2 as 88; constants subtract just like numbers, so it must be 44. The choice 2z−y=42z-y=4 swaps the coefficients of yy and zz or misapplies the signs when distributing the subtraction; keep the terms in the same order and combine like terms. A strong habit: always subtract both left sides and right sides separately, writing each step vertically. That prevents sign errors and helps you confirm that one variable truly disappears.

Question 11

At a concert, student tickets cost 5 dollars, adult tickets cost 8 dollars, and senior tickets cost 6 dollars. A total of 200 tickets were sold for 1440 dollars. If there were 20 more student tickets than senior tickets, how many adult tickets were sold?

  1. 40
  2. 140 (correct answer)
  3. 120
  4. 20
Explanation: Whenever you see a ticket-price problem with several ticket types, your first move should be to define variables and translate each sentence into an equation. Let ss, aa, and rr be the numbers of student, adult, and senior tickets. The condition "20 more student tickets than senior tickets" becomes s=r+20s = r + 20. Total tickets gives (r+20)+a+r=200(r+20)+a+r=200, so a+2r=180a+2r=180. Revenue gives 5(r+20)+8a+6r=14405(r+20)+8a+6r=1440, or 8a+11r=13408a+11r=1340. Now solve: from a+2r=180a+2r=180, get a=180−2ra=180-2r. Substitute into the revenue equation: 8(180−2r)+11r=13408(180-2r)+11r=1340. Simplify to 1440−16r+11r=13401440-16r+11r=1340, so −5r=−100-5r=-100, and r=20r=20. Therefore a=180−40=140a=180-40=140. So adult tickets sold were 140. . The other choices are traps. The value 40 is the number of student tickets, not adult tickets — it appears if you stop after finding $s$. The value 20 is the number of senior tickets; it is tempting if you misread the question as asking about seniors instead of adults. The value 120 could come from subtracting the extra 20 students from the 140 adults, or from treating student and senior counts as though they were equal; butthats ignorestherelationship $s=r+20$. Always check your variables against the actual question before selecting.

Question 12

Which statement correctly describes the system $$ \begin{cases} x+2y-z=4 \ 2x+4y-2z=7 \ x-y+z=1 \end{cases}

  1. It has no solution; the first two equations describe parallel planes. (correct answer)
  2. It has infinitely many solutions; the first two equations are multiples of each other.
  3. It has exactly one solution, (2,1,0)(2,1,0).
  4. It has exactly one solution, (1,2,0)(1,2,0).
Explanation: Whenever you see a system of three equations in three variables, first check whether any two equations have proportional coefficients. Here, the first equation x+2y−z=4x+2y-z=4 multiplied by 2 gives 2x+4y−2z=82x+4y-2z=8, but the second equation is 2x+4y−2z=72x+4y-2z=7. The left sides are identical while the right sides differ, so those two planes are parallel and distinct: they never intersect. Since every solution must satisfy both equations, the system has no solution — matching the statement that it has no solution because the first two equations describe parallel planes. The choice claiming infinitely many solutions because the first two equations are multiples of each other is a trap: the coefficients are proportional, but the full equations are not multiples, because a true multiple would require the constant term to be 8, not 7. The proposed exact solutions also fail substitution: (2,1,0)(2,1,0) makes the second equation 8=78=7, and (1,2,0)(1,2,0) makes the first equation 5=45=4. Neither can be a solution, and the inconsistency of the first two equations already rules out any single intersection point. When analyzing systems, compare entire equations: proportional coefficients with different constants means parallel planes and no solution; proportional coefficients with the same constant means identical planes and infinitely many solutions. Quick substitution is also a reliable way to test any proposed point.