Three-Variable Systems - Algebra 3

1. Which statement correctly describes the solution set of the system $x-y+2z=4, 2x-2y+4z=8, 3x-3y+6z=12$?
   Answer: Infinitely many ordered triples lying on one plane Each equation is a nonzero multiple of the first, so all three equations represent the same plane. The solution set is that plane, which has two free variables. A line would occur if only two equations were dependent; here all three are identical.
2. For what value of $k$ does the system $x+2y-z=1, 2x+5y+z=3, 3x+7y+kz=4$ have infinitely many solutions?
   Answer: $0$ Subtracting twice the first equation from the second gives $y+3z=1$. Subtracting three times the first from the third gives $y+(k+3)z=1$. Subtracting these results gives $kz=0$. For infinitely many solutions, this must be $0=0$, so $k=0$. If $k\neq0$, the system has a unique solution.
3. A student solves the first equation in $\begin{cases} x-2y+3z=7 \\ 2x+y-z=4 \\ -x+3y+2z=5 \end{cases}$ for $x$ and substitutes into the second equation. Which equation results?
   Answer: $5y-7z=-10$ 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 $x$: $$x = 7 + 2y - 3z$$ Now substitute this expression into the second equation: $$2(7 + 2y - 3z) + y - z = 4$$ Distribute the 2: $$14 + 4y - 6z + y - z = 4$$ Combine like terms: $$5y - 7z = -10$$ So the resulting equation is indeed $5y - 7z = -10$. Why the others fail: - $5y + 7z = -10$ comes from incorrectly adding the z-terms: you must combine $-6z$ and $-z$ to get $-7z$, not $+7z$. - $5y - 7z = 10$ has the correct left side but the wrong constant; remember $14 + \text{(stuff)} = 4$ means the stuff equals $4 - 14 = -10$, not $10$. - $3y - 4z = -3$ likely results from a distribution error, such as not multiplying the 2 through every term, or miscombining $4y$ and $y$. 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.
4. What is the solution of the system $\begin{cases} x+y+z=6 \\ 2x-y+z=3 \\ x+2y-z=3 \end{cases}$?
   Answer: $\left(\frac{9}{7}, \frac{15}{7}, \frac{18}{7}\right)$ 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, $x+y+z=6$, so the first equation alone cannot distinguish them. For the correct triple $\left(\frac{9}{7}, \frac{15}{7}, \frac{18}{7}\right)$, substitute into each equation. The sum is $\frac{9+15+18}{7}=\frac{42}{7}=6$. The second equation gives $2\cdot\frac{9}{7}-\frac{15}{7}+\frac{18}{7}=\frac{18-15+18}{7}=\frac{21}{7}=3$. The third gives $\frac{9}{7}+2\cdot\frac{15}{7}-\frac{18}{7}=\frac{9+30-18}{7}=\frac{21}{7}=3$. All three equations hold, so it is the solution. The choice $(1,2,3)$ works in the first equation and in the second, $2-2+3=3$, but in the third it gives $1+4-3=2$, not $3$. The choice $\left(\frac{9}{7}, \frac{18}{7}, \frac{15}{7}\right)$ also sums to $6$, but its second equation gives $\frac{18-18+15}{7}=\frac{15}{7}$, not $3$. Finally, $(3,1,2)$ fails the second equation immediately: $6-1+2=7$, not $3$. 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.
5. The system $\begin{cases} x-y+2z=5 \\ 2x+y-z=1 \\ 3x+z=6 \end{cases}$ has infinitely many solutions. If $x=t$, which gives the complete solution set?
   Answer: $(t, 7-5t, 6-3t)$ When a system has infinitely many solutions, substituting $x=t$ reduces it to finding $y$ and $z$ in terms of $t$. Start with the third equation, $3x+z=6$: plugging in $x=t$ gives $3t+z=6$, so $z=6-3t$. Then use the first equation: $t-y+2(6-3t)=5$, which simplifies to $-5t-y+12=5$, so $-y=-7+5t$, hence $y=7-5t$. Thus any solution has the form $(t,\,7-5t,\,6-3t)$. Checking the second equation, $2t+(7-5t)-(6-3t)=1$, confirms the triple also satisfies the middle equation, so this parameterization is complete. The option $(t,7+5t,6-3t)$ uses the wrong sign for $y$: when solving $-y=-7+5t$, you must multiply both sides by $-1$, not drop the minus sign. The option $(t,7-5t,6+3t)$ also has a sign error, for $z$: $3t+z=6$ means $z=6-3t$, not $6+3t$. The option $(t,5t-7,3t-6)$ 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.
6. What is the $x$-coordinate of the solution of the system $x=2y-z+1, 2x+y+z=8, x-y+2z=1$?
   Answer: $4$ Substitute $x=2y-z+1$ into the other equations. The second becomes $5y-z=6$, and the third becomes $y+z=0$. Thus $z=-y$, so $5y+y=6$, giving $y=1$ and $z=-1$. Then $x=2(1)-(-1)+1=4$.
7. The system $x+2y-z=3, 2x+4y-2z=6, x-y+z=0$ has infinitely many solutions. If $y=t$, which expression gives $x$?
   Answer: $\frac{3-t}{2}$ The second equation is twice the first, so it is redundant. From $x-y+z=0$, $x=y-z$. Substituting into $x+2y-z=3$ gives $3y-2z=3$, so with $y=t$, $z=\frac{3t-3}{2}$. Therefore $x=t-\frac{3t-3}{2}=\frac{3-t}{2}$. The expression $\frac{3t-3}{2}$ is $z$, while $\frac{t-3}{2}$ and $\frac{3+t}{2}$ have sign errors.
8. The sum of three numbers is 12. Twice the first plus the second is 10, and the second minus the third is $-1$. What is the second number?
   Answer: $4$ Let the numbers be $x,y,z$. Then $x+y+z=12$, $2x+y=10$, and $y-z=-1$. From the third, $z=y+1$; from the second, $x=\frac{10-y}{2}$. Substituting into the sum gives $\frac{10-y}{2}+y+(y+1)=12$, so $y=4$. The first number is 3 and the third is 5; the value 1 could result from incorrectly writing $x=10-y$ instead of $x=\frac{10-y}{2}$.
9. Which of the following systems has exactly one solution?
   Answer: $x+y+z=1, x-y+z=2, x+2y-z=0$ System $x+y+z=1, 2x+2y+2z=2, x-y=0$: the first two equations represent the same plane, and the third plane intersects it in a line, so there are infinitely many solutions. System $x+y+z=1, 2x+2y+2z=3, x-y=0$: the first two planes are parallel, so there is no solution. System $x+y+z=1, x-y+z=2, 2x+2z=3$: the third equation is the sum of the first two, and the first two are independent, so the system is dependent and has infinitely many solutions. Solving $x+y+z=1, x-y+z=2, x+2y-z=0$ gives $\left(\frac{5}{4},-\frac{1}{2},\frac{1}{4}\right)$, a unique solution.
10. Consider the system below, where $a$ is a real constant: $$\begin{cases} x-2y+z&=1\\ 2x+y-3z&=-1\\ 8x-y+az&=0 \end{cases}$$ For which value of $a$ does the system have no solution?
    Answer: $a=-7$ Solve the first two equations in terms of $z$. From $x-2y+z=1$, $x=1+2y-z$. Substitute into $2x+y-3z=-1$: $2(1+2y-z)+y-3z=-1$, so $5y-5z=-3$, giving $y=z-\frac35$. Then $x=1+2(z-\frac35)-z=z-\frac15$. Substituting into $8x-y+az=0$ gives $8(z-\frac15)-(z-\frac35)+az=0$, which simplifies to $(a+7)z-1=0$. This equation has no solution only when the coefficient of $z$ is zero and the constant is nonzero: $a+7=0$, so $a=-7$. For $a=-1$, $a=1$, or $a=7$, the coefficient $a+7$ is not zero, so $z=\frac{1}{a+7}$ and the system has a unique solution; those values reflect sign or combination errors.
11. If $(1,-2,3)$ is the solution of the system $x+ay-z=4, 2x-y+3z=b, x+2y+cz=0$, what is $a+b+c$?
    Answer: $11$ Substituting $(1,-2,3)$ gives $1-2a-3=4$, so $a=-3$; $2+2+9=b$, so $b=13$; and $1-4+3c=0$, so $c=1$. Therefore $a+b+c=-3+13+1=11$. The value 9 comes from using $c=-1$, 17 from using $a=3$, and 15 from using both $a=3$ and $c=-1$.
12. 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?
    Answer: 21 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 $m$. Then the largest number is $2m - 5$, and the smallest number is $m - 7$. Since the sum is 40: $$m + (2m - 5) + (m - 7) = 40$$ Combine like terms: $$4m - 12 = 40$$ $$4m = 52$$ $$m = 13$$ So the middle number is 13. The largest number is: $$2(13) - 5 = 26 - 5 = 21$$ 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 $13 - 7 = 6$. And the choice 26 comes from computing $2 \times 13$ 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.
13. Which ordered triple satisfies the system $\begin{cases} 2x+y-z=4 \\ x-2y+3z=-1 \\ 3x+y+z=6 \end{cases}$?
    Answer: $\left(\frac{4}{3}, \frac{5}{3}, \frac{1}{3}\right)$ 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 $\left(\frac{4}{3}, \frac{5}{3}, \frac{1}{3}\right)$, verify: - First: $2\left(\frac43\right)+\frac53-\frac13=\frac{8+5-1}{3}=4$ ✓ - Second: $\frac43-2\left(\frac53\right)+3\left(\frac13\right)=\frac{4-10+3}{3}=-1$ ✓ - Third: $3\left(\frac43\right)+\frac53+\frac13=\frac{12+5+1}{3}=6$ ✓ So this triple works everywhere. The choice $\left(\frac{5}{3}, \frac{4}{3}, \frac{1}{3}\right)$ is a trap: it swaps the roles of $x$ and $y$. It fails the first equation because $2(\frac53)+\frac43-\frac13=\frac{13}{3}\neq 4$. The choice $\left(\frac{4}{3}, \frac{1}{3}, \frac{5}{3}\right)$ also fails the first equation, giving $\frac{4}{3}$ instead of $4$; only the third equation happens to work. Finally, $(1,2,0)$ satisfies the first equation, $2(1)+2-0=4$, but it fails the second: $1-2(2)+3(0)=-3$, not $-1$. 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.
14. Which ordered triple is NOT a solution of the system $\begin{cases} x+y+z=6 \\ 2x-y+z=3 \\ 3x+2z=9 \end{cases}$?
    Answer: $(2,2,2)$ 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 $(1,2,3)$: $1+2+3=6$, $2(1)-2+3=3$, and $3(1)+2(3)=9$, so it works. Next, $(3,3,0)$: $3+3+0=6$, $2(3)-3+0=3$, and $3(3)+2(0)=9$, so it also works. The fractional triple $\left(2,\frac{5}{2},\frac{3}{2}\right)$: $2+\frac{5}{2}+\frac{3}{2}=6$, $4-\frac{5}{2}+\frac{3}{2}=3$, and $6+3=9$, so it satisfies all three equations. But check $(2,2,2)$: the first equation holds since $2+2+2=6$, but the second gives $4-2+2=4$, not $3$, and the third gives $6+4=10$, not $9$. 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.
15. For the system $x+2y-z=4, 2x-y+2z=7, 3x+y-z=9$, what is $x+y-z$?
    Answer: $3$ Solving the system gives $(x,y,z)=(3,1,1)$. Therefore $x+y-z=3+1-1=3$. The value 4 is the right side of the first equation; 1 is one of the coordinates; and 2 is a coordinate combination such as $y+z$ or $x-z$.
16. After correctly eliminating variables in a system of three equations in three variables, a student obtains $x=2, y+3z=4, 0=0$. Which statement about the original system is true?
    Answer: It has infinitely many solutions. The equation $0=0$ means the system is consistent and dependent. Letting $z=t$ gives $y=4-3t$ and $x=2$, so there are infinitely many solutions. The ordered triple $(2,4,0)$ is only one of those solutions, and a linear system cannot have exactly two solutions.
17. For what value of $k$ does the system $x+y+z=6, 2x-y+3z=9, 4x+y+kz=22$ have no solution?
    Answer: $5$ Twice the first equation plus the second gives $4x+y+5z=21$. Thus when $k=5$, the third equation has the same left side as this combination but a different right side, $22$, so the system is inconsistent. For $k\neq 5$, the coefficient determinant is nonzero and the system has a unique solution.
18. A student is solving the system $\begin{cases} x+y+z=6 \\ x-y+2z=2 \\ 2x+3y-z=5 \end{cases}$ by elimination. She eliminates $x$ from the first two equations. Which equation results?
    Answer: $2y-z=4$ Whenever you solve by elimination, ask: "What can I add or subtract so one variable cancels?" Here, the first two equations both have $x$-coefficient $1$, so subtracting them eliminates $x$. Write it vertically: $$(x+y+z)-(x-y+2z)=6-2$$ Distribute carefully: $$x-x + y-(-y) + z-2z = 2y - z$$ and the right side is $4$. So the resulting equation is $2y-z=4$. The choice $2y+z=8$ comes from adding the first two equations instead of subtracting, but adding gives $2x+3z=8$, not an $x$-free equation. The choice $2y-z=8$ makes the same correct subtraction but treats $6-2$ as $8$; constants subtract just like numbers, so it must be $4$. The choice $2z-y=4$ swaps the coefficients of $y$ and $z$ 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.
19. What is the value of $z$ in the solution of the system $\begin{cases} x+y+z=9 \\ 2x-y+z=5 \\ x+2y-z=4 \end{cases}$?
    Answer: 4 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 $z$: $2x+3y=13.$ Adding the second and third equations also eliminates $z$: $3x+y=9.$ Now solve the two-variable system. From $3x+y=9$, you get $y=9-3x$. Substitute into $2x+3y=13$: $2x+3(9-3x)=13 \Rightarrow -7x=-14 \Rightarrow x=2.$ Then $y=9-3(2)=3$. Finally, use $x+y+z=9$: $2+3+z=9 \Rightarrow z=4.$ So the value of $z$ is $4$. The wrong choices are traps built from the same numbers: $2$ is the $x$-value, not $z$, and $3$ is the $y$-value. Choosing $5$ likely comes from computing $x+y=2+3=5$ and mistaking that sum for $z$. But the equation $x+y+z=9$ shows $z$ must be $9-5=4$. Strategy: in a $3\times3$ system, look for equation pairs that eliminate the same variable — here, $z$ cancels by adding first+third and second+third. Then solve the resulting $2\times2$ system. Finally, always label which variable your answer represents; exam questions often make the other variables distractors.
20. A system has solution set $\{(2-3t, t, 4+t): t\in\mathbb{R}\}$. Which equation is satisfied by every solution in this set?
    Answer: $x+3y=2$ Substitute $x=2-3t$, $y=t$, $z=4+t$. Then $x+3y=2-3t+3t=2$, so $x+3y=2$ holds for every $t$. The other equations simplify to $x-y+z=6-3t$, $x+3y-z=-2-t$, and $2x+y-z=-6t$, which depend on $t$.

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