Algebra 2 Quiz: Solving Systems Of Linear Equations
20 questions · exam conditions
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Solving Systems Of Linear EquationsQuestion 1 of 20

Solve the system by both substitution and elimination (they should give the same result). What is the solution?

{x−y=22x+y=10\begin{cases} x - y = 2\\ 2x + y = 10 \end{cases}
(4,2)(4, 2)
(2,4)(2, 4)
(3,1)(3, 1)
(6,−2)(6, -2)
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Algebra 2 Quiz

Algebra 2 Quiz: Solving Systems Of Linear Equations

Practice Solving Systems Of Linear Equations in Algebra 2 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 Solving Systems Of Linear Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

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

Solve the system by both substitution and elimination (they should give the same result). What is the solution?

{x−y=22x+y=10\begin{cases} x - y = 2\\ 2x + y = 10 \end{cases}
  1. (4,2)(4, 2) (correct answer)
  2. (2,4)(2, 4)
  3. (3,1)(3, 1)
  4. (6,−2)(6, -2)
Explanation: This question tests your ability to solve systems using multiple methods and verify they give the same result, reinforcing that all valid methods lead to the same solution. The system x - y = 2 and 2x + y = 10 is well-suited for both substitution and elimination, so let's solve it both ways. By substitution: from x - y = 2, we get x = y + 2. Substituting into 2x + y = 10: 2(y + 2) + y = 10, which gives 2y + 4 + y = 10, so 3y = 6, thus y = 2. Then x = 2 + 2 = 4. Solution: (4, 2). By elimination: notice that the first equation has -y and the second has +y, so adding eliminates y: (x - y) + (2x + y) = 2 + 10, giving 3x = 12, so x = 4. Substituting back: 4 - y = 2, so y = 2. Same solution: (4, 2)! Verification: first equation: 4 - 2 = 2 ✓; second equation: 2(4) + 2 = 8 + 2 = 10 ✓. Choice A correctly identifies (4, 2) as the solution obtained by both methods, confirming that different approaches yield the same answer when properly executed. Choice B makes the common error of writing the ordered pair as (y, x) instead of (x, y)—always write x-coordinate first! Both substitution and elimination are valid methods that must give the same solution: (1) Choose the method that looks easier for the specific system, (2) Work carefully through the algebra, (3) Verify your answer works in both original equations, (4) If solving by multiple methods, they should all give the same (x, y). This redundancy helps catch errors—if methods give different answers, recheck your work!

Question 2

Determine whether the system has one solution, no solution, or infinitely many solutions:

{2x−4y=10x−2y=5\begin{cases} 2x - 4y = 10 \\ x - 2y = 5 \end{cases}
  1. No solution (parallel lines)
  2. One solution: (0,5)(0, 5)
  3. Infinitely many solutions (same line) (correct answer)
  4. One solution: (5,0)(5, 0)
Explanation: This question tests your ability to solve systems of linear equations using substitution, elimination, or graphing methods—all leading to the same solution when one exists. A system can have one solution (lines intersect), no solution (parallel lines), or infinitely many solutions (same line). To analyze 2x−4y=102x - 4y = 10 and x−2y=5x - 2y = 5: Notice the second equation can be multiplied by 2 to get 2x−4y=102x - 4y = 10, which is exactly the first equation! This means both equations represent the same line. When two equations describe the same line, every point on that line is a solution—infinitely many solutions. Choice C correctly identifies that the system has infinitely many solutions because the equations represent the same line. Choice B incorrectly suggests no solution (parallel lines), but these aren't parallel—they're identical! To determine the number of solutions: (1) Put both equations in the same form, (2) Compare coefficients—if one equation is a multiple of the other, they're the same line (infinitely many solutions), (3) If coefficients of x and y are proportional but constants aren't, lines are parallel (no solution), (4) Otherwise, lines intersect at one point (one solution). Here, multiplying x−2y=5x - 2y = 5 by 2 gives exactly 2x−4y=102x - 4y = 10, confirming the same line!

Question 3

Determine whether the system has one solution, no solution, or infinitely many solutions:

{2x+y=54x+2y=12\begin{cases} 2x + y = 5\\ 4x + 2y = 12 \end{cases}
  1. One solution: (2,1)(2, 1)
  2. Infinitely many solutions
  3. One solution: (1,3)(1, 3)
  4. No solution (correct answer)
Explanation: This question tests your ability to determine if a system of linear equations has one solution, no solution, or infinitely many solutions. A system of linear equations has three possibilities: (1) one unique solution (lines intersect at one point), (2) no solution (parallel lines never meet), or (3) infinitely many solutions (same line, every point on it works). To solve algebraically, we use substitution (solve one equation for a variable, plug into other) or elimination (align coefficients, add or subtract to eliminate a variable). Graphically, we plot both lines and find where they cross (if they do). All methods should give the same answer! For the system 2x+y=52x + y = 5 and 4x+2y=124x + 2y = 12: multiply the first by 2 to get 4x+2y=104x + 2y = 10, but the second is 4x+2y=124x + 2y = 12—same left side, different constants, so inconsistent equations, parallel lines, no solution. No point satisfies both. You can also see the slopes are the same (−2-2) but different y-intercepts (55 vs. 66). Choice B correctly identifies there is no solution due to the inconsistency. Choice A might come from solving incorrectly, like ignoring the doubled equation—always compare after aligning! To classify systems: Make coefficients match; if constants differ, no solution; if identical, infinite; else, solve for unique. Example: 2x+y=52x + y = 5 and 4x+2y=104x + 2y = 10 is infinite (same line). You're getting the hang of it!

Question 4

Solve the system by graphing. (If you read the intersection from the graph, use the exact ordered pair.)

{y=x+2y=−2x+8\begin{cases} y = x + 2 \\ y = -2x + 8 \end{cases}
  1. (2,4)(2, 4) (correct answer)
  2. (4,2)(4, 2)
  3. (3,5)(3, 5)
  4. (2,6)(2, 6)
Explanation: This question tests your ability to solve systems of linear equations using substitution, elimination, or graphing methods—all leading to the same solution when one exists. To solve by graphing, we plot both lines and find their intersection point. For y=x+2y = x + 2 and y=−2x+8y = -2x + 8: The first line has slope 11 and y-intercept 22, passing through (0,2)(0, 2) and (1,3)(1, 3). The second line has slope −2-2 and y-intercept 88, passing through (0,8)(0, 8) and (1,6)(1, 6). To find the intersection algebraically (to verify): set x+2=−2x+8x + 2 = -2x + 8, so 3x=63x = 6, giving x=2x = 2. Then y=2+2=4y = 2 + 2 = 4. The lines intersect at (2,4)(2, 4). Choice A correctly identifies (2,4)(2, 4) as the intersection point where both lines meet. Choice B reverses the coordinates—remember that ordered pairs are always written as (x,y)(x, y), not (y,x)(y, x)! When graphing to solve systems, the intersection point gives the x-coordinate first, then the y-coordinate. To graph accurately: (1) Identify slope and y-intercept for each line, (2) Plot the y-intercept, (3) Use slope to find another point, (4) Draw the line through both points, (5) Repeat for second line, (6) Find where they cross, (7) Write intersection as (x,y)(x, y). Always verify your graphical solution by substituting back into both original equations!

Question 5

Determine whether the system has no solution, one solution, or infinitely many solutions:

{2x+y=64x+2y=12\begin{cases} 2x + y = 6\\ 4x + 2y = 12 \end{cases}
  1. No solution
  2. Exactly one solution
  3. The solution is (3,0)(3, 0)
  4. Infinitely many solutions (correct answer)
Explanation: This question tests your ability to identify when a system of linear equations has no solution, one solution, or infinitely many solutions by analyzing the relationship between the equations. A system has infinitely many solutions when both equations represent the same line, no solution when they represent parallel lines, and one solution when they intersect at exactly one point. Looking at the system 2x + y = 6 and 4x + 2y = 12, notice that the second equation is exactly 2 times the first equation: 2(2x + y) = 2(6) gives 4x + 2y = 12. This means both equations represent the same line! We can verify by putting both in slope-intercept form: from 2x + y = 6, we get y = -2x + 6; from 4x + 2y = 12, we get 2y = -4x + 12, so y = -2x + 6. Same equation! Choice C correctly identifies that the system has infinitely many solutions because the two equations are equivalent—they describe the same line, so every point on that line is a solution. Choice A incorrectly assumes the lines are parallel because they have the same slope, but parallel lines must have different y-intercepts to never meet—these have the same y-intercept too, making them identical. To determine the number of solutions: (1) Check if one equation is a multiple of the other (infinitely many solutions), (2) Put in slope-intercept form y = mx + b to compare slopes and y-intercepts, (3) Same slope but different y-intercepts means parallel lines (no solution), (4) Different slopes means one intersection point (one solution). Quick trick: if you can multiply one equation by a constant to get the other equation exactly, you have infinitely many solutions!

Question 6

Solve the system graphically by finding the intersection point of the two lines. Choose the best answer (an exact ordered pair).

{y=x+2y=−2x+8\begin{cases} y = x + 2\\ y = -2x + 8 \end{cases}
  1. (2,4)(2,4) (correct answer)
  2. (4,2)(4,2)
  3. (3,5)(3,5)
  4. (2,6)(2,6)
Explanation: This question tests your ability to solve systems of linear equations using the graphical method—all leading to the same solution when one exists. A system of linear equations has three possibilities: (1) one unique solution (lines intersect at one point), (2) no solution (parallel lines never meet), or (3) infinitely many solutions (same line, every point on it works). To solve algebraically, we use substitution (solve one equation for a variable, plug into other) or elimination (align coefficients, add or subtract to eliminate a variable). Graphically, we plot both lines and find where they cross (if they do). All methods should give the same answer! To solve the system y=x+2y = x + 2 and y=−2x+8y = -2x + 8 graphically: the intersection is at (2,4)(2,4), or algebraically set equal: x+2=−2x+8x + 2 = -2x + 8, 3x=63x = 6, x=2x=2, y=4y=4; solution: (2,4)(2,4); verify in both (it checks out). Choice A correctly identifies the solution (2,4)(2,4) that satisfies both equations—excellent work spotting the intersection! Choice B gives (4,2)(4,2), which might come from a sign error in solving, like x+2=2x+8x + 2 = 2x + 8 leading to −x=6-x=6, but always double-check your algebra! Transferable strategy: For graphing, plot y-intercepts and use slopes to draw lines, then estimate intersection; confirm algebraically if needed. You're building strong skills—keep practicing these methods!

Question 7

Solve the system graphically (estimate from the graph). The lines are y=−x+4andy=12x+1.y = -x + 4 \quad \text{and} \quad y = \tfrac{1}{2}x + 1. What is the approximate intersection point (x,y)(x, y)?

  1. (2,2)(2, 2) (correct answer)
  2. (4,0)(4, 0)
  3. (3,1)(3, 1)
  4. (1,3)(1, 3)
Explanation: This question tests your ability to solve systems of linear equations using the graphing method—all leading to the same solution when one exists. A system of linear equations has three possibilities: (1) one unique solution (lines intersect at one point), (2) no solution (parallel lines never meet), or (3) infinitely many solutions (same line, every point on it works). To solve algebraically, we use substitution (solve one equation for a variable, plug into other) or elimination (align coefficients, add or subtract to eliminate a variable). Graphically, we plot both lines and find where they cross (if they do). All methods should give the same answer! To solve graphically, plot y = -x + 4 (intercept 4, slope -1) and y = (1/2)x + 1 (intercept 1, slope 1/2); they intersect at (2, 2), confirmed algebraically: -x + 4 = (1/2)x + 1, - (3/2)x = -3, x = 2, y = 2. The approximate intersection from the graph is (2, 2). Verify: both equations give y = 2 when x = 2 (check!). Excellent observation! Choice B correctly identifies the intersection point (2, 2) that satisfies both equations. Choice A might come from misreading the graph or an arithmetic error in solving, like incorrect slope calculation—always check the intersection carefully! Graphing tip: Use graph paper, plot at least two points per line, and extend to find the cross; confirm with algebra if possible. Example: y = x and y = -x + 4 intersect at (2, 2). Keep graphing, you're doing awesome!

Question 8

Solve the system using elimination and give the exact solution.

{5x−y=92x+y=12\begin{cases} 5x - y = 9 \\ 2x + y = 12 \end{cases}
  1. (3,6)(3, 6) (correct answer)
  2. (1,10)(1, 10)
  3. (2,8)(2, 8)
  4. (4,4)(4, 4)
Explanation: This question tests your ability to solve systems using elimination when the coefficients are already set up for easy elimination. When one variable has opposite coefficients (like -y and +y), we can add the equations directly to eliminate that variable. Adding the equations: (5x - y) + (2x + y) = 9 + 12, which gives 7x = 21, so x = 3. Substituting x = 3 into the second equation: 2(3) + y = 12, so 6 + y = 12, giving y = 6. The solution is (3, 6). Choice A correctly identifies this solution. Let's verify: First equation: 5(3) - 6 = 15 - 6 = 9 ✓. Second equation: 2(3) + 6 = 6 + 6 = 12 ✓. Both equations are satisfied! The elimination method shines when coefficients are already opposites or when one coefficient is 1 or -1. Here, having -y and +y made the elimination immediate. Always look for the easiest elimination path—sometimes it's x, sometimes y, depending on the coefficients given.

Question 9

Solve the system using the elimination method:

{2x+3y=124x−3y=6\begin{cases} 2x + 3y = 12\\ 4x - 3y = 6 \end{cases}

What is the solution (x,y)(x, y)?

  1. (3,2)(3, 2) (correct answer)
  2. (2,3)(2, 3)
  3. (1,4)(1, 4)
  4. (4,1)(4, 1)
Explanation: This question tests your ability to solve systems of linear equations using the elimination method—all leading to the same solution when one exists. A system of linear equations has three possibilities: (1) one unique solution (lines intersect at one point), (2) no solution (parallel lines never meet), or (3) infinitely many solutions (same line, every point on it works). To solve algebraically, we use substitution (solve one equation for a variable, plug into other) or elimination (align coefficients, add or subtract to eliminate a variable). Graphically, we plot both lines and find where they cross (if they do). All methods should give the same answer! To solve the system 2x+3y=122x + 3y = 12 and 4x−3y=64x - 3y = 6 by elimination: the y-coefficients are opposites (3 and -3), so add the equations: 6x=186x = 18, x=3x = 3. Back-substitute into the first: 2(3)+3y=122(3) + 3y = 12, 6+3y=126 + 3y = 12, 3y=63y = 6, y=2y = 2. Solution: (3,2)(3, 2). Verify: second equation 4(3)−3(2)=12−6=64(3) - 3(2) = 12 - 6 = 6 (check!). Both satisfied—great job! Choice A correctly solves by the elimination method to find the solution (3,2)(3, 2) that satisfies both equations. Choice B gives only the swapped pair, but solves correctly for x and y then mixes them up—after finding values, label them properly as (x,y)(x, y)! Elimination method recipe: (1) Align equations vertically, (2) Multiply one or both to make coefficients opposites, (3) Add or subtract to eliminate, (4) Solve for remaining variable, (5) Back-substitute, (6) Write solution, (7) Verify. Example: x+y=9x + y = 9 and 2x−y=32x - y = 3, add: 3x=123x = 12, x=4x = 4, y=5y = 5, solution (4,5)(4, 5). You're building strong skills here!

Question 10

Determine whether the system has one solution, no solution, or infinitely many solutions:

{3x−y=66x−2y=12\begin{cases} 3x - y = 6\\ 6x - 2y = 12 \end{cases}
  1. One solution: (2,0)(2, 0)
  2. No solution
  3. Infinitely many solutions (correct answer)
  4. One solution: (0,−6)(0, -6)
Explanation: This question tests your ability to determine if a system of linear equations has one solution, no solution, or infinitely many solutions. A system of linear equations has three possibilities: (1) one unique solution (lines intersect at one point), (2) no solution (parallel lines never meet), or (3) infinitely many solutions (same line, every point on it works). To solve algebraically, we use substitution (solve one equation for a variable, plug into other) or elimination (align coefficients, add or subtract to eliminate a variable). Graphically, we plot both lines and find where they cross (if they do). All methods should give the same answer! For the system 3x−y=63x - y = 6 and 6x−2y=126x - 2y = 12: multiply the first by 2 to get 6x−2y=126x - 2y = 12, which matches the second exactly—same line, infinitely many solutions. Every point on the line works. Slopes and intercepts match too. Choice C correctly identifies infinitely many solutions as the equations are dependent. Choice A might result from a calculation error, like thinking it's unique without checking—always multiply and compare! Classification tip: If equations are multiples with same constants, infinite; different constants, none; otherwise unique. Example: x−y=2x - y = 2 and 2x−2y=42x - 2y = 4 is infinite. Awesome progress!

Question 11

Solve the system graphically by finding the intersection point. Choose the best answer (exact if possible).

{y=−x+6y=2x\begin{cases} y = -x + 6 \\ y = 2x \end{cases}
  1. (2,4)(2, 4) (correct answer)
  2. (4,2)(4, 2)
  3. (6,0)(6, 0)
  4. (0,6)(0, 6)
Explanation: This question tests your ability to solve systems of linear equations graphically by finding where two lines intersect on the coordinate plane. A graphical solution involves plotting both lines and identifying their intersection point—this gives the same answer as algebraic methods when a unique solution exists. To solve graphically, we need to find where y = -x + 6 and y = 2x intersect. Since both expressions equal y, we can set them equal: -x + 6 = 2x, which gives 6 = 3x, so x = 2. Substituting back into either equation (using y = 2x): y = 2(2) = 4. The intersection point is (2, 4). Choice A correctly identifies this solution. Let's verify: First equation: 4 = -(2) + 6 = 4 ✓. Second equation: 4 = 2(2) = 4 ✓. Both equations are satisfied! When solving graphically, you can also plot points: for y = -x + 6, when x = 0, y = 6; when x = 6, y = 0. For y = 2x, when x = 0, y = 0; when x = 1, y = 2. Drawing these lines shows they intersect at (2, 4). All three methods—substitution, elimination, and graphing—give the same answer when properly executed!

Question 12

Determine whether the system is consistent. If it has a solution, give it.

{3x−2y=66x−4y=5\begin{cases} 3x - 2y = 6\\ 6x - 4y = 5 \end{cases}
  1. No solution (parallel lines) (correct answer)
  2. Infinitely many solutions
  3. One solution: (1,32)(1, \tfrac{3}{2})
  4. One solution: (2,0)(2,0)
Explanation: This question tests your ability to determine if a system is consistent (has solutions) and find it if so, using algebraic methods. A system of linear equations has three possibilities: (1) one unique solution (lines intersect at one point), (2) no solution (parallel lines never meet), or (3) infinitely many solutions (same line, every point on it works). To solve algebraically, we use substitution (solve one equation for a variable, plug into other) or elimination (align coefficients, add or subtract to eliminate a variable). Graphically, we plot both lines and find where they cross (if they do). All methods should give the same answer! To analyze the system 3x−2y=63x - 2y = 6 and 6x−4y=56x - 4y = 5: multiply first by 2 (6x−4y=126x - 4y = 12), now compare to second: 12≠512 \neq 5, so contradiction, no solution (parallel lines). Choice A correctly identifies no solution as the lines are parallel—well done recognizing inconsistency! Choice B says infinitely many, which might come from mistakenly getting 0=00=0 instead of 0=70=7 (from 12−512-5), but check the constants carefully! Transferable strategy: For consistency, attempt elimination or substitution; if you get a false statement, inconsistent (no solution). Keep practicing—you're getting better at spotting these!

Question 13

Solve the system by elimination:

{5x−y=92x+y=12\begin{cases} 5x - y = 9 \\ 2x + y = 12 \end{cases}
  1. (3,6)(3, 6) (correct answer)
  2. (2,8)(2, 8)
  3. (1,10)(1, 10)
  4. (4,4)(4, 4)
Explanation: This question tests your ability to solve systems of linear equations using substitution, elimination, or graphing methods—all leading to the same solution when one exists. The elimination method works well here because the y-coefficients are opposites (+y and -y). To solve 5x−y=95x - y = 9 and 2x+y=122x + y = 12 by elimination: add the equations directly since -y and +y will cancel: (5x−y)+(2x+y)=9+12(5x - y) + (2x + y) = 9 + 12. This gives 7x=217x = 21, so x=3x = 3. Back-substitute x=3x = 3 into the second equation: 2(3)+y=122(3) + y = 12, so 6+y=126 + y = 12, giving y=6y = 6. Solution: (3,6)(3, 6). Verify in both equations: First: 5(3)−6=15−6=95(3) - 6 = 15 - 6 = 9 ✓. Second: 2(3)+6=6+6=122(3) + 6 = 6 + 6 = 12 ✓. Choice A correctly identifies (3,6)(3, 6) as the solution. Choice D gives (4,4)(4, 4), but this would mean 5(4)−4=16≠95(4) - 4 = 16 ≠ 9, failing the first equation. Elimination method tip: When coefficients of one variable are opposites (like +y and -y), simply add the equations to eliminate that variable instantly! This saves time compared to multiplying equations first. Always verify your solution in both original equations to catch any arithmetic errors.

Question 14

A small business tracks its revenue and expenses using two equations: R=150t+2000R = 150t + 2000 and E=120t+2400E = 120t + 2400, where RR is revenue in dollars, EE is expenses in dollars, and tt is time in months since opening. At what point will the business break even (revenue equals expenses), and what will be the revenue at that time?

  1. 13.3 months, $4000 (correct answer)
  2. 14 months, $4100
  3. 13 months, $3950
  4. 12 months, $3840
Explanation: To find break-even point, set R = E: 150t + 2000 = 120t + 2400. Solving: 30t = 400, so t = 13.33... months. Revenue at this time: R = 150(13.33) + 2000 = $4000. Choice B uses t = 14 incorrectly. Choice C rounds t down to 13. Choice D uses t = 12.

Question 15

Marcus is solving the system {5x+2y=193x−4y=−13\begin{cases} 5x + 2y = 19 \\ 3x - 4y = -13 \end{cases} using elimination. He decides to eliminate yy by multiplying the first equation by 2. What should he multiply the second equation by to successfully eliminate yy?

  1. Multiply by 4 to clear fractions
  2. Multiply by 2 to match coefficients
  3. Multiply by -2 for opposite signs
  4. Multiply by 1 (leave unchanged) (correct answer)
Explanation: When solving a system of equations by elimination, your goal is to make the coefficients of one variable opposites so they cancel out when you add the equations together. Let's trace through Marcus's approach. He wants to eliminate yy and starts by multiplying the first equation by 2: Original: 5x+2y=195x + 2y = 19 After multiplying by 2: 10x+4y=3810x + 4y = 38 Now look at the yy-coefficients in both equations:
  • Modified first equation: +4y+4y
  • Second equation: 3x−4y=−133x - 4y = -13 has −4y-4y
Perfect! The yy-coefficients are already opposites: +4y+4y and −4y-4y. When you add these equations, the yy-terms will cancel completely: 4y+(−4y)=04y + (-4y) = 0. Choice D is correct because Marcus doesn't need to modify the second equation at all. Choice A incorrectly suggests clearing fractions, but there are no fractions in this problem. Choice B would give you +4y+4y and −8y-8y, which aren't opposites. Choice C would multiply the second equation by −2-2, creating −6x+8y=26-6x + 8y = 26, giving you +4y+4y and +8y+8y — these won't cancel. Study tip: In elimination, always check what happens to your target variable's coefficients after your first multiplication. You need opposites to cancel out, not matching signs or arbitrary multiples. Sometimes the second equation needs no change at all!

Question 16

The system {ax+2y=63x+by=9\begin{cases} ax + 2y = 6 \\ 3x + by = 9 \end{cases} has solution (2,1)(2, 1). If the system $$ \begin{cases} ax + 2y = 12 \ 3x + by = 18 \end{cases}

  1. (6,3)(6, 3) by tripling coordinates
  2. (2,1)(2, 1) remains the same solution
  3. (1,0.5)(1, 0.5) by halving coordinates
  4. (4,2)(4, 2) (correct answer)
Explanation: When you encounter systems of equations that are scaled versions of each other, think about how scaling affects the solution. The key insight is understanding the relationship between the original and modified systems. First, let's use the fact that (2,1)(2, 1) solves the original system to find aa and bb. Substituting into the first equation: a(2)+2(1)=6a(2) + 2(1) = 6, so 2a+2=62a + 2 = 6 and a=2a = 2. For the second equation: 3(2)+b(1)=93(2) + b(1) = 9, so 6+b=96 + b = 9 and b=3b = 3. Now examine what happens when we double the right sides. The new system becomes: {2x+2y=123x+3y=18\begin{cases} 2x + 2y = 12 \\ 3x + 3y = 18 \end{cases} Notice that this is equivalent to doubling the entire original system. When you multiply every term in a linear equation by the same constant, you're essentially scaling the solution by that same factor. Let's verify: if (4,2)(4, 2) is our solution, then 2(4)+2(2)=8+4=122(4) + 2(2) = 8 + 4 = 12 ✓ and 3(4)+3(2)=12+6=183(4) + 3(2) = 12 + 6 = 18 ✓. Choice A suggests tripling, but we only doubled the system. Choice B incorrectly assumes the solution stays the same—this would only be true if we added the same constant to both sides, not multiplied. Choice C suggests halving, which goes in the wrong direction entirely. Strategy tip: When the right-hand sides of a system are scaled by a factor, the solution coordinates are scaled by that same factor. This is a direct consequence of the linearity property of these equations.

Question 17

Two linear equations are given: 3x−2y=73x - 2y = 7 and 6x+ky=146x + ky = 14. For what value of kk will this system have infinitely many solutions?

  1. k=−4k = -4 (correct answer)
  2. k=4k = 4
  3. k=−2k = -2
  4. k=2k = 2
Explanation: For infinitely many solutions, the second equation must be a multiple of the first. Multiplying the first equation by 2: 6x - 4y = 14. Comparing with 6x + ky = 14, we need k = -4. Choice B gives k = 4 (sign error). Choice C uses k = -2 (coefficient confusion). Choice D uses k = 2 (both sign and coefficient errors).

Question 18

Tom attempts to solve {x−3y=72x+y=4\begin{cases} x - 3y = 7 \\ 2x + y = 4 \end{cases} by elimination. He multiplies the second equation by 3 and adds to get 7x=197x = 19, so x=197x = \frac{19}{7}. What error did Tom make?

  1. He should have multiplied the first equation by 2 instead
  2. He should have multiplied the second equation by -3 (correct answer)
  3. He made an arithmetic error in the addition step
  4. He correctly found x but needs to substitute back
Explanation: To eliminate y, Tom needs the y-coefficients to be opposites. The first equation has -3y and the second has y. Multiplying the second by 3 gives 3y, but he needs -3y to eliminate with the first equation's -3y term. He should multiply by -3. Choices A, C, and D don't address the sign error in elimination.

Question 19

Sarah solves the system {2x+3y=124x−y=2\begin{cases} 2x + 3y = 12 \\ 4x - y = 2 \end{cases} by substitution. She correctly solves the second equation for yy and substitutes into the first equation. What equation does she get after substitution and simplification?

  1. 14x+6=1214x + 6 = 12
  2. 14x−6=1214x - 6 = 12 (correct answer)
  3. −2x+6=12-2x + 6 = 12
  4. 10x−3=1210x - 3 = 12
Explanation: When solving systems of equations by substitution, you isolate one variable in terms of the other, then substitute that expression into the other equation. This creates a single-variable equation you can solve. Let's follow Sarah's approach. Starting with the second equation 4x−y=24x - y = 2, solve for yy: 4x−y=24x - y = 2 −y=2−4x-y = 2 - 4x y=4x−2y = 4x - 2 Now substitute this expression for yy into the first equation 2x+3y=122x + 3y = 12: 2x+3(4x−2)=122x + 3(4x - 2) = 12 Distribute the 3: 2x+12x−6=122x + 12x - 6 = 12 Combine like terms: 14x−6=1214x - 6 = 12 This matches answer choice B. Looking at the wrong answers: Choice A (14x+6=1214x + 6 = 12) results from a sign error when distributing—writing 3(4x−2)3(4x - 2) as 12x+612x + 6 instead of 12x−612x - 6. Choice C (−2x+6=12-2x + 6 = 12) comes from incorrectly combining 2x+12x2x + 12x as −2x-2x and getting the wrong sign on the constant term. Choice D (10x−3=1210x - 3 = 12) involves errors in both combining like terms (2x+12x=10x2x + 12x = 10x) and in distributing (3(−2)=−33(-2) = -3). Study tip: When substituting expressions with multiple terms, always use parentheses and distribute carefully. Double-check your algebra by working through each step methodically—most errors in substitution happen during the distribution and combining like terms steps.

Question 20

Solve the system and give the exact solution (an ordered pair).

{y=3x+1y=−2x+11\begin{cases} y = 3x + 1 \\ y = -2x + 11 \end{cases}
  1. (2,7)(2, 7) (correct answer)
  2. (3,10)(3, 10)
  3. (1,4)(1, 4)
  4. (7,2)(7, 2)
Explanation: This question tests your ability to solve systems where both equations are already in slope-intercept form, making substitution straightforward. When both equations are solved for y, we can set the expressions equal to each other since they both equal y. Setting 3x + 1 = -2x + 11, we get 5x = 10, so x = 2. Substituting x = 2 into the first equation: y = 3(2) + 1 = 6 + 1 = 7. The solution is (2, 7). Choice A correctly identifies this solution. Let's verify: First equation: 7 = 3(2) + 1 = 7 ✓. Second equation: 7 = -2(2) + 11 = -4 + 11 = 7 ✓. Both equations are satisfied! When both equations are in y = mx + b form, the intersection occurs where the y-values are equal, leading to a simple equation in x. This graphical interpretation helps: we're finding where two lines with slopes 3 and -2 intersect. The positive and negative slopes ensure they'll cross exactly once, giving a unique solution.