Algebra 2 Quiz: Solving Systems Of Linear Equations
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Solving Systems Of Linear EquationsQuestion 1 of 20

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}
No solution (parallel lines)
One solution: (0,5)(0, 5)
Infinitely many solutions (same line)
One solution: (5,0)(5, 0)
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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

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 2

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 3

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 4

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 5

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 6

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 7

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 8

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 9

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 10

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.

Question 11

Using the table shown, determine which system of equations has the solution that corresponds to the intersection point of Line A and Line B.

  1. {y=0.5x+3y=−2x+12\begin{cases} y = 0.5x + 3 \\ y = -2x + 12 \end{cases} (correct answer)
  2. {y=0.5x+2y=−2x+12\begin{cases} y = 0.5x + 2 \\ y = -2x + 12 \end{cases}
  3. {y=2x+3y=−0.5x+12\begin{cases} y = 2x + 3 \\ y = -0.5x + 12 \end{cases}
  4. {y=0.5x+3y=−x+10\begin{cases} y = 0.5x + 3 \\ y = -x + 10 \end{cases}
Explanation: From the table, Line A passes through (0,3) and (2,4), giving slope 0.5 and y-intercept 3: y = 0.5x + 3. Line B passes through (0,12) and (2,8), giving slope -2 and y-intercept 12: y = -2x + 12. These intersect at (3.6, 4.8). Choice B has wrong y-intercept for Line A. Choice C swaps the slopes. Choice D has wrong equation for Line B.

Question 12

Based on the graph shown, which system of equations best represents the two lines?

  1. {y=2x−1y=−x+5\begin{cases} y = 2x - 1 \\ y = -x + 5 \end{cases} (correct answer)
  2. {y=2x+1y=−x+5\begin{cases} y = 2x + 1 \\ y = -x + 5 \end{cases}
  3. {y=2x−1y=x−5\begin{cases} y = 2x - 1 \\ y = x - 5 \end{cases}
  4. {y=−2x+1y=−x+5\begin{cases} y = -2x + 1 \\ y = -x + 5 \end{cases}
Explanation: The first line has positive slope 2 and y-intercept -1. The second line has negative slope -1 and y-intercept 5. These lines intersect at (2,3). Choice B has wrong y-intercept for first line. Choice C has wrong slope and y-intercept for second line. Choice D has wrong slope for first line.

Question 13

Solve the system using the substitution method:

{x−y=1y=3x−5\begin{cases} x - y = 1\\ y = 3x - 5 \end{cases}

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

  1. (2,1)(2, 1) (correct answer)
  2. (3,4)(3, 4)
  3. (1,−2)(1, -2)
  4. (2,−1)(2, -1)
Explanation: This question tests your ability to solve systems of linear equations using substitution when one equation is already solved for a variable. A system of linear equations can be solved by substitution: replace one variable with its expression from one equation into the other equation. We have x - y = 1 and y = 3x - 5. Since the second equation already gives y in terms of x, substitute this into the first equation: x - (3x - 5) = 1. Carefully distribute the negative sign: x - 3x + 5 = 1, which gives -2x + 5 = 1, so -2x = -4, thus x = 2. Now back-substitute into y = 3x - 5: y = 3(2) - 5 = 6 - 5 = 1. Choice A correctly gives the solution (2, 1), which we verify: first equation: 2 - 1 = 1 ✓; second equation: 1 = 3(2) - 5 = 6 - 5 = 1 ✓. Choice D might tempt students who get x = 2 correctly but make a sign error in back-substitution, getting y = -1 instead of y = 1. The substitution method is powerful when one equation is already solved for a variable: (1) substitute the expression into the other equation, (2) solve for the remaining variable, (3) back-substitute to find the other variable, (4) write as an ordered pair (x, y), (5) always verify in both original equations. Remember to use parentheses when substituting to avoid sign errors—this is where many mistakes happen!

Question 14

Determine whether the system is consistent. If it has a solution, find it; otherwise, state no solution.

{3x+2y=106x+4y=25\begin{cases} 3x + 2y = 10 \\ 6x + 4y = 25 \end{cases}
  1. No solution (correct answer)
  2. Infinitely many solutions
  3. (5,−2.5)(5, -2.5)
  4. (0,5)(0, 5)
Explanation: This question tests your ability to determine if a system is consistent and find the solution or state no solution. 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 or elimination, and graphically, we plot both lines and find intersection. For 3x+2y=103x + 2y = 10 and 6x+4y=256x + 4y = 25: multiply first by 2 to get 6x+4y=206x + 4y = 20, which has same coefficients but 20≠2520 ≠ 25, so parallel lines, no solution; verify: inconsistent. Choice A correctly states there is no solution. Choice B (infinitely many) fails by not checking if the constants scale proportionally; always compare after scaling! Substitution method recipe: (1) Pick the easier equation to solve for one variable, (2) Solve for that variable, (3) Substitute the expression into the other equation, (4) Solve the resulting equation, (5) Back-substitute, (6) Write as ordered pair, (7) Verify. Elimination method recipe: (1) Align equations, (2) Multiply to make coefficients opposites, (3) Add or subtract to eliminate a variable, (4) Solve, (5) Substitute back, (6) Write solution, (7) Verify; example: 2x+3y=132x + 3y = 13 and x−3y=−4x - 3y = -4, multiply second by 2 to get 2x−6y=−82x - 6y = -8, subtract from first: 9y=219y = 21, y=7/3y = 7/3, then x=3x = 3.

Question 15

Determine whether the system is consistent. If it has a solution, give it; otherwise state no solution:

{3x−y=26x−2y=5\begin{cases} 3x - y = 2\\ 6x - 2y = 5 \end{cases}
  1. Infinitely many solutions
  2. No solution (correct answer)
  3. (1,1)(1, 1)
  4. (2,4)(2, 4)
Explanation: This question tests your ability to determine whether a system is consistent (has at least one solution) or inconsistent (has no solution), requiring analysis of the relationship between equations. A system is inconsistent when the equations represent parallel lines that never intersect, which happens when they have the same slope but different y-intercepts. Looking at 3x - y = 2 and 6x - 2y = 5, let's check if the second equation is a multiple of the first: if we multiply the first equation by 2, we get 2(3x - y) = 2(2), which gives 6x - 2y = 4. But our second equation has 6x - 2y = 5, not 4! This means the left sides are proportional but the right sides are not (4 ≠ 5), indicating parallel lines. We can verify by converting to slope-intercept form: from 3x - y = 2, we get y = 3x - 2; from 6x - 2y = 5, we get 2y = 6x - 5, so y = 3x - 5/2. Same slope (3) but different y-intercepts (-2 vs -5/2), confirming parallel lines. Choice A correctly identifies that the system has no solution because the lines are parallel—they have the same slope but different y-intercepts, so they never intersect. Choice B incorrectly assumes that having proportional left sides means infinitely many solutions, but the right sides must also be proportional for that to be true. To check consistency: (1) See if one equation is a constant multiple of the other (consistent with infinitely many solutions), (2) Check if left sides are proportional but right sides are not (inconsistent, no solution), (3) Otherwise, the system has exactly one solution. Remember: parallel lines (same slope, different y-intercepts) mean no solution, while the same line written two ways means infinitely many solutions!

Question 16

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 17

Solve the system and state the solution set.

{x+2y=102x+4y=20\begin{cases} x + 2y = 10 \\ 2x + 4y = 20 \end{cases}
  1. No solution
  2. Exactly one solution: (10,0)(10, 0)
  3. Infinitely many solutions: all (x,y)(x, y) such that x+2y=10x + 2y = 10 (correct answer)
  4. Exactly one solution: (0,5)(0, 5)
Explanation: This question tests your ability to recognize when a system has infinitely many solutions and express the solution set properly. A system has infinitely many solutions when both equations represent the same line, just written differently. Looking at x + 2y = 10 and 2x + 4y = 20, notice that the second equation is exactly 2 times the first: 2(x + 2y) = 2(10) gives 2x + 4y = 20. This means both equations describe the same line! Choice C correctly identifies this as infinitely many solutions and properly expresses the solution set as all (x, y) such that x + 2y = 10. This is the correct mathematical way to describe all points on the line. We can verify by solving for y: y = (10 - x)/2, showing that for any x-value, there's a corresponding y-value on the line. For example: (0, 5), (2, 4), (4, 3), (6, 2), (8, 1), (10, 0) all satisfy both equations. When you have infinitely many solutions, express the solution set using one of the original equations (they're equivalent) or in parametric form. The key recognition pattern: when all coefficients and constants are proportional by the same factor, you have one line written two ways.

Question 18

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

{2x−4y=8x−2y=4\begin{cases} 2x - 4y = 8 \\ x - 2y = 4 \end{cases}
  1. One solution: (4,0)(4,0)
  2. No solution (parallel lines)
  3. Infinitely many solutions (same line) (correct answer)
  4. One solution: (0,−2)(0,-2)
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 or elimination, and graphically, we plot both lines and find intersection. For 2x−4y=82x - 4y = 8 and x−2y=4x - 2y = 4: multiply the second by 2 to get 2x−4y=82x - 4y = 8, which is identical to the first, so same line, infinitely many solutions; verify: equations are dependent. Choice C correctly identifies that there are infinitely many solutions (same line). Choice B (no solution) might come from thinking they are parallel but not checking if constants match after scaling; always scale and compare both sides! Substitution method recipe: (1) Pick the easier equation to solve for one variable, (2) Solve for that variable, (3) Substitute the expression into the other equation, (4) Solve the resulting equation, (5) Back-substitute, (6) Write as ordered pair, (7) Verify. Elimination method recipe: (1) Align equations, (2) Multiply to make coefficients opposites, (3) Add or subtract to eliminate a variable, (4) Solve, (5) Substitute back, (6) Write solution, (7) Verify; example: 2x+3y=132x + 3y = 13 and x−3y=−4x - 3y = -4, multiply second by 2 to get 2x−6y=−82x - 6y = -8, subtract from first: 9y=219y = 21, y=7/3y = 7/3, then x=3x = 3.

Question 19

Solve the system by elimination. Give the solution as an ordered pair (x,y)(x,y).

{x+y=72x−y=8\begin{cases} x + y = 7 \\ 2x - y = 8 \end{cases}
  1. (5,2)(5,2) (correct answer)
  2. (2,5)(2,5)
  3. (3,4)(3,4)
  4. (4,3)(4,3)
Explanation: This question tests your ability to solve systems of linear equations using the elimination method, giving the ordered pair solution. 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 or elimination, and graphically, we plot both lines and find intersection. To solve x+y=7x + y = 7 and 2x−y=82x - y = 8 by elimination: add the equations to eliminate y, 3x=153x = 15, x=5x = 5, then y=7−5=2y = 7 - 5 = 2, solution (5,2)(5, 2); verify: both hold. Choice A correctly identifies the solution (5,2)(5, 2) that satisfies both equations. Choice B (2,5)(2,5) swaps x and y, perhaps from solving for y first incorrectly; remember the ordered pair is (x, y)! Substitution method recipe: (1) Pick the easier equation to solve for one variable, (2) Solve for that variable, (3) Substitute the expression into the other equation, (4) Solve the resulting equation, (5) Back-substitute, (6) Write as ordered pair, (7) Verify. Elimination method recipe: (1) Align equations, (2) Multiply to make coefficients opposites, (3) Add or subtract to eliminate a variable, (4) Solve, (5) Substitute back, (6) Write solution, (7) Verify; example: 2x+3y=132x + 3y = 13 and x−3y=−4x - 3y = -4, multiply second by 2 to get 2x−6y=−82x - 6y = -8, subtract from first: 9y=219y = 21, y=73y = \frac{7}{3}, then x=3x = 3.

Question 20

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.