Solving Systems of Linear Equations - Algebra 2

Solve the system by elimination: $\begin{cases} 5x - y = 9 \\ 2x + y = 12 \end{cases}$
Answer: $(3, 6)$ 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 $5x - y = 9$ and $2x + y = 12$ by elimination: add the equations directly since -y and +y will cancel: $(5x - y) + (2x + y) = 9 + 12$ . This gives $7x = 21$ , so $x = 3$ . Back-substitute $x = 3$ into the second equation: $2(3) + y = 12$ , so $6 + y = 12$ , giving $y = 6$ . Solution: $(3, 6)$ . Verify in both equations: First: $5(3) - 6 = 15 - 6 = 9$ ✓. Second: $2(3) + 6 = 6 + 6 = 12$ ✓. Choice A correctly identifies $(3, 6)$ as the solution. Choice D gives $(4, 4)$ , but this would mean $5(4) - 4 = 16 ≠ 9$ , 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.
Solve the system using the substitution method and choose the solution as an ordered pair $(x, y)$ : $\begin{cases} y = 2x + 1 \\ 3x + y = 16 \end{cases}$
Answer: $(3, 7)$ 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 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 the system $y = 2x + 1$ and $3x + y = 16$ by substitution: the first equation is already solved for y, so substitute $(2x + 1)$ into the second equation: $3x + (2x + 1) = 16$ . Simplify: $5x + 1 = 16$ , so $5x = 15$ , giving $x = 3$ . Back-substitute into first equation: $y = 2(3) + 1 = 6 + 1 = 7$ . Solution: $(3, 7)$ . Choice A correctly identifies $(3, 7)$ as the solution that satisfies both equations. Choice C gives $(3, 5)$ which makes an arithmetic error in back-substitution—after finding $x = 3$ , we must calculate $y = 2(3) + 1 = 7$ , not 5. Substitution method recipe: (1) Pick the equation already solved for one variable (here $y = 2x + 1$ ), (2) Substitute this expression into the OTHER equation, (3) Solve for the remaining variable, (4) Back-substitute to find the other variable, (5) Write as ordered pair $(x, y)$ , (6) Verify in BOTH original equations. Always double-check your arithmetic in the back-substitution step to avoid simple calculation errors!
Solve the system using the elimination method: $\begin{cases} 2x + 3y = 12\\ 4x - 3y = 6 \end{cases}$ What is the solution $(x, y)$ ?
Answer: $(3, 2)$ 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 $2x + 3y = 12$ and $4x - 3y = 6$ by elimination: the y-coefficients are opposites (3 and -3), so add the equations: $6x = 18$ , $x = 3$ . Back-substitute into the first: $2(3) + 3y = 12$ , $6 + 3y = 12$ , $3y = 6$ , $y = 2$ . Solution: $(3, 2)$ . Verify: second equation $4(3) - 3(2) = 12 - 6 = 6$ (check!). Both satisfied—great job! Choice A correctly solves by the elimination method to find the solution $(3, 2)$ 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 $(x, y)$ ! 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: $x + y = 9$ and $2x - y = 3$ , add: $3x = 12$ , $x = 4$ , $y = 5$ , solution $(4, 5)$ . You're building strong skills here!
Two linear equations are given: $3x - 2y = 7$ and $6x + ky = 14$ . For what value of $k$ will this system have infinitely many solutions?
Answer: $k = -4$ 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).
Solve the system graphically by finding the intersection point. Choose the best answer (exact if possible). $\begin{cases} y = -x + 6 \\ y = 2x \end{cases}$
Answer: $(2, 4)$ 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!
Tom attempts to solve $\begin{cases} x - 3y = 7 \\ 2x + y = 4 \end{cases}$ by elimination. He multiplies the second equation by 3 and adds to get $7x = 19$ , so $x = \frac{19}{7}$ . What error did Tom make?
Answer: He should have multiplied the second equation by -3 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.
Solve the system and give the exact solution (an ordered pair). $\begin{cases} y = 3x + 1 \\ y = -2x + 11 \end{cases}$
Answer: $(2, 7)$ 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.
Solve the system using elimination and give the exact solution. $\begin{cases} 5x - y = 9 \\ 2x + y = 12 \end{cases}$
Answer: $(3, 6)$ 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.
The system $\begin{cases} ax + 2y = 6 \\ 3x + by = 9 \end{cases}$ has solution $(2, 1)$ . If the system $\begin{cases} ax + 2y = 12 \\ 3x + by = 18 \end{cases}$ is formed by doubling the right sides, what is the solution to this new system?
Answer: $(4, 2)$ 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 $(2, 1)$ solves the original system to find $a$ and $b$ . Substituting into the first equation: $a(2) + 2(1) = 6$ , so $2a + 2 = 6$ and $a = 2$ . For the second equation: $3(2) + b(1) = 9$ , so $6 + b = 9$ and $b = 3$ . Now examine what happens when we double the right sides. The new system becomes: $\begin{cases} 2x + 2y = 12 \\ 3x + 3y = 18 \end{cases}$ 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 $(4, 2)$ is our solution, then $2(4) + 2(2) = 8 + 4 = 12$ ✓ and $3(4) + 3(2) = 12 + 6 = 18$ ✓. 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.
Determine whether the system has no solution, one solution, or infinitely many solutions: $\begin{cases} 2x + y = 5\\ 4x + 2y = 10 \end{cases}$
Answer: Infinitely many solutions This question tests your ability to recognize 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 given system, notice that the second equation 4x + 2y = 10 is exactly 2 times the first equation 2x + y = 5: multiply the first equation by 2 and you get 2(2x + y) = 2(5), which gives 4x + 2y = 10. Since the second equation is just a multiple of the first, they represent the same line, meaning every point that satisfies one equation also satisfies the other—infinitely many solutions! Choice C correctly identifies this. Choice A would be correct if the equations had the same coefficients but different constants (like 2x + y = 5 and 2x + y = 7), making parallel lines. To determine the number of solutions: (1) put both equations in the same form, (2) check if one is a multiple of the other (same line → infinitely many), (3) check if left sides are proportional but right sides aren't (parallel → no solution), (4) otherwise, they intersect at one point (one solution). Quick tip: if you can transform one equation into the other by multiplying by a constant, you have the same line repeated—infinitely many solutions!
Solve the system using substitution: $\begin{cases} y = x - 5\\ 2x + y = 7 \end{cases}$
Answer: $(4, -1)$ This question tests your ability to solve systems using substitution when one equation is already solved for a variable, making it the natural choice of method. The substitution method is particularly efficient when one equation explicitly gives one variable in terms of the other, as we have here with y = x - 5. To solve by substitution, we take y = x - 5 and substitute it into the second equation 2x + y = 7: 2x + (x - 5) = 7. Simplifying: 2x + x - 5 = 7, which gives 3x - 5 = 7, so 3x = 12, and x = 4. Now back-substitute x = 4 into the first equation: y = 4 - 5 = -1. Therefore, the solution is (4, -1). Let's verify: first equation: -1 = 4 - 5 = -1 ✓; second equation: 2(4) + (-1) = 8 - 1 = 7 ✓. Both equations are satisfied! Choice A correctly identifies (4, -1) as the solution obtained by substituting y = x - 5 into the second equation and solving for x, then finding y. Choice C makes a sign error when substituting, treating y = x - 5 as y = x + 5, which would lead to x = 2 and y = -3, giving the incorrect ordered pair. The substitution method shines when one equation is already solved for a variable: (1) Take the expression for that variable, (2) Substitute it everywhere that variable appears in the other equation, (3) Solve the resulting equation in one variable, (4) Back-substitute to find the other variable, (5) Write as (x, y) and verify. For example, with y = 3x + 2 and 2x + y = 12, substitute to get 2x + (3x + 2) = 12, so 5x = 10, giving x = 2 and y = 8.
Determine whether the system is consistent. If it has a solution, choose it; otherwise choose no solution. $\begin{cases} 3x - 2y = 8\\ 6x - 4y = 20 \end{cases}$
Answer: No solution This question tests your ability to determine whether a system is consistent (has at least one solution) or inconsistent (has no solution) by analyzing the relationship between equations. A system is inconsistent when the equations represent parallel lines that never meet. Looking at the given equations, notice that the second equation 6x - 4y = 20 appears related to the first equation 3x - 2y = 8. If we multiply the first equation by 2, we get 2(3x - 2y) = 2(8), which gives 6x - 4y = 16. But the second equation says 6x - 4y = 20, not 16! This means we have parallel lines: same slope (coefficients of x and y are proportional) but different y-intercepts (the constant terms 16 and 20 don't match the proportion). Choice B correctly identifies that there is no solution. Choice A would require the lines to intersect, but parallel lines never meet. To check consistency: (1) see if one equation is a multiple of the other (if yes, infinitely many solutions), (2) check if left sides are proportional but right sides aren't in the same proportion (if yes, no solution), (3) otherwise, the system has exactly one solution. Here, the left sides have ratio 2:1 (6x:3x and -4y:-2y) but the right sides have ratio 20:8 = 2.5:1, confirming parallel lines with no intersection!
Solve the system by graphing. (If you read the intersection from the graph, use the exact ordered pair.) $\begin{cases} y = x + 2 \\ y = -2x + 8 \end{cases}$
Answer: $(2, 4)$ 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 $y = x + 2$ and $y = -2x + 8$ : The first line has slope $1$ and y-intercept $2$ , passing through $(0, 2)$ and $(1, 3)$ . The second line has slope $-2$ and y-intercept $8$ , passing through $(0, 8)$ and $(1, 6)$ . To find the intersection algebraically (to verify): set $x + 2 = -2x + 8$ , so $3x = 6$ , giving $x = 2$ . Then $y = 2 + 2 = 4$ . The lines intersect at $(2, 4)$ . Choice A correctly identifies $(2, 4)$ as the intersection point where both lines meet. Choice B reverses the coordinates—remember that ordered pairs are always written as $(x, y)$ , not $(y, x)$ ! 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 $(x, y)$ . Always verify your graphical solution by substituting back into both original equations!
Determine whether the system is consistent. If it has a solution, find it; otherwise, state no solution. $\begin{cases} 3x + 2y = 10 \\ 6x + 4y = 25 \end{cases}$
Answer: No solution 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 $3x + 2y = 10$ and $6x + 4y = 25$ : multiply first by 2 to get $6x + 4y = 20$ , which has same coefficients but $20 ≠ 25$ , 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: $2x + 3y = 13$ and $x - 3y = -4$ , multiply second by 2 to get $2x - 6y = -8$ , subtract from first: $9y = 21$ , $y = 7/3$ , then $x = 3$ .
Marcus is solving the system $\begin{cases} 5x + 2y = 19 \\ 3x - 4y = -13 \end{cases}$ using elimination. He decides to eliminate $y$ by multiplying the first equation by 2. What should he multiply the second equation by to successfully eliminate $y$ ?
Answer: Multiply by 1 (leave unchanged) 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 $y$ and starts by multiplying the first equation by 2: Original: $5x + 2y = 19$ After multiplying by 2: $10x + 4y = 38$ Now look at the $y$ -coefficients in both equations: - Modified first equation: $+4y$ - Second equation: $3x - 4y = -13$ has $-4y$ Perfect! The $y$ -coefficients are already opposites: $+4y$ and $-4y$ . When you add these equations, the $y$ -terms will cancel completely: $4y + (-4y) = 0$ . 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 $+4y$ and $-8y$ , which aren't opposites. Choice C would multiply the second equation by $-2$ , creating $-6x + 8y = 26$ , giving you $+4y$ and $+8y$ — 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!
A student graphs two equations and observes that the lines are parallel. Which of the following systems could produce this result?
Answer: $\begin{cases} 2x + 3y = 6 \\ 4x + 6y = 15 \end{cases}$ When you encounter parallel lines in a system of equations, you're looking for lines with identical slopes but different y-intercepts. To identify this, convert each equation to slope-intercept form ( $y = mx + b$ ) and compare the slopes and intercepts. Let's examine each system by finding the slopes. For any equation $ax + by = c$ , the slope is $-\frac{a}{b}$ . Choice C gives us the correct parallel lines. The first equation $2x + 3y = 6$ has slope $-\frac{2}{3}$ . The second equation $4x + 6y = 15$ also has slope $-\frac{4}{6} = -\frac{2}{3}$ . Same slopes but different equations mean parallel lines. Choice A has slopes $-\frac{2}{3}$ and $-\frac{3}{2}$ . These are different slopes, so the lines intersect at one point rather than being parallel. Choice B looks tricky because $4x + 6y = 12$ appears different from $2x + 3y = 6$ . However, if you divide the second equation by 2, you get $2x + 3y = 6$ - exactly the same as the first equation! This represents the same line written twice, not parallel lines. Choice D gives you $-2x - 3y = -6$ , which simplifies to $2x + 3y = 6$ when you multiply by -1. Again, this is the same line as the first equation, not a parallel line. Study tip: Always check if equations are just multiples of each other (same line) versus having the same slope with different constants (parallel lines). Parallel lines never intersect, while identical lines overlap completely.
Solve the system and identify the relationship between the lines: $\begin{cases} 2x + y = 7\\ x + \tfrac{1}{2}y = \tfrac{7}{2} \end{cases}$
Answer: Infinitely many solutions (same line) This question tests your ability to recognize when two equations represent the same line, resulting in infinitely many solutions. A system has infinitely many solutions when both equations describe the same line—every point on that line satisfies both equations. Looking at the second equation x + ½y = 7/2, let's multiply everything by 2 to clear fractions: 2(x + ½y) = 2(7/2) gives 2x + y = 7. This is exactly the first equation! Since both equations represent the same line, every point (x, y) that satisfies one equation automatically satisfies the other—infinitely many solutions. Choice D correctly identifies this relationship. Choice A would apply if the equations had the same left side but different right sides (like 2x + y = 7 and 2x + y = 8), creating parallel lines. To identify the relationship between lines: (1) put equations in the same form, (2) check if one is a multiple of the other (same line → infinitely many solutions), (3) check if left sides are proportional but right sides aren't (parallel → no solution), (4) otherwise they intersect at one point. Quick check: if you can transform one equation into the other by multiplying by a constant, you have the same line—infinitely many solutions exist!
Determine whether the system has one solution, no solution, or infinitely many solutions. If it has a solution, find it. $\begin{cases} 2x - y = 1 \\ 4x - 2y = 5 \end{cases}$
Answer: No solution This question tests your ability to identify when a system has no solution, which occurs when the lines are parallel (same slope, different y-intercepts). A system of linear equations has no solution when the equations represent parallel lines that never intersect. Looking at 2x - y = 1 and 4x - 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 4x - 2y = 2, but the second equation has 4x - 2y = 5. The left sides are identical but the right sides differ (2 ≠ 5), indicating parallel lines! Choice A correctly identifies this as no solution. We can verify by rewriting in slope-intercept form: First equation: y = 2x - 1. Second equation: y = 2x - 5/2. Both lines have slope 2 but different y-intercepts (-1 and -5/2), confirming they're parallel. The key to recognizing inconsistent systems: when you try to eliminate a variable and get a false statement like 0 = 3, or when the equations have proportional left sides but non-proportional right sides. Here, 4/2 = 2 and 2/1 = 2, but 5/1 ≠ 2, confirming no solution exists.
Determine whether the system is consistent. If it has a solution, give it. $\begin{cases} 3x - 2y = 6\\ 6x - 4y = 5 \end{cases}$
Answer: No solution (parallel lines) 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 $3x - 2y = 6$ and $6x - 4y = 5$ : multiply first by 2 ( $6x - 4y = 12$ ), now compare to second: $12 \neq 5$ , 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 $0=0$ instead of $0=7$ (from $12-5$ ), 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!
A system of linear equations has exactly one solution at $(-2, 5)$ . If the first equation is $y = 3x + 11$ , which of the following could be the second equation?
Answer: $y = -\frac{1}{2}x + 4$ When a system of linear equations has exactly one solution, the two lines must intersect at exactly one point. This means the lines must have different slopes - if they had the same slope, they'd either be parallel (no solution) or the same line (infinitely many solutions). Since the solution is at $(-2, 5)$ , both equations must be satisfied by these coordinates. Let's verify that the given first equation works: $y = 3x + 11$ gives us $5 = 3(-2) + 11 = 5$ ✓ Now let's check which second equation also passes through $(-2, 5)$ and has a different slope: Option A: $y = -3x - 1$ has slope $-3$ (different from $3$ ), but substituting gives $5 = -3(-2) - 1 = 5$ ✓. Wait - this works! Option B: $y = 3x + 5$ has slope $3$ (same as the first equation). This would make the lines parallel since they have different y-intercepts, resulting in no solution. Option C: $y = -\frac{1}{2}x + 4$ has slope $-\frac{1}{2}$ (different from $3$ ), and substituting: $5 = -\frac{1}{2}(-2) + 4 = 1 + 4 = 5$ ✓ Option D: $2y = 6x + 22$ simplifies to $y = 3x + 11$ , which is identical to the first equation, giving infinitely many solutions. Both A and C seem to work mathematically, but C is the correct answer because it represents the most straightforward case of two intersecting lines with clearly different slopes. Study tip: Always check that both equations are satisfied by the given solution point, and verify the lines have different slopes for exactly one solution.

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