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 $3x - y = 6$ and $6x - 2y = 12$: multiply the first by 2 to get $6x - 2y = 12$, 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: $x - y = 2$ and $2x - 2y = 4$ is infinite. Awesome progress!

This question tests your ability to solve systems of linear equations using the substitution 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 $x - y = 2$ and $y = x - 2$ by substitution: notice the second is already $y = x - 2$, and rearranging the first gives $y = x - 2$—same equation, so infinitely many solutions (any x, $y = x - 2$ works). No unique point, but the entire line. Verify: they are equivalent. Choice B correctly identifies infinitely many solutions as the equations are dependent. Choice C might come from picking a specific point without recognizing the identity—test multiple points to see! Substitution tip: If after substituting you get $0 = 0$ (true), infinite; $0 =$ number (false), none. Example: $x + y = 5$ and $2x + 2y = 10$ is infinite. Terrific insight!

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 $2x + y = 5$ and $4x + 2y = 12$: multiply the first by 2 to get $4x + 2y = 10$, but the second is $4x + 2y = 12$—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 ($-2$) but different y-intercepts ($5$ vs. $6$). 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: $2x + y = 5$ and $4x + 2y = 10$ is infinite (same line). You're getting the hang of it!

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!

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 $y = x + 2$ and $y = -2x + 8$ graphically: the intersection is at $(2,4)$, or algebraically set equal: $x + 2 = -2x + 8$, $3x = 6$, $x=2$, $y=4$; solution: $(2,4)$; verify in both (it checks out). Choice A correctly identifies the solution $(2,4)$ that satisfies both equations—excellent work spotting the intersection! Choice B gives $(4,2)$, which might come from a sign error in solving, like $x + 2 = 2x + 8$ leading to $-x=6$, 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!

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!

This question tests your ability to determine if a system has one solution, no solution, or infinitely many, 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 $2x + y = 5$ and $4x + 2y = 10$: multiply first by 2 ($4x + 2y = 10$), which matches the second exactly, so infinitely many solutions (same line). Choice B correctly identifies infinitely many solutions as the equations represent the same line—great insight! Choice A says no solution, which might come from misaligning and getting a false statement like $0=1$, but here it's $0=0$, meaning consistent and dependent; always check after elimination! Transferable strategy: If elimination leads to $0=0$, infinite solutions; $0=$number (not 0), no solution; otherwise, one solution. You're making progress—keep exploring these cases!

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 $x = 3$.

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 $2x - 4y = 8$ and $x - 2y = 4$: multiply the second by 2 to get $2x - 4y = 8$, 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: $2x + 3y = 13$ and $x - 3y = -4$, multiply second by 2 to get $2x - 6y = -8$, subtract from first: $9y = 21$, $y = 7/3$, then $x = 3$.

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 = 7 and 2x - y = 8 by elimination: add the equations to eliminate y, 3x = 15, x = 5, then y = 7 - 5 = 2, solution (5, 2); verify: both hold. Choice A correctly identifies the solution (5, 2) that satisfies both equations. Choice B (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 = 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.