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 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!

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 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.

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.

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 $2x - 4y = 10$ and $x - 2y = 5$: Notice the second equation can be multiplied by 2 to get $2x - 4y = 10$, 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 $x - 2y = 5$ by 2 gives exactly $2x - 4y = 10$, confirming the same line!

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: $9y = 21$, $y = \frac{7}{3}$, 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 (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 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.