This question tests your ability to solve systems of linear equations—finding the \((x, y)\) pair that makes both equations true at the same time. The elimination method (also called addition method) works by adding or subtracting the equations to make one variable disappear: if you have \(x + y = 7\) and \(x - y = 1\), adding them gives \(2x = 8\) because the y terms cancel out. Then solve for x, and use that to find y! Let's add the equations: \((x + y) + (x - y) = 7 + 1\), which gives us \(2x = 8\), so \(x = 4\). Now substitute \(x = 4\) into the first equation: \(4 + y = 7\), so \(y = 3\). Choice A is correct because \((4, 3)\) satisfies both equations when you substitute back to check: \(4 + 3 = 7\) ✓ and \(4 - 3 = 1\) ✓. Choice B would give us \(3 + 4 = 7\) ✓ but \(3 - 4 = -1\) ✗, not 1, so it only works for one equation. Always check your answer by plugging both x and y into BOTH original equations. If you get true statements (like \(7 = 7\) and \(1 = 1\)), you're correct! If even one equation doesn't work, there's an error somewhere. This check habit catches almost all mistakes and builds confidence!

This question tests your ability to solve systems of linear equations—finding the $ (x, y) $ pair that makes both equations true at the same time. The elimination method (also called addition method) works by adding or subtracting the equations to make one variable disappear: if you have $ x + y = 5 $ and $ x - y = 1 $, adding them gives $ 2x = 6 $ because the y terms cancel out. Then solve for x, and use that to find y! Add the equations: $ (3x - y) + (x + y) = 7 + 5 $, which simplifies to $ 4x = 12 $, so $ x = 3 $; then substitute into $ x + y = 5 $ to get $ 3 + y = 5 $, so $ y = 2 $. Choice A is correct because it gives the $ (x, y) $ pair $ (3, 2) $ that satisfies both equations when you substitute back to check: $ 3(3) - 2 = 9 - 2 = 7 $ and $ 3 + 2 = 5 $. Switching x and y might lead to choice B $ (2, 3) $, but remember to solve step by step. Here's how to choose a method: if one equation is already solved for a variable (like $ y = 3x + 1 $), use substitution—it's set up perfectly! If the coefficients of one variable are opposites (like $ 2x $ and $ -2x $) or the same (like $ 3y $ and $ 3y $), use elimination—one variable will cancel nicely. And you can always graph both lines to see where they cross! With practice, you'll spot the easiest method for each system.

This question tests your ability to solve systems of linear equations—finding the $ (x, y) $ pair that makes both equations true at the same time. The elimination method (also called addition method) works by adding or subtracting the equations to make one variable disappear: if you have $ x + y = 5 $ and $ x - y = 1 $, adding them gives $ 2x = 6 $ because the y terms cancel out. Then solve for x, and use that to find y! Notice both equations have 2x, so if we subtract the second from the first: $ (2x + 3y) - (2x - y) = 12 - 4 $. This gives us $ 4y = 8 $, so $ y = 2 $. Now substitute $ y = 2 $ into either equation; using the second: $ 2x - 2 = 4 $, so $ 2x = 6 $, giving us $ x = 3 $. Choice A is correct because $ (3, 2) $ satisfies both equations when you substitute back to check: $ 2(3) + 3(2) = 6 + 6 = 12 $ ✓ and $ 2(3) - 2 = 6 - 2 = 4 $ ✓. If you got $ (2, 3) $, you might have switched the x and y values—remember that ordered pairs are always written as $ (x, y) $! Always check your answer by plugging both x and y into BOTH original equations. If you get true statements (like $ 5 = 5 $ and $ 7 = 7 $), you're correct! If even one equation doesn't work, there's an error somewhere. This check habit catches almost all mistakes and builds confidence!

This question tests your ability to solve systems of linear equations—finding the (x, y) pair that makes both equations true at the same time. A system of equations is like a puzzle where you need to find values that work for both equations simultaneously: the solution (x, y) must make the first equation true AND make the second equation true. Graphically, this is where the two lines intersect—that one point where both equations are satisfied! Looking at the system 2x + 4y = 10 and x + 2y = 5, notice that if we multiply the second equation by 2, we get 2x + 4y = 10—exactly the same as the first equation! This means both equations represent the same line, so every point on that line is a solution. The system has infinitely many solutions. Choice C is correct because the two equations are actually the same line in disguise—divide the first equation by 2 and you get the second equation exactly. When two equations represent the same line, every point on that line satisfies both equations! Always check your answer by plugging both x and y into BOTH original equations. If you get true statements (like 10 = 10 and 5 = 5), you're correct! If even one equation doesn't work, there's an error somewhere. This check habit catches almost all mistakes and builds confidence!

This question tests your ability to solve systems of linear equations—finding the (x, y) pair that makes both equations true at the same time. A system of equations is like a puzzle where you need to find values that work for both equations simultaneously: the solution (x, y) must make the first equation true AND make the second equation true. Graphically, this is where the two lines intersect—that one point where both equations are satisfied! Let's examine these equations: 2x + 4y = 10 and x + 2y = 5. If we multiply the second equation by 2, we get 2x + 4y = 10—that's exactly the same as the first equation! This means both equations represent the same line, so every point on that line is a solution. Choice C is correct because when two equations represent the same line, they have infinitely many solutions—every point on the line works! You can verify this by dividing the first equation by 2: (2x + 4y)/2 = 10/2 gives x + 2y = 5, which is exactly the second equation. Always check your answer by plugging both x and y into BOTH original equations. If you get true statements (like 5 = 5 and 10 = 10), you're correct! If even one equation doesn't work, there's an error somewhere. This check habit catches almost all mistakes and builds confidence!

This question tests your ability to solve systems of linear equations—finding the (x, y) pair that makes both equations true at the same time. The elimination method (also called addition method) works by adding or subtracting the equations to make one variable disappear: if you have 2x + y = 8 and x - y = 1, adding them gives 3x = 9 because the y terms cancel out. Then solve for x, and use that to find y! Let's add the equations: (2x + y) + (x - y) = 8 + 1, which gives us 3x = 9, so x = 3. Now substitute x = 3 into the second equation: 3 - y = 1, so y = 2. Choice A is correct because (3, 2) satisfies both equations when you substitute back to check: 2(3) + 2 = 6 + 2 = 8 ✓ and 3 - 2 = 1 ✓. If you picked (2, 3), you might have mixed up which value is x and which is y—remember that ordered pairs are always written as (x, y), not (y, x)! Here's how to choose a method: if one equation is already solved for a variable (like y = 3x + 1), use substitution—it's set up perfectly! If the coefficients of one variable are opposites (like +y and -y here), use elimination—one variable will cancel nicely. And you can always graph both lines to see where they cross! With practice, you'll spot the easiest method for each system.

This question tests your ability to solve systems of linear equations—finding the (x, y) pair that makes both equations true at the same time. The elimination method (also called addition method) works by adding or subtracting the equations to make one variable disappear: if you have x + y = 5 and x - y = 1, adding them gives 2x = 6 because the y terms cancel out. Then solve for x, and use that to find y! To solve, subtract the second from the first: (2x + 3y) - (2x - y) = 12 - 4, which simplifies to 4y = 8, so y = 2; then substitute into 2x - y = 4 to get 2x - 2 = 4, so 2x = 6 and x = 3. Choice A is correct because it gives the (x, y) pair (3, 2) that satisfies both equations when you substitute back to check: 2(3) + 3(2) = 6 + 6 = 12 and 2(3) - 2 = 6 - 2 = 4. A possible error is adding instead of subtracting, which might give choice D (2, 2), but ensure the operation cancels the x terms properly. Here's how to choose a method: if one equation is already solved for a variable (like y = 3x + 1), use substitution—it's set up perfectly! If the coefficients of one variable are opposites (like 2x and -2x) or the same (like 3y and 3y), use elimination—one variable will cancel nicely. And you can always graph both lines to see where they cross! With practice, you'll spot the easiest method for each system.

This question tests your ability to solve systems of linear equations—finding the (x, y) pair that makes both equations true at the same time. A system of equations is like a puzzle where you need to find values that work for both equations simultaneously: the solution (x, y) must make the first equation true AND make the second equation true. Graphically, this is where the two lines intersect—that one point where both equations are satisfied! Plug in (3, 2): for x + y = 5, 3 + 2 = 5 true; for 2x - y = 4, 6 - 2 = 4 true. Choice A is correct because it confirms (3, 2) satisfies both equations perfectly. A distractor like Choice C might come from checking only one equation, but remember, it must work for both! Always check your answer by plugging both x and y into BOTH original equations. If you get true statements (like 5 = 5 and 7 = 7), you're correct! If even one equation doesn't work, there's an error somewhere. This check habit catches almost all mistakes and builds confidence! Here's how to choose a method: if one equation is already solved for a variable (like y = 3x + 1), use substitution—it's set up perfectly! If the coefficients of one variable are opposites (like 2x and -2x) or the same (like 3y and 3y), use elimination—one variable will cancel nicely. And you can always graph both lines to see where they cross! With practice, you'll spot the easiest method for each system.

This question tests your ability to solve systems of linear equations—finding the (x, y) pair that makes both equations true at the same time. A system of equations is like a puzzle where you need to find values that work for both equations simultaneously: the solution (x, y) must make the first equation true AND make the second equation true. To check if (3, 2) is a solution, we substitute x = 3 and y = 2 into both equations. For the first equation $x + y = 5$: we get $3 + 2 = 5$ ✓. For the second equation $2x - y = 4$: we get $2(3) - 2 = 6 - 2 = 4$ ✓. Choice A is correct because (3, 2) satisfies both equations—we get true statements ($5 = 5$ and $4 = 4$) for both! Choice C would be wrong because we just showed that (3, 2) satisfies both equations, not just the first one. Always check your answer by plugging both x and y into BOTH original equations. If you get true statements (like $5 = 5$ and $4 = 4$), you're correct! If even one equation doesn't work, there's an error somewhere. This check habit catches almost all mistakes and builds confidence!

This question tests your ability to solve systems of linear equations—finding the (x, y) pair that makes both equations true at the same time. The elimination method (also called addition method) works by adding or subtracting the equations to make one variable disappear: if you have 3x - y = 7 and x + y = 5, adding them gives 4x = 12 because the y terms cancel out. Then solve for x, and use that to find y! Let's add the equations: (3x - y) + (x + y) = 7 + 5, which gives us 4x = 12, so x = 3. Now substitute x = 3 into the second equation: 3 + y = 5, so y = 2. Therefore, the solution is (3, 2). Choice A is correct because when we check (3, 2) in both equations, we get 3(3) - 2 = 9 - 2 = 7 ✓ and 3 + 2 = 5 ✓, confirming both equations are satisfied. If you got (2, 3), you might have mixed up which value was x and which was y—remember that in ordered pairs, (x, y) means x comes first! Always check your answer by plugging both x and y into BOTH original equations. If you get true statements (like 7 = 7 and 5 = 5), you're correct! If even one equation doesn't work, there's an error somewhere. This check habit catches almost all mistakes and builds confidence!