Solving Systems of Linear Equations - Algebra 1

The sum of two numbers is 15 and their difference is 3. Let $x$ be the larger number and $y$ be the smaller number. What are $(x,y)$ ?
Answer: $(9,6)$ 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. For word problems: first define your variables clearly ('let x = the larger number, y = the smaller number'), then translate each sentence into an equation. The word 'and' usually signals you have two separate equations that form your system! From "The sum of two numbers is 15," we get x + y = 15. From "their difference is 3" with x being larger, we get x - y = 3. Using elimination by adding the equations: (x + y) + (x - y) = 15 + 3, which gives 2x = 18, so x = 9. Substituting back: 9 + y = 15, so y = 6. Choice A is correct because (9, 6) represents the larger number (9) and smaller number (6) that satisfy both conditions: 9 + 6 = 15 ✓ and 9 - 6 = 3 ✓. If you got (6, 9), you might have confused which variable represents which number—the problem specifically states x is the larger number! Always check your answer by plugging both x and y into BOTH original equations. If you get true statements (like 15 = 15 and 3 = 3), 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!
How many solutions does the system have? $\begin{cases}2x+4y=10\\x+2y=5\end{cases}$
Answer: Infinitely many solutions 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!
The sum of two numbers is 15 and their difference is 3. If $x$ is the larger number and $y$ is the smaller number, what are $(x, y)$ ? $\begin{cases} x + y = 15 \\ x - y = 3 \end{cases}$
Answer: $(9, 6)$ 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! Adding the two equations: $(x + y) + (x - y) = 15 + 3$ , which gives us $2x = 18$ , so $x = 9$ . Now substitute $x = 9$ into the first equation: $9 + y = 15$ , so $y = 6$ . Let's verify: $9 + 6 = 15$ ✓ and $9 - 6 = 3$ ✓. Choice A is correct because (9, 6) represents the larger number (9) and smaller number (6) that satisfy both conditions: their sum is 15 and their difference is 3. If you got (6, 9), you might have confused which variable represents which number—remember x is the larger number! 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!
Solve the system of equations: $\begin{cases} x+y=7 \\ x-y=1 \end{cases}$ What is the solution $(x,y)$ that satisfies both equations?
Answer: $(4,3)$ 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, add the equations: $(x + y) + (x - y) = 7 + 1$ , which simplifies to $2x = 8$ , so $x = 4$ ; then substitute into $x + y = 7$ to get $4 + y = 7$ , so $y = 3$ . Choice B is correct because it gives the (x, y) pair (4, 3) that satisfies both equations when you substitute back to check: $4 + 3 = 7$ and $4 - 3 = 1$ . A common mistake is subtracting instead of adding, which might lead to choice A (3, 4), but remember to choose the operation that eliminates a variable effectively. 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.
The graphs of the lines $y = -x + 6$ and $y = x + 2$ intersect at what point?
Answer: $(2, 4)$ 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! To find the intersection, set -x + 6 = x + 2, so -x - x = 2 - 6, -2x = -4, x = 2; then y = 2 + 2 = 4. Choice A is correct because it gives the (2, 4) pair that satisfies both equations when you substitute back to check: 4 = -2 + 6 and 4 = 2 + 2. A distractor like Choice B (4, 2) might come from swapping x and y, but it fails because 2 ≠ -4 + 6? Wait, 2 = -4 + 6 is true, but 2 ≠ 4 + 2, so it only works for one. 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. 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!
The sum of two numbers is $15$ and their difference is $3$ (larger minus smaller). What are the two numbers?
Answer: $9$ and $6$ 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. For word problems: first define your variables clearly ('let x = larger number, y = smaller number'), then translate each sentence into an equation. The word 'and' usually signals you have two separate equations that form your system! From the problem: x + y = 15 (sum) and x - y = 3 (difference). Using elimination by adding the equations: (x + y) + (x - y) = 15 + 3, which gives us 2x = 18, so x = 9. Substituting back: 9 + y = 15, so y = 6. The two numbers are 9 and 6. Choice A is correct because 9 + 6 = 15 ✓ and 9 - 6 = 3 ✓, satisfying both conditions. Choice B would give us 8 + 7 = 15 ✓ but 8 - 7 = 1, not 3, so it only satisfies one condition. Always check your answer by plugging both values back into the original context to ensure they make sense!
How many solutions does the system have? $\begin{cases}y=2x+3\\y=2x-1\end{cases}$
Answer: No solution 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 these equations, y = 2x + 3 and y = 2x - 1, notice that both lines have the same slope (2) but different y-intercepts (3 and -1). This means the lines are parallel—they never intersect! Since there's no point where both equations are true simultaneously, there is no solution. Choice A is correct because parallel lines never meet, so there's no (x, y) pair that satisfies both equations. If you thought there was one solution, you might have tried to solve by setting 2x + 3 = 2x - 1, but this gives 3 = -1, which is impossible! 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.
Is $(3,2)$ a solution to the system $\begin{cases}x+y=5\\2x-y=4\end{cases}$ ?
Answer: Yes, because it satisfies both equations. 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! To check if (3, 2) is a solution, we substitute x = 3 and y = 2 into both equations. First equation: 3 + 2 = 5 ✓. Second equation: 2(3) - 2 = 6 - 2 = 4 ✓. Both equations give true statements! Choice A is correct because (3, 2) satisfies both equations—when we substitute x = 3 and y = 2, we get 5 = 5 and 4 = 4, which are both true. If you thought it didn't work, you might have made a calculation error—double-check your arithmetic when substituting! 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!
Is $(3, 2)$ a solution to the system? $\begin{cases} x + y = 5 \\ 2x - y = 4 \end{cases}$
Answer: Yes, it satisfies both equations. 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.
Sarah is solving $\begin{cases} 3x + 2y = 16 \\ x - y = 2 \end{cases}$ and finds that $x = 4$ . She wants to check her work by substituting back into both original equations. If $x = 4$ is correct, what value of $y$ should make both equations true?
Answer: $y = 2$ When you're checking solutions to systems of equations, you need to substitute your values into both original equations to verify they work. This is a crucial step that catches calculation errors and confirms your solution is correct. Since Sarah found $x = 4$ , let's find the corresponding $y$ value. Using the second equation $x - y = 2$ , substitute $x = 4$ : $4 - y = 2$ . Solving for $y$ : $y = 4 - 2 = 2$ . Now verify this works in both equations. First equation: $3(4) + 2(2) = 12 + 4 = 16$ ✓. Second equation: $4 - 2 = 2$ ✓. Both equations are satisfied, confirming $y = 2$ . Looking at the wrong answers: Choice A gives $y = 1$ . Checking the first equation: $3(4) + 2(1) = 14 \neq 16$ , so this fails. Choice B gives $y = 4$ . In the second equation: $4 - 4 = 0 \neq 2$ , which doesn't work. Choice C gives $y = 3$ . In the first equation: $3(4) + 2(3) = 18 \neq 16$ , and in the second: $4 - 3 = 1 \neq 2$ , so this fails both equations. The answer is D) $y = 2$ . Study tip: Always check your solutions in both original equations, not just one. A value might satisfy one equation by coincidence but fail the other, indicating an error in your work. This verification step is essential for catching mistakes in systems problems.
Which ordered pair satisfies both equations in the system? $\begin{cases} 2x + y = 9 \\ x + y = 6 \end{cases}$
Answer: $(3, 3)$ 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! Subtract the second from the first: $(2x + y) - (x + y) = 9 - 6$ , $x = 3$ ; then into $x + y = 6$ : $3 + y = 6$ , $y = 3$ . Choice A is correct because it gives the $(3, 3)$ pair that satisfies both equations when you substitute back to check: $2(3) + 3 = 6 + 3 = 9$ and $3 + 3 = 6$ . A distractor like Choice B $(2, 4)$ might come from an arithmetic mistake, but it fails because $2(2) + 4 = 4 + 4 = 8 ≠ 9$ . 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.
Is $(3,2)$ a solution to the system $\begin{cases} x+y=5 \\ 2x-y=4 \end{cases}$ ?
Answer: Yes, it satisfies both equations. 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 x=3, y=2: $3 + 2 = 5$ (true) and $2(3) - 2 = 6 - 2 = 4$ (true). Choice A is correct because (3, 2) satisfies both equations perfectly. Miscalculating one equation might lead to choice C, but verify both to confirm. 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!
A system of two linear equations has no solution. If one equation is $2x + 5y = 10$ , which of the following could be the other equation?
Answer: $4x + 10y = 15$ When a system of linear equations has no solution, the equations represent parallel lines that never intersect. This happens when the equations have the same slope but different y-intercepts. To identify this situation, look for equations that are equivalent except for their constant terms. Let's examine the given equation $2x + 5y = 10$ and check each option. The key is to see if any choice can be rewritten to have the same left side as our given equation but a different right side. Choice D, $4x + 10y = 15$ , is correct because when you divide the entire equation by 2, you get $2x + 5y = 7.5$ . This has identical coefficients to our given equation ( $2x + 5y = 10$ ) but a different constant term. Since $7.5 \neq 10$ , these equations represent parallel lines with no intersection point. Choice A ( $4x + 10y = 20$ ) is wrong because dividing by 2 gives $2x + 5y = 10$ , which is identical to the given equation. This creates infinitely many solutions, not no solution. Choice B ( $6x + 15y = 30$ ) is wrong for the same reason as A. Dividing by 3 yields $2x + 5y = 10$ , creating the same line. Choice C ( $2x - 5y = 8$ ) is wrong because it has different coefficients for the y-term. These lines would intersect at exactly one point, giving the system a unique solution. Study tip: For no solution systems, look for equations that become identical when simplified, except for the constant term on the right side.
The system $\begin{cases} ax + 3y = 12 \\ 2x - y = 4 \end{cases}$ has the solution $x = 3$ , $y = 2$ . What is the value of $a$ ?
Answer: $a = 2$ Since $(3, 2)$ is the solution to the system, these values must satisfy both equations. First, let's verify they satisfy the second equation: $2(3) - 2 = 6 - 2 = 4$ ✓. Now substitute into the first equation: $a(3) + 3(2) = 12$ , which gives $3a + 6 = 12$ . Solving: $3a = 6$ , so $a = 2$ .
The solution to the system $\begin{cases} 3x + 2y = k \\ 6x + 4y = 18 \end{cases}$ depends on the value of $k$ . For what value of $k$ does this system have infinitely many solutions?
Answer: $k = 9$ For a system to have infinitely many solutions, the equations must be equivalent (one is a scalar multiple of the other). Notice that the second equation $6x + 4y = 18$ is exactly $2$ times the first equation $3x + 2y = k$ on the left side. For the equations to be equivalent, the right side must also be multiplied by the same factor: $2k = 18$ , so $k = 9$ . When $k = 9$ , both equations represent the same line: $3x + 2y = 9$ .
Solve the system: $\begin{cases} x+y=7 \\ x-y=1 \end{cases}$ What is the solution $(x,y)$ that satisfies both equations?
Answer: $(4,3)$ 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. Therefore, the solution is (4, 3). Choice A is correct because when we check (4, 3) in both equations, we get 4 + 3 = 7 ✓ and 4 - 3 = 1 ✓, confirming both equations are satisfied. If you picked (3, 4), you might have switched the x and y values—remember that in ordered pairs, x always 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 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!
The graphs of the lines $y=x+2$ and $y=-x+4$ intersect at one point. What is the intersection point (the solution to the system)?
Answer: $(1,3)$ 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! To find where y = x + 2 and y = -x + 4 intersect, we can set the equations equal: x + 2 = -x + 4. Adding x to both sides gives 2x + 2 = 4, so 2x = 2, and x = 1. Now substitute x = 1 into either equation to find y: y = 1 + 2 = 3. Therefore, the intersection point is (1, 3). Choice A is correct because when we check (1, 3) in both equations, we get y = 1 + 2 = 3 ✓ and y = -1 + 4 = 3 ✓, confirming that both lines pass through this point. If you picked (3, 1), you might have switched the x and y coordinates—remember that the intersection point is 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 = x + 2), 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.
Based on the graph shown, what is the solution to the system of linear equations?
Answer: $(1, -2)$ The solution to a system of linear equations graphically is the point where the two lines intersect. Looking at the graph, the two lines intersect at the point $(1, -2)$ . This means $x = 1$ and $y = -2$ satisfy both equations simultaneously.
Is $(3,2)$ a solution to the system $\begin{cases} x+y=5\\ 2x-y=4 \end{cases}$ ?
Answer: Yes, because it satisfies both equations. 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!
Solve the system of equations (you may use any method): $\begin{cases} 2x + y = 8\\ x - y = 1 \end{cases}$ What is the solution $(x, y)$ that satisfies both equations?
Answer: $(3, 2)$ 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! Here, we have 2x + y = 8 and x - y = 1. Adding these equations: (2x + y) + (x - y) = 8 + 1, which gives us 3x = 9, so x = 3. Substituting x = 3 into x - y = 1: 3 - y = 1, so y = 2. Choice B is correct because (3, 2) satisfies both equations when you substitute back to check: 2(3) + 2 = 8 ✓ and 3 - 2 = 1 ✓. Choice A would give 2(2) + 4 = 8 ✓ but 2 - 4 = -2 ≠ 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 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!

Start this quiz