Algebra Quiz: Solving Systems Of Linear Equations

20 questions · exam conditions

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Solving Systems Of Linear EquationsQuestion 1 of 20

Solve using elimination: $\begin{cases} 2x + 3y = 12 \\ 2x - y = 4 \end{cases}$ What is the solution $(x, y)$ ?

(3, 2)

(2, 3)

(4, 1)

(1, 4)

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Algebra Quiz

Algebra Quiz: Solving Systems Of Linear Equations

Practice Solving Systems Of Linear Equations in Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Solving Systems Of Linear Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Solve using elimination: $\begin{cases} 2x + 3y = 12 \\ 2x - y = 4 \end{cases}$ What is the solution $(x, y)$ ?

$(3, 2)$ (correct answer)
$(2, 3)$
$(4, 1)$
$(1, 4)$

Explanation: 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 into

2x - y = 4

2x - 2 = 4

, so

2x = 6

and x = 3. Choice A is correct because it gives the

(3, 2)

pair 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 distractor like Choice B

(2, 3)

might come from mixing up addition and subtraction, but it fails because

2(2) + 3(3) = 4 + 9 = 13 ≠ 12

. 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!

Question 2

Solve using substitution: $\begin{cases} y = 2x + 1 \\ x + y = 7 \end{cases}$ Find the solution $(x, y)$ .

$(4, 3)$
$(2, 5)$ (correct answer)
$(3, 4)$
$(2, 3)$

Explanation: 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 substitution method works by solving one equation for one variable (like getting

y = 2x + 1

), then plugging that expression into the other equation everywhere you see that variable. This gives you one equation with one unknown that you can solve! Since

y = 2x + 1

is given, substitute into

x + y = 7

x + (2x + 1) = 7

, which simplifies to

3x + 1 = 7

, so

3x = 6

and

x = 2

; then

y = 2(2) + 1 = 5

. Choice A is correct because it gives the

(2, 5)

pair that satisfies both equations when you substitute back to check:

5 = 2(2) + 1

and

2 + 5 = 7

. A distractor like Choice B

(3, 4)

might result from an arithmetic error, such as

3x = 9

instead of 6, but it fails because

4 ≠ 2(3) + 1

. 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

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!

Question 3

Solve the system (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?

$(3, 2)$ (correct answer)
$(2, 2)$
$(2, 3)$
$(2, 4)$

Explanation: 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! Let's add these 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 B is correct because

(3, 2)

satisfies both equations when you substitute back to check:

2(3) + 2 = 8

✓ and

3 - 2 = 1

✓. If you picked

(2, 4)

, you might have made an arithmetic error when solving—always double-check your calculations! 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!

Question 4

Solve the system: $\begin{cases} x+2y=10\\ x-y=1 \end{cases}$ What is the solution $(x,y)$ ?

$(2,4)$
$(5,2)$
$(3,4)$
$(4,3)$ (correct answer)

Explanation: 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: notice that the second equation has

-y

while the first has

+2y

, so we can manipulate them to eliminate a variable. Let's multiply the second equation by 2:

2(x - y) = 2(1)

, which gives us

2x - 2y = 2

. Now we have

x + 2y = 10

and

2x - 2y = 2

. Adding these equations:

(x + 2y) + (2x - 2y) = 10 + 2

, which gives us

3x = 12

, so

x = 4

. Substituting

x = 4

into the second original equation:

4 - y = 1

, so

y = 3

. Choice A is correct because

(4, 3)

satisfies both equations when you substitute back to check:

4 + 2(3) = 4 + 6 = 10

✓ and

4 - 3 = 1

✓. Choice B would give us

3 + 2(4) = 3 + 8 = 11

, not 10, so it fails the first equation. 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.

Question 5

Which ordered pair satisfies both equations in the system? $$ \begin{cases} 2x + y = 9 \ x + y = 6 \end{cases}

$(3, 3)$ (correct answer)
$(2, 4)$
$(1, 5)$
$(6, 0)$

Explanation: 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.

Question 6

Solve the system: $\begin{cases} 3x-y=7 \\ x+y=5 \end{cases}$ What is the solution $(x, y)$ ?

$(3,2)$ (correct answer)
$(2,3)$
$(4,1)$
$(1,4)$

Explanation: 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.

Question 7

Is $(3, 2)$ a solution to the system?

\begin{cases} x + y = 5\\ 2x - y = 4 \end{cases} $$​

Yes, because it makes both equations true. (correct answer)
No, because it makes neither equation true.
No, because it makes only $x + y = 5$ true.
No, because it makes only $2x - y = 4$ true.

Explanation: 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: x + y = 5 becomes 3 + 2 = 5 ✓ True! Second equation: 2x - y = 4 becomes 2(3) - 2 = 6 - 2 = 4 ✓ True! Choice A is correct because (3, 2) makes both equations true. When you substitute these values, you get 5 = 5 and 4 = 4, which are both true statements. This is the hallmark of a solution to a system—it must satisfy ALL equations in the system, not just one. 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!

Question 8

Solve using the elimination method:

\begin{cases} 2x + 3y = 12 \\ 2x - y = 4 \end{cases}

What is the solution $(x, y)$ ?

$(2, 3)$
$(3, 2)$ (correct answer)
$(1, 2)$
$(2, 1)$

Explanation: 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!

Question 9

Solve the system: $\begin{cases} x+y=7 \\ x-y=1 \end{cases}$ What is the solution $(x,y)$ that satisfies both equations?

$(4,3)$ (correct answer)
$(3,4)$
$(7,1)$
$(4,-3)$

Explanation: 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!

Question 10

What is the intersection point of the lines $y=x+1$ and $y=-x+5$ ?

$(4,1)$
$(1,4)$
$(3,2)$
$(2,3)$ (correct answer)

Explanation: 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! Since both equations are already solved for y, we can set them equal: x + 1 = -x + 5. Adding x to both sides: 2x + 1 = 5, so 2x = 4, and x = 2. Now substitute x = 2 into either equation to find y: y = 2 + 1 = 3. Choice B is correct because (2, 3) is the intersection point where both lines meet: checking y = x + 1 gives 3 = 2 + 1 ✓ and checking y = -x + 5 gives 3 = -2 + 5 = 3 ✓. If you picked (3, 2), you might have switched the x and y values—remember that the intersection point is written as (x, y), not (y, x)! Always check your answer by plugging both x and y into BOTH original equations. If you get true statements (like 3 = 3 for both), 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!

Question 11

How many solutions does the system have? $$ \begin{cases} 2x+4y=10 \ x+2y=5 \end{cases}

Exactly two solutions
Exactly one solution
Infinitely many solutions (correct answer)
No solution

Explanation: 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!

Question 12

Solve using elimination: $\begin{cases} 2x+3y=12\\ 2x-y=4 \end{cases}$ What is the solution $(x, y)$ ?

$(3, 2)$ (correct answer)
$(3, -2)$
$(2, 3)$
$(4, 0)$

Explanation: This question tests your ability to solve systems of linear equations—finding the

(x, y)

2x

, so if we subtract the second from the first, the x terms will cancel out! Let's subtract:

(2x + 3y) - (2x - y) = 12 - 4

. This gives us

2x + 3y - 2x + y = 8

, which simplifies to

4y = 8

, so

y = 2

. Now substitute

y = 2

into either original equation; using the second:

2x - 2 = 4

, so

2x = 6

, and

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

✓. Choice B would give us

2(3) + 3(-2) = 6 - 6 = 0

, not 12, so it fails the first equation. Always check your answer by plugging both x and y into BOTH original equations. If you get true statements (like

12 = 12

and

4 = 4

), you're correct! If even one equation doesn't work, there's an error somewhere.

Question 13

Solve using elimination: $\begin{cases} 3x-y=11\\ x+y=5 \end{cases}$ Find the solution $(x, y)$ .

$(4,1)$ (correct answer)
$(1,4)$
$(3,2)$
$(5,0)$

3x - y = 11

and

x + y = 5

, adding them gives

4x = 16

because the y terms cancel out (

-y + y = 0

). Then solve for x, and use that to find y! Adding the equations:

(3x - y) + (x + y) = 11 + 5

, which gives us

4x = 16

, so

x = 4

. Now substitute

x = 4

into the second equation:

4 + y = 5

, so

y = 1

. Choice A is correct because

(4, 1)

satisfies both equations when you substitute back to check:

3(4) - 1 = 12 - 1 = 11

✓ and

4 + 1 = 5

✓. Choice B would give us

3(1) - 4 = 3 - 4 = -1

, not 11, so it fails the first equation. Always check your answer by plugging both x and y into BOTH original equations. If you get true statements (like

11 = 11

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!

Question 14

The graphs of $y=x+1$ and $y=-x+5$ intersect at what point?

$(2,3)$ (correct answer)
$(3,2)$
$(1,4)$
$(-2,3)$

Explanation: 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 the equations equal: x + 1 = -x + 5, so 2x = 4 and x = 2; then y = 2 + 1 = 3. Choice A is correct because it gives the (x, y) pair (2, 3) that satisfies both equations when you substitute back to check: 3 = 2 + 1 and 3 = -2 + 5. Mixing up the signs might lead to choice D (-2, 3), but carefully combine like terms to avoid that. 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!

Question 15

Solve the system: $\begin{cases} x+y=7 \\ x-y=1 \end{cases}$ What is the solution $(x, y)$ ?

$(3,4)$
$(4,-3)$
$(3,-4)$
$(4,3)$ (correct answer)

Explanation: 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!

Question 16

How many solutions does the system have? $$ \begin{cases}2x+4y=10\x+2y=5\end{cases}

Exactly one solution
Infinitely many solutions (correct answer)
Exactly two solutions
No solution

Explanation: 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!

Question 17

Is $(3, 2)$ a solution to the system?$$ \begin{cases} x + y = 5 \ 2x - y = 4 \end{cases}

Yes, it satisfies both equations. (correct answer)
No, it satisfies neither equation.
No, it satisfies only $x + y = 5$ .
No, it satisfies only $2x - y = 4$ .

Explanation: 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

✓ (True!). Second equation:

2(3) - 2 = 6 - 2 = 4

✓ (True!). Choice A is correct because (3, 2) satisfies both equations—when we substitute these values, we get true statements for both equations. If you got a different answer, make sure you're substituting correctly: replace

x

with 3 and

y

with 2 in each 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!

Question 18

Solve the system: $\begin{cases}2x+y=8\\x-y=1\end{cases}$ What is $(x,y)$ ?

$(1,6)$
$(3,2)$ (correct answer)
$(6,1)$
$(2,3)$

Explanation: 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.

Question 19

Solve using elimination: $\begin{cases} 2x+3y=12 \\ 2x-y=4 \end{cases}$ What is the solution $(x,y)$ ?

$(3,2)$ (correct answer)
$(1,3)$
$(2,3)$
$(2,2)$

Explanation: 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.

Question 20

What is the intersection point of the lines $y = x + 1$ and $y = -x + 5$ ? (This point is the solution to the system.)

$(2, 3)$ (correct answer)
$(3, 2)$
$(1, 4)$
$(4, 1)$

Explanation: 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! Since both equations are solved for y, we can set them equal: x + 1 = -x + 5. Adding x to both sides: 2x + 1 = 5. Subtracting 1: 2x = 4, so x = 2. Now substitute x = 2 into either equation to find y: y = 2 + 1 = 3. Choice A is correct because (2, 3) is where these lines intersect: checking y = x + 1 gives 3 = 2 + 1 ✓ and y = -x + 5 gives 3 = -2 + 5 = 3 ✓. Choice B would give y = 3 + 1 = 4 and y = -3 + 5 = 2, which are different values—this point can't be on both lines! 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!