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Algebra Help: Solving Systems Of Linear Equations

Review real example questions for Solving Systems Of Linear Equations in Algebra.

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

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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 $3y$ ), 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 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)
$(2, 2)$
$(2, 3)$
$(2, 4)$

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!