6-3 Solving Systems By Elimination Quiz

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1. The admission fee at a small fair is $1.50 for children and $4.00 for adults. On a certain day, 2200 people enter the fair and $5050 is collected. How many adults attended the fair? (Just type in the number--no words!)

Let's assume that x represents the number of children attending the fair and y represents the number of adults attending the fair. We can set up a system of equations based on the given information. The first equation is x + y = 2200, representing the total number of people attending the fair. The second equation is 1.5x + 4y = 5050, representing the total amount of money collected. By solving this system of equations, we find that y = 700, which represents the number of adults attending the fair.

Explanation

Let's assume that x represents the number of children attending the fair and y represents the number of adults attending the fair. We can set up a system of equations based on the given information. The first equation is x + y = 2200, representing the total number of people attending the fair. The second equation is 1.5x + 4y = 5050, representing the total amount of money collected. By solving this system of equations, we find that y = 700, which represents the number of adults attending the fair.

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About This Quiz

6-3 Solving Systems By Elimination Quiz - Quiz

This quiz titled '6-3 Solving Systems by Elimination' tests the ability to solve linear equations using the elimination method. It includes problems that require finding specific coordinates and... see morethe number of attendees at an event, enhancing algebraic skills and problem-solving abilities. see less

2. Use elimination to solve the following system.2x - y = 322x + y = 60What is the y-coordinate of the solution?

To solve the system of equations using elimination, we can add the two equations together. This eliminates the variable "x" and leaves us with 2y = 82. Dividing both sides of the equation by 2, we find that y = 41. Therefore, the y-coordinate of the solution is 41.

Explanation

To solve the system of equations using elimination, we can add the two equations together. This eliminates the variable "x" and leaves us with 2y = 82. Dividing both sides of the equation by 2, we find that y = 41. Therefore, the y-coordinate of the solution is 41.

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3. Use elimination to solve the following system.3x - 5y = 84x - 7y = 10What is the x-coordinate of the solution?

To solve the given system of equations using elimination, we can multiply the first equation by 4 and the second equation by 3 to make the coefficients of x in both equations the same. This results in the equations 12x - 20y = 336 and 12x - 21y = 30. By subtracting the second equation from the first, we eliminate the x variable and are left with -y = 306. Dividing both sides by -1 gives us y = -306. Substituting this value of y into either of the original equations, we find that x = 6. Therefore, the x-coordinate of the solution is 6.

Explanation

To solve the given system of equations using elimination, we can multiply the first equation by 4 and the second equation by 3 to make the coefficients of x in both equations the same. This results in the equations 12x - 20y = 336 and 12x - 21y = 30. By subtracting the second equation from the first, we eliminate the x variable and are left with -y = 306. Dividing both sides by -1 gives us y = -306. Substituting this value of y into either of the original equations, we find that x = 6. Therefore, the x-coordinate of the solution is 6.

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4. Use elimination to solve the following system.-3x + 2y = 10-2x - y = -5What is the y-coordinate of the solution?

To solve the system of equations, we can use the method of elimination. By multiplying the second equation by 2, we can eliminate the x term when we add the two equations together. This results in -3x + 2y = 10 and -4x - 2y = -10. When we add these equations, the y term cancels out, leaving us with -7x = 0. Solving for x, we find that x = 0. Substituting this value back into one of the original equations, we can solve for y. Plugging in x = 0 into -3x + 2y = 10, we get 2y = 10, and therefore y = 5.

Explanation

To solve the system of equations, we can use the method of elimination. By multiplying the second equation by 2, we can eliminate the x term when we add the two equations together. This results in -3x + 2y = 10 and -4x - 2y = -10. When we add these equations, the y term cancels out, leaving us with -7x = 0. Solving for x, we find that x = 0. Substituting this value back into one of the original equations, we can solve for y. Plugging in x = 0 into -3x + 2y = 10, we get 2y = 10, and therefore y = 5.

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5. An exam worth 145 points contains 50 questions. Some of the questions are worth two points and some are worth five points. How many two point questions are on the test? (Just write the number--no words!)

Since the exam is worth a total of 145 points and there are 50 questions, we can assume that each question is worth either 2 or 5 points. Let's assume there are x number of 2-point questions on the test. So, the number of 5-point questions would be (50 - x).

Now, we can set up an equation to represent the total points on the test:
2x + 5(50 - x) = 145

Simplifying the equation, we get:
2x + 250 - 5x = 145
-3x = -105
x = 35

Therefore, there are 35 two-point questions on the test.

Explanation

Since the exam is worth a total of 145 points and there are 50 questions, we can assume that each question is worth either 2 or 5 points. Let's assume there are x number of 2-point questions on the test. So, the number of 5-point questions would be (50 - x).

Now, we can set up an equation to represent the total points on the test:
2x + 5(50 - x) = 145

Simplifying the equation, we get:
2x + 250 - 5x = 145
-3x = -105
x = 35

Therefore, there are 35 two-point questions on the test.

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6. Use elimination to solve the following system.2x + 5y = -32x + 2y = 6What is the x-coordinate of the solution?

-6

-3

3

6

To solve the system of equations using elimination, we can multiply the first equation by 2 and the second equation by -2, which will allow us to eliminate the x variable. The resulting equations are 4x + 10y = -64 and -4x - 4y = -12. Adding these two equations together will eliminate the x variable, leaving us with 6y = -76. Solving for y, we get y = -12. Substituting this value back into either of the original equations, we find that x = 6. Therefore, the x-coordinate of the solution is 6.

Explanation

To solve the system of equations using elimination, we can multiply the first equation by 2 and the second equation by -2, which will allow us to eliminate the x variable. The resulting equations are 4x + 10y = -64 and -4x - 4y = -12. Adding these two equations together will eliminate the x variable, leaving us with 6y = -76. Solving for y, we get y = -12. Substituting this value back into either of the original equations, we find that x = 6. Therefore, the x-coordinate of the solution is 6.

7. Use elimination to solve the following system.2x + 5y = 134x - 3y = -13

(-3, 1)

(-1, 3)

(1, 3)

No solution

By substituting the values of x and y into the given system of equations, we can determine which solution satisfies both equations. When we substitute x = -1 and y = 3 into the system, we get 2*(-1) + 5*(3) = 13 and 4*(-1) - 3*(3) = -13. Therefore, the solution (-1, 3) is the correct answer.

Explanation

By substituting the values of x and y into the given system of equations, we can determine which solution satisfies both equations. When we substitute x = -1 and y = 3 into the system, we get 2*(-1) + 5*(3) = 13 and 4*(-1) - 3*(3) = -13. Therefore, the solution (-1, 3) is the correct answer.