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Introduction

Elimination is a fundamental concept in mathematics used in solving systems of linear equations. It is a technique that involves manipulating a system of equations algebraically to eliminate one of the variables, leaving only one variable to be solved. This technique is useful in various areas of mathematics, including algebra, calculus, and linear algebra. In this article, we will delve deeper into the concept of elimination, explore its applications, provide examples, and answer frequently asked questions.

Definitions

To understand the concept of elimination, it is essential to define the terms involved.

System of Linear Equations: A system of linear equations is a set of two or more equations that involve linear expressions of variables.

Elimination: Elimination is a technique used to solve systems of linear equations by adding or subtracting the equations to eliminate one of the variables.

Solution: A solution is a set of values that satisfy all the equations in a system of linear equations.

Coefficients: Coefficients are numerical values that multiply variables in an equation.

Variable: A variable is an unknown quantity represented by a letter, usually x, y, or z.

Elimination is a powerful tool used in solving systems of linear equations, particularly when there are more than two equations. When we have two equations, the standard method of solving them is substitution. However, when the number of equations increases, substitution becomes cumbersome and may not always yield a solution.

Consider the following system of linear equations:

2x + 3y = 5 4x – 5y = 1

To solve this system of equations using elimination, we must eliminate one of the variables, either x or y, by adding or subtracting the equations. In this case, we can eliminate x by multiplying the first equation by 2 and the second equation by -1, as shown below:

4x + 6y = 10 -4x + 5y = -1

When we add the two equations, we get:

11y = 9

Therefore, y = 9/11. We can then substitute this value of y into one of the equations to find the value of x, as shown below:

2x + 3(9/11) = 5 2x = (55/11) – (27/11) x = (28/11)

Therefore, the solution to the system of linear equations is (x, y) = (28/11, 9/11).

Another example of using elimination to solve a system of linear equations is shown below:

3x + 2y – z = 8 2x – 4y + 2z = -4 -x + 0.5y – z = 0

To eliminate z, we can add the first and second equations to get:

5x – 2y = 4

We can then add the second and third equations to eliminate z again, as shown below:

5x – 3.5y = 4

Finally, we can multiply the third equation by 2 and add it to the second equation to eliminate y, as shown below:

5x – 2z = 4 2x + z = 0 7x = 4

Therefore, x = 4/7. We can then substitute this value of x into one of the equations to find the values of y and z, as shown below:

3(4/7) + 2y – z = 8 2(4/7) – 4y + 2z = -4 -(4/7) + 0.5y – z = 0

Simplifying the equations yields:

y = 4/7 z = -10/7

Therefore, the solution to the system of linear equations is (x, y, z) = (4/7, 4/7, -10/7).

Examples

Here are ten more examples of using elimination to solve systems of linear equations:

Example 1:

3x + 4y = 5 5x – 2y = 3

To eliminate y, we can multiply the first equation by 2 and the second equation by 4, as shown below:

6x + 8y = 10 20x – 8y = 12

Adding the two equations yields:

26x = 22

Therefore, x = 22/26 = 11/13. Substituting this value of x into one of the equations gives us:

3(11/13) + 4y = 5

Solving for y gives us:

y = 1/13

Therefore, the solution to the system of linear equations is (x, y) = (11/13, 1/13).

Example 2:

x – y + 2z = 8 2x + y – 3z = -1 3x + 2y + z = 18

To eliminate y, we can multiply the first equation by 2 and subtract the second equation from it, as shown below:

0x – 3y + 7z = 17

We can then add this equation to the third equation to eliminate y and z, as shown below:

4x + 5y = 34

To eliminate y again, we can multiply the first equation by 5 and subtract the third equation from it, as shown below:

-4x – 3y = -22

Adding this equation to the equation we obtained earlier yields:

-2x = 12

Therefore, x = -6. Substituting this value of x into one of the equations gives us:

-6 – y + 2z = 8

Simplifying the equation yields:

y – 2z = -14

We can then substitute the value of x into the equation we obtained earlier to get:

-20 – 3y + 7z = 17

Simplifying the equation yields:

3y – 7z = -37

Solving these two equations simultaneously gives us:

y = -5 z = 3

Therefore, the solution to the system of linear equations is (x, y, z) = (-6, -5, 3).

Example 3:

x + y + z = 1 2x + 2y + 3z = 2 3x + 5y + 7z = 3

To eliminate x, we can subtract the first equation from the second equation and the first equation from the third equation, as shown below:

x + y + z = 1 0x + 0y + 2z = 0 2x + 4y + 6z = 2

We can then divide the second equation by 2 and subtract it from the third equation to eliminate z, as shown below:

x + y + z = 1 0x + 0y + 2z = 0 2x + 4y + 0z = 2

We can then divide the third equation by 2 and subtract it from the first equation to eliminate x, as shown below:

x + y + 0z = -1 0x + 0y + 2z = 0 0x + 2y + 0z = -1

Solving these equations simultaneously gives x = -1, y = -1/2, z = 0. Therefore, the solution to the system of linear equations is (x, y, z) = (-1, -1/2, 0).

Example 4:

x – 2y + z = 3 2x – 3y + 2z = 7 3x – 4y + 3z = 10

To eliminate x, we can subtract the first equation from the second equation and the first equation from the third equation, as shown below:

x – 2y + z = 3 0x + y + z = 1 2x – 2y + 2z = 7

We can then subtract twice the second equation from the third equation to eliminate y, as shown below:

x – 2y + z = 3 0x + y + z = 1 0x – 2y + 0z = 5

Solving the last equation gives us y = -5/2. Substituting this value of y into the second equation gives us:

z = 3/2

Substituting the values of y and z into the first equation gives us:

x = 5/2

Therefore, the solution to the system of linear equations is (x, y, z) = (5/2, -5/2, 3/2).

Example 5:

x + y + z = 1 2x + 3y + 4z = 5 4x + 5y + 7z = 10

To eliminate x, we can subtract twice the first equation from the second equation and four times the first equation from the third equation, as shown below:

x + y + z = 1 y + 2z = 3 y + 3z = 6

We can then subtract the second equation from the third equation to eliminate y, as shown below:

x + y + z = 1 y + 2z = 3 0y + z = 3

Solving the last equation gives us z = 3. Substituting this value of z into the second equation gives us:

y = -3

Substituting the values of y and z into the first equation gives us:

x = 7

Therefore, the solution to the system of linear equations is (x, y, z) = (7, -3, 3).

Example 6:

x – 2y + z = 1 2x – 4y + 2z = 2 3x – 6y + 3z = 3

To eliminate x, we can subtract twice the first equation from the second equation and three times the first equation from the third equation, as shown below:

x – 2y + z = 1 0x + 0y + 0z = 0 0x – 0y + 0z = 0

Since the second and third equations are both equal to 0, they do not provide any additional information. Therefore, we can solve the first equation for x:

x = 2y – z + 1

Substituting this expression for x into the second equation gives us:

2(2y – z + 1) – 4y + 2z = 2

Simplifying the equation yields:

-2y + 2z = -2

Dividing both sides by -2 gives us:

y – z = 1

Substituting the expression for x into the third equation gives us:

3(2y -z + 1) – 6y + 3z = 3

Simplifying the equation yields:

-3y + 2z = 0

Dividing both sides by -3 gives us:

y – (2/3)z = 0

Solving the system of linear equations y – z = 1 and y – (2/3)z = 0 yields:

y = (2/3), z = (1/3)

Substituting the values of y and z into the expression for x gives us:

x = 2(2/3) – (1/3) + 1 = (7/3)

Therefore, the solution to the system of linear equations is (x, y, z) = (7/3, 2/3, 1/3).

Example 7:

x + 2y + 3z = 1 3x + 4y + 5z = 2 5x + 6y + 7z = 3

To eliminate x, we can subtract three times the first equation from the second equation and five times the first equation from the third equation, as shown below:

x + 2y + 3z = 1 -3x – 2y – z = -1 0x – 4y – 8z = -2

We can then divide the third equation by -4 to simplify it:

0x + y + 2z = (1/2)

We can now use the second equation to eliminate y. Adding two times the second equation to the first equation yields:

7x + 13z = 5

Substituting the expression for y in terms of z from the third equation into the first equation and solving for x yields:

x = 2 – 2z

Substituting this expression for x into the equation 7x + 13z = 5 yields:

-7z = -9

Solving for z yields:

z = (9/7)

Substituting this value of z into the expression for x yields:

x = 2 – 2(9/7) = (5/7)

Substituting the values of x and z into the equation y + 2z = (1/2) and solving for y yields:

y = -(5/7)

Therefore, the solution to the system of linear equations is (x, y, z) = (5/7, -(5/7), 9/7).

Example 8:

x + y + z = 6 2x + 3y + 4z = 14 4x + 5y + 6z = 21

To eliminate x, we can subtract twice the first equation from the second equation and four times the first equation from the third equation, as shown below:

x + y + z = 6 y + 2z = 2 y + 2z = 3

The third equation contradicts the second equation, which means that the system of linear equations has no solution.

Example 9:

x – 2y + z = 4 3x – 4y – z = -3 2x – 3y + 2z = 5

To eliminate x, we can subtract two times the first equation from the third equation and add three times the first equation to the second equation, as shown below:

x – 2y + z = 4 9x – 14y + 2z = 9 0x – y + 4z = 15

We can then solve the third equation for y: y = -15 + 4z

Substituting this expression for y into the second equation and solving for x yields:

x = (3/2) + (7/2)z

Substituting these expressions for x and y into the first equation and solving for z yields:

z = (1/3)

Substituting this value of z into the expression for x yields:

x = (4/3)

Substituting this value of z into the expression for y yields:

y = -(47/3)

Therefore, the solution to the system of linear equations is (x, y, z) = (4/3, -(47/3), 1/3).

Example 10:

2x + 3y – z = 1 3x + 5y – 2z = 3 4x + 7y – 3z = 5

To eliminate x, we can subtract two times the first equation from the second equation and three times the first equation from the third equation, as shown below:

2x + 3y – z = 1 -y + 4z = 1 y – 3z = 1

We can then add the second and third equations to eliminate y:

-y + 4z = 1 y – 3z = 1 0y + z = 2

Substituting this value of z into the expression for y yields:

y = -5

Substituting the values of y and z into the first equation and solving for x yields:

x = 2

Therefore, the solution to the system of linear equations is (x, y, z) = (2, -5, 2).

FAQs:

Q: What is the main idea behind elimination in math? A: The main idea behind elimination in math is to use algebraic manipulations to eliminate one variable from a system of linear equations, thereby reducing the number of variables in the system. This makes it easier to solve for the remaining variables using techniques such as substitution or back-substitution.

Q: What are some common strategies for eliminating variables in a system of linear equations? A: Some common strategies for eliminating variables in a system of linear equations include adding or subtracting equations, multiplying or dividing equations by constants, and rearranging equations to create cancellation terms.

Q: Can elimination always be used to solve a system of linear equations? A: No, elimination cannot always be used to solve a system of linear equations. In some cases, elimination may not be possible or may lead to an inconsistent or indeterminate system.

Q: What is an inconsistent system of linear equations? A: An inconsistent system of linear equations is a system in which there are no solutions that satisfy all of the equations simultaneously. This can happen when the equations are contradictory or when they do not have enough information to determine a unique solution.

Q: What is an indeterminate system of linear equations? A: An indeterminate system of linear equations is a system in which there are infinitely many solutions that satisfy all of the equations simultaneously. This can happen when the equations are redundant or when they only provide constraints on some of the variables.

Quiz:

1. What is elimination in math? a) The process of finding all possible solutions to a system of linear equations. b) The process of reducing a system of linear equations to a single equation. c) The process of using algebraic manipulations to eliminate one variable from a system of linear equations.
2. Which of the following is a common strategy for eliminating variables in a system of linear equations? a) Adding or subtracting equations b) Multiplying or dividing equations by constants c) Rearranging equations to create cancellation terms d) d) All of the above
3. What is the main advantage of using elimination to solve a system of linear equations? a) It always leads to a unique solution. b) It is always faster than other methods. c) It reduces the number of variables in the system, making it easier to solve for the remaining variables.
4. Which of the following systems of linear equations is inconsistent? a) 2x + 3y = 7 4x + 6y = 14 b) 2x + 3y = 7 4x + 6y = 8 c) 2x + 3y = 7 4x + 6y = 21
5. Which of the following systems of linear equations is indeterminate? a) 2x + 3y = 7 4x + 6y = 14 b) 2x + 3y = 7 4x + 6y = 8 c) 2x + 3y = 7 4x + 6y = 28
6. What is the first step in using elimination to solve a system of linear equations? a) Solve one of the equations for a variable. b) Multiply one or more equations by constants. c) Rearrange the equations to create cancellation terms.
7. What is the second step in using elimination to solve a system of linear equations? a) Add or subtract equations to eliminate a variable. b) Multiply or divide equations to eliminate a variable. c) Rearrange the equations again to create cancellation terms.
8. What is the third step in using elimination to solve a system of linear equations? a) Solve for one of the variables using one of the equations. b) Solve for one of the variables using the other equation. c) Solve for one of the variables using a combination of the equations.
9. What is the fourth and final step in using elimination to solve a system of linear equations? a) Check your solution by substituting it into each of the original equations. b) Check your solution by graphing the equations and verifying that they intersect at the correct point. c) Check your solution by using a different method to solve the system and verifying that you get the same answer.
10. Which of the following systems of linear equations can be solved using elimination? a) x + y = 3 x – y = 1 b) x + y = 4 2x – y = 5 c) 2x + 3y = 6 4x + 6y = 9

Answers:

1. c
2. d
3. c
4. b
5. c
6. a
7. a
8. c
9. a
10. b and c