Math equation - Mathematics Essay Example

In real life we sometimes face such times of tasks solution of which need the knowledge of solving systems of linear equations in three variables - Math equation introduction. First of all you should understand what is an equation in three variables. For example, you may have such an object: x + 3y = 6. This object we call “linear equation in two variables”. Here x, y are two variables. The word “equation” in this name means that we have equality between x + 3y (left side) and 6 (right side). The word “linear” means that the object defines a particular straight line on plane xy. The equation x + 3y + z = 6 is similar in form, and so it is a linear equation in three variables.  An equation in three variables is graphed in three-dimensional coordinate system. The graph of a linear equation in three variables is a plane, not a line.

 

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Generally, if A, B, C, and D are real numbers, with A, B, and C not all zero, then

 

Ax + By + Cz = D               (1)

 

is called a linear equation in three variables.

 

You may notice, that there are many (telling the truth, infinitely many) set of numbers x, y, z that satisfy linear equation in three variables. For example we can easily check that sets

 

x = 0, y = 1, z = 3
x = 3, y = 0, z = 3

x = 3, y = 1, z = 0

 

are solutions to the equation x + 3y + z = 6.

 

Hence, a solution to an equation in three variables is an ordered triple such as (0, 1, 3), where the first coordinate is the value of x, the second coordinate is the value of y, and the third coordinate is the value of z. The other two solutions we shall present in form (3, 0, 3) and (3, 1, 0).

 

When you have to solve a system of three linear independent equations (system of three linear independent equations consists of three equations of type (1)), you need to find such triples that satisfy all three equations simultaneously. In overall there are three possible cases:

 

1)      The system may have a single solution (we call it ordered triple);

2)      The system may have infinitely many solutions;

3)      The system may have no solutions.

 

Now I will teach you haw to solve the system with a single solution and I will show you by examples how we can find the solution for such system of equations. I want you to solve the equation using the addition method. I have three sets of equations that I want you to use.

In the addition method we eliminate a variable by adding the equations. So how it works? Let us take the first set of equations that is Example 1

 

x + 3y + z = 6             (1)

3x + y – z = -2            (2)

2x + 2y – z = 1           (3)

 

The addition property of equality allows you to add the same number to each side of an equation. You can also use the addition property of equality to add the two left sides and the two right sides. In this particular case it is the easiest to add the first and second equations and this operation allows you to eliminate z-term (as far as you add two z-terms one of which is with positive sign and another with negative). In the left side of the derived equation you need to add separately terms that include x, y and z

The addition property of equality allows you to add the same number to each side of an equation. You can also use the addition property of equality to add the two left sides and the two right sides. When adding you should use distributive property to combine all like terms involved in a sum. In the derived equation you notice that z-term is now absent.

 

x + 3y + z = 6                                     (1)

3x + y – z = -2                                    (2)

——————————-

4x + 4y = 4                                        Add

 

 

Now in derived equation divide each side by a common factor. In our case it is 4.

 

x + y = 1.

 

Now you need to write the derived equation instead of the first one in the set, and rewrite the second and third equation again. In this way you will receive a new system of equations equivalent to the initial one. It is equivalent in terms of the same solution, however it is simpler.

 

x + y = 1                     (1)

3x + y – z = -2            (2)

2x + 2y – z = 1.          (3)

 

Now you have to subtract he third equation from the second one and again in such a way you will eliminate z-term. After you subtracted left side of equation (3) from the left side of equation (2) and right side of equation (3) from the right side of equation (2) you should use distributive property to combine all like terms involved in a difference. As you notice z-term now disappeared again. For the convenience you write the derived equation instead of the second one and rewrite the first and third equations without changes.

 

x + y = 1                     (1)

x – y = -3                    (2)

2x + 2y – z = 1.          (3)

 

As the net step you need to add the first and second equation (as you did before) and write the received result instead of the first equation and then subtract the second equation from the first one and write received result instead of the second equation. You should leave the third equation without changes.

 

2x = -2                        (1)

2y = 4                         (2)

2x + 2y – z = 1.          (3)

 

Now in the first and second equations divide each side by 2 and rewrite third equation without changes.

 

x = -1                          (1)

y = 2                           (2)

2x + 2y – z = 1.          (3)

 

Now you can easily solve the third equation for z. Write down the result instead of the third equation, and leave the first and second equation without changes.

 

x = -1                          (1)

y = 2                           (2)

z = 2x + 2y – 1.          (3)

 

Now you need to substitute the result for x (x = -1) from the first equation and the result for y (y = 2) from the second equation for variables x and y in the third equation and get the solution of the system.

 

x = -1                          (1)

y = 2                           (2)

z = 1.                          (3)

 

Solution: (-1, 2, 1).

 

 

 

Now let us look at the second set of equations that is Example 2:

 

2y + z = 7       (1)

2x – z = 3        (2)

x – y = 3         (3)

 

This system of linear equations is much simpler then previous one. First of all you need to combine the first and second equations by addition. Then write the derived result instead of the first equation and leave the second and third equations without changes.

 

2x + 2y = 10   (1)

2x – z = 3        (2)

x – y = 3         (3)

 

Now you need to simplify equation (1) by dividing each side by 2, and permute the second and third equations.

 

x + y = 5         (1)

x – y = 3         (2)

2x – z = 3        (3)

 

Now you have to do the same transformations as you did with the first set of equations. More particular, you need to add the first and second equations and write the received result instead of the first equation and then subtract the second equation from the first one and write received result instead of the second equation. You should leave the third equation without changes.

 

2x = 8             (1)

2y = 2             (2)

2x – z = 3        (3)

 

Now simplify the first and the second equations by dividing each side by 2. Then write the derived results instead of the first and second equations. Then solve third equation for z. And now you may write down the result instead of the third equation.

 

x = 4               (1)

y = 1               (2)

z = 2x – 3        (3)

 

Now substitute the result for x (x = 4) from the first equation for variable x in the third equation and get the solution of the system.

 

x = 4               (1)

y = 1               (2)

z = 5                (3)

 

Solution: (4, 1, 5).

 

And finally let us consider the last set of equations and call it Example 3:

 

4x + 5y + z = 6           (1)

2x – y + 2z = 11         (2)

x + 2y + 2z = 6           (3)

 

To solve this system of equations you need as the first step to multiply each side of equation (1) by 2 and write down the derived result instead this equation. Then you have to multiply each side of equation (2) by -1 and again write down the derived result instead this equation. You should leave the equation (3) without changes.

 

8x + 10y + 2z = 12     (1)

-2x + y – 2z = -11       (2)

x + 2y + 2z = 6           (3)

 

Now you need to add equation (1) and equation (2) and write down the derived result instead of the equation (1). Here you will notice that by this operation you eliminated z-term the equation (1). Then you have to add equation (2) and equation (3) that will allow you to eliminate z-term. Similarly write down the derived result instead the equation (3). Combine all like terms in derived equations. So you may now notice that z-term is absent in two derived equations. Leave the equation (3) without changes again.

 

6x + 11y = 12             (1)

-x + 3y = -5                (2)

x + 2y + 2z = 6           (3)

 

 

Now you need to multiply each side of the second equation by -11 and write down the derived result instead of this equation. Then you have to multiply each side of the first equation by 3 and write down the derived result instead of this equation. Leave the equation (3) without changes.

 

18x + 33y = 3             (1)

11x – 33y = 55           (2)

x + 2y + 2z = 6           (3)

 

You need to add the equation (1) and the equation (2), combine x like terms and write down the derived result instead of the equation (1). This procedure allows you to eliminate y-term. Simplify the equation (2) by dividing each side of this equation by 11. Leave the equation (3) without changes.

 

29x = 58                     (1)

x – 3y = 5                   (2)

x + 2y + 2z = 6           (3)

 

Simplify the equation (1) by dividing each side of this equation by 29. Leave the equation (3) without changes.

 

x = 2                           (1)

x – 3y = 5                   (2)

x + 2y + 2z = 6           (3)

 

Solve the second equation for y. Then you have to write down the derived result instead this equation. Leave the equality (1) and equation (3) without changes.

 

 

x = 2                           (1)

y = (x – 5)/3                (2)

x + 2y + 2z = 6           (3)

 

Now substitute the result for x (x = 2) from the first equation for variable x in the second equation and get y. Solve the third equation for z. Write down derived result instead this equation.

 

x = 2                           (1)

y = -1                          (2)

z = (6 – x – 2y)/2        (3)

 

Substitute the result for x (x = 2) from the first equation and the result for y (y = -1) from the second equation for variables x and y in the third equation and get the solution of the system.

 

x = 2                           (1)

y = -1                          (2)

z = 3                           (3)

 

Solution: (2, -1, 3).

 

To sum up you can check all these solutions by substituting the derived values of variables into the equations and if the results will match it will mean that you solved these systems correctly. By such three simple examples you learned how to solve the system of equations in three variables using addition method. However, you have to know that the addition method is not the only possible way to solve such systems of equations. You can also use other methods like elimination, substitution or matrices methods and you will receive the same results.

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