3 Systems of Linear Equations

Introduction to 3 systems of linear equations:

In mathematics, a system of linear equations (or linear system) is a collection of linear equations involving the same set of variables.  A solution to a linear system is an assignment of numbers to the variables such that all the equations are simultaneously satisfied.

In 3 systems of linear equations, there are 3 unknown variables. We have to find all 3 unknown variables. The example problems for 3 systems of linear equations are given below which helps you to learn solving system of 3 equations.

(Source: Wikipedia)

Example Problem of Solving 3 Systems of Linear Equations: 1

Solve the following system of 3 equations:

x + 2y +3 z = 1

x + 3y + 4z = 3

x + 4y + 6z = 5

Solution:

Step 1: Given equations

x + 2y + 3z = 1 .............. (1)

x + 3y + 4z = 3 ........... (2)

x + 4y + 6z = 5 ............(3)

Step 2: Subtract equation (2) from equation (1) to eliminate x

x  +  2y  +  3z  = 1

x  +  3y  +  4z  = 3     ( - )

---------------------------------------

0  -    y    -   z  = - 2

---------------------------------------

We get,

- y - z = - 2

Multiply the above equation by -1,

y + z = 2 .................. (4)

Step 3: Subtract equation (3) from equation (2) to eliminate x

x  +  3y  +  4z   =  3

x  +  4y  +  6z   =  5     ( - )

---------------------------------------

0  -   y    -  2z  =  - 2

---------------------------------------

We get,

- y - 2z = - 2

Multiply the above equation by -1,

y + 2z = 2 .................. (5)

Step 4: Subtract equation (4) from equation (5) to z value

y  +  2z  =  2

y  +    z  =  2     ( - )

---------------------------------------

0  +   z   =  0

---------------------------------------

Therefore,

z = 0

Step 5: Plug z = 0 in equation (4) to get y value

y + z = 2 .................. (4)

y + 0 = 2

y = 2

Step 6: Plug y = 2 and z = 0  in equation (1) to get x value

x + 2y + 3z = 1 .............. (1)

x + 2(2) + 3(0) = 1

x +  4 + 0 = 1

x = 1 - 4

x = - 3

Step 7: Solution

x = - 3, y = 2, z = 0

Please express your views of this topic Algebraic Equations by commenting on blog.

Example Problem of Solving 3 Systems of Linear Equations: 2

Solve the following system of 3 equations:

2x + 3y + z = 2

4x + 5y + z = 3

3x + 2y + z = 5

Solution:

Step 1: Given equations

2x + 3y + z = 2 .............. (1)

4x + 5y + z = 3 ........... (2)

3x + 2y + z = 5 ............(3)

Step 2: Subtract equation (2) from equation (1) to eliminate z

2x  +  3y  +  z  = 2

4x  +  5y  +  z  = 3     ( - )

---------------------------------------

- 2x  -  2y    +  0  = - 1

---------------------------------------

We get,

- 2x - 2y = - 1

Multiply the above equation by -1,

2x + 2y = 1 .................. (4)

Step 3: Subtract equation (3) from equation (2) to eliminate z

4x  +  5y  +  z   =  3

3x  +  2y  +  z   =  5     ( - )

---------------------------------------

x   +  3y  +  0  =  - 2

---------------------------------------

We get,

x + 3y = - 2 .................. (5)

Step 4: Multiply the equation (5) by 2 and subtract from equation (4) to get y value

2x  +  2y  =  1

2x  +  6y  = - 4

---------------------------------------

0   -  4y   =  5

---------------------------------------

Therefore,

y = - 1.25

Step 5: Plug y = - 1.25 in equation (4) to get x value

2x + 2y = 1 .................. (4)

2x + 2(-1.25) = 1

2x - 2.5 = 1

2x = 3.5

x = 1.75

Step 6: Plug x = 1.75 and y = - 1.25  in equation (1) to get z value

2x + 3y + z = 2 .............. (1)

2(1.75)x + 3(- 1.25) + z = 2

3.5 - 3.75 + z = 2

- 0.25 + z = 2

z = 2.25

Step 7: Solution

x = 1.75, y = - 1.25, z = 2.25