# Linear combination

A linear combination it is a series of methods to solve systems of linear equations of two or more unknowns.

## Methods for solving systems of linear equations

### Elimination or addition and subtraction method

To carry out this type of linear combination we must follow the following steps:

1. Choose an unknown quantity to eliminate.
2. Find the least common multiple (lcm) of the quotients that accompany the unknown. Having this value we will look for the way that these quotients become the value of the lcm, one positive and the other negative.
3. The equations are added vertically and one unknown is removed.
4. The second unknown is cleared. Its value is obtained.
5. The value of the second unknown is substituted in any of the equations, to obtain the value of the first unknown removed. And so finally the solution values to the system of equations are obtained.

Example: Solve the following system of equations:

5x + 6y = 20 (1)

3x + 8y = 34 (2)

Step 1: we will choose to eliminate the x's.

Step 2: The lcm of 5 and 3 is 15, since it is their product.

To convert coefficients of the x's from equation (1) and (2) to the lcm value, we must multiply equation (1) by 3 and equation (2) by - 5, so that this is the negative value:

(5x + 6y = 20) × 3 → 15x + 18y = 60 (3)

(3x + 8y = 34) × (- 5) → -15x - 40y = - 170 (4)

Step 3: We add equations (3) and (4)

0x - 22y = - 110

- 22y = - 110

Step 4: We clear and:

y = 110/22

y = 5

Step 5: We substitute the previous value in equation (1) and we solve for x:

5x + 6 (5) = 20

5x + 30 = 20

5x = 20 - 30

x = - 10/5

x = - 2

With this method we managed to find the solution values to any of the equations. If we substitute x = - 2 and y = 5 in any of the equations, we verify that they are a solution to the system of equations.

### Method of equalization

To carry out this type of linear combination we must follow the following steps:

1. Clear the same unknown from the equations.
2. We match the unknown unknown in each equation.
3. The second unknown is removed from equality.
4. The value of the second unknown is substituted in one of the equations solved in step 2 and with this we obtain the solution values to the system of equations.

Example: Solve the following system of equations:

3x + 2y = 3 (A)

- x + 5y = 16 (B)

Step 1: let's clear “and” in both equations:

3x + 2y = 3 → y = (3 - 3x) / 2 (C)

-x + 5y = 16 → y = (16 + x) / 5 (D)

Step 2: we equalize (C) and (D):

(3 - 3x) / 2 = (16 + x) / 5

Step 3: we clear x:

5 (3 - 3x) = 2 (16 + x)

15 - 15x = 32 + 2x

- 15x - 2x = 32 - 15

- 17x = 17

x = - 17/17

x = - 1

Step 4: we substitute the value of x in equation (B):

y = [16 + (-1)] / 5

y = 15/5

y = 3

With this method we managed to find the solution values to any of the equations. If we substitute x = - 1 and y = 3 in any of the equations, we verify that they are a solution to the system of equations.

### Substitution method

To carry out this type of linear combination we must follow the following steps:

1. We choose an unknown to be solved from one of the equations.
2. It is substituted in the other equation.
3. The second unknown is cleared and its value is obtained.
4. Finally, the value of the second unknown is substituted in one of the main equations, with this the value of the first unknown is obtained and thus the solution to the system of equations is obtained.

Example: Solve the following system of equations:

2x + 3y = 12 (α)

x - y = 1 (β)

Step 1: we will clear “x” from the equation (β):

x = 1 + y

Step 2: we substitute x in (α):

2 (1 + y) + 3y = 12

Step 3: we clear and:

2 + 2y + 3y = 12

5y = 12 - 2

y = 10/5

y = 2

Step 4: We substitute y = 2 in the equation (β):

x - (2) = 1

x = 1 + 2

x = 3

With this method we managed to find the solution values to any of the equations. If we substitute x = 3 and y = 2 in any of the equations, we verify that they are a solution to the system of equations.