What is x if 2x 6 24

Solve systems of linear equations

In this section we are concerned with solving systems of linear equations with more than one variable. First of all, we will show you what it is all about and how to successfully master these tasks.

Before we start solving systems of equations here, one thing must be made clear: If you still have problems solving equations with an unknown (e.g. 5x + 2 = 3), then you should definitely read our chapter on solving Find equations and read this. Everyone else can get started with linear systems of equations right away and ignore the following link.

Systems of equations video:
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Systems of equations with 2 variables: Introduction

First of all, you should know what is meant by a system of equations with two variables. First of all, a small example: You go shopping and you know that 6 apples and 12 pears of particularly good quality cost 30 euros. And you know that 3 apples and 3 pears cost 9 euros. The question now is: what does an apple or a pear cost? Since the terms apples and pears are too long, we substitute "x" for the price of an apple and "y" for the price of a pear. This results in the following equations (compare these with the information in the text!):

Table scrollable to the right
6Applesand12Pearscosts30 euro
3Applesand3Pearscosts9 euros

Of course, that doesn't look very clear yet. For this reason, the following notation has been introduced in mathematics to provide a better overview:

Table scrollable to the right
| 6x + 12y =30 | Equation No. 1
| 3x + 3y= 9 |Equation No. 2

Such a system of equations indicates: These equations belong to one another. This is also the reason why you have to solve them together. The goal is to get a number for x and y that satisfies both equations. And we'll take care of that now.


System of equations with 2 variables: insertion method

I would like to introduce two methods here to solve such a linear system of equations: the insertion method and the Gaussian elimination method. I would like to start with the substitution procedure, which still works well with 2 variables in 2 equations (with more variables this procedure is too time-consuming). The substitution method involves solving one of the equations for a variable and inserting it into the other equation.

Let's do this using our example just now. The first equation should be solved for x:

Table scrollable to the right
6x + 12y = 30| -12y
6x = 30-12y| :6
x = 5-2y

At the end we have now received an equation that is solved for x. We now insert what is to the right of the "=" in the second equation of our system of equations from above, whereby what we insert is always put in brackets! It will look like that:

Table scrollable to the right
3x + 3y = 9| Deploy
3 (5 - 2y) + 3y = 9| Multiply out
15 - 6y + 3y = 9| Sum up
15 - 3y = 9| - 15
-3y = -6| ·(-1)
3y = 6| : 3
y = 2

We have now calculated a solution for y. This solution is now inserted into one of the equations that contains both x and y.

Table scrollable to the right
x = 5-2y| Insert y = 2
x = 5 - 2 (2) | Multiply out
x = 5 - 4| Sum up
x = 1

This gives us the solutions x = 1 and y = 2 for our equations. Thus, an apple costs 1 euro and a pear 2 euros. If we have not made a calculation error, both equations must now be correct. We're just doing a rehearsal here.

Table scrollable to the right
6x + 12 y = 30| Insert x = 1 and y = 2
6·(1) + 12·(2) = 30 | Multiply out
6 + 24 = 30| Add
30 = 30 | Right!
Table scrollable to the right
3x + 3y = 9| Insert x = 1 and y = 2
3·(1) + 3·(2) = 9| Multiply out
3 + 6 = 9| Add
9 = 9| Right !

The equations add up at the end: 30 = 30 and 9 =. So the solution is right. For example, if we got 20 = 30, you'd have to Check the complete invoice again from the beginning! So it is worthwhile not to make any miscalculations.

System of equations with 2 unknowns: Gauss method

We have just learned about the installation process. Now I'll show you another way to solve these equations. It is called the Gauss elimination method and was named after the mathematician Carl Friedrich Gauß. As the name suggests, variables are eliminated here. We want to do this using the same equations that we gave above. In memory of:

Table scrollable to the right
| 6x + 12y =30 | Equation No. 1
| 3x + 3y= 9 |Equation No. 2

The goal is now to get rid of either x or y, i.e. to eliminate them. We have a 6x in the first equation and a 3x in the second equation. If we also had a 6x in the second equation at the beginning, we could subtract the two so that the variable x drops out. It looks like this:

Table scrollable to the right
3x + 3y = 9| ·2
6x + 6y = 18

Our equation system now looks like this:

Table scrollable to the right
| 6x + 12y =30 | Equation No. 1
| 6x + 6y= 18 |Equation No. 2

We now subtract these two equations from each other:

Table scrollable to the right
| 6x + 12 y = 30 |Equation No. 1
- | 6x + 6y = 18 |Equation No. 2
6y = 12

If you divide this last equation by 6, you get y = 2. This y = 2 is simply inserted into one of the two starting equations, as in the substitution procedure, and you get x = 1.

Which method you use to solve a linear system of equations is ultimately up to you (unless the teacher tells you to). The tastes diverge there. However, especially with systems of equations with more than two variables, the substitution procedure can be very time-consuming. For this reason my tip: Always use the Gaussian method from 3 variables.

Linear systems of equations with 3 unknowns: Gauss method

The Gaussian method works for 2 unknowns, why shouldn't it also work for 3 unknowns? And that is exactly what it does. In the following we show you an example for solving a linear system of equations with 3 variables. Unfortunately it becomes more confusing as the number of unknowns increases. So try to follow the example carefully:

Table scrollable to the right
| -x + y + z = 0 |1st equation
| x - 3y -2z = 5 |2nd equation
| 5x + y + 4z = 3 |3rd equation

In the first equation we have -x and in the second we have + x. If we add the two together, the x is thrown out. We'll do that right away:

Table scrollable to the right
| -x + y + z = 0 |
| x - 3y -2z = 5 |Add
-2y - z = 5

Now we've turned the first two equations into an equation with two unknowns. Stupidly, the 3rd equation also has an "x" in it. This must now also be eliminated. For this we take the 3rd equation and one of the other two starting equations. I now take the 1st equation and multiply it by 5. This results in: -5x + 5y + 5z = 0. This transformed 1st equation is added to the 3rd equation.

Table scrollable to the right
| -5x + 5y + 5z = 0 |1st equation
| 5x + y + 4z = 3 |3rd equation
6y + 9z = 3 Adding the equations

We have now "created" two equations that have only two unknowns. These two equations are now:

Table scrollable to the right
| -2y -z = 5 |First new equation
| 6y + 9z = 3 |Second new equation

Now we have a system of equations with 2 unknowns and 2 equations. Now the same game starts as we have already discussed in the previous sections. In order to eliminate the y here, the second new equation is divided by 3. This yields: 2y + 3z = 1. Now you can add again:

Table scrollable to the right
-2y - z = 51. new equation
2y + 3z = 12. New equation, is now added
2z = 6| : 2
z = 3

We get z = 3. We plug this into the equation -2y - z = 5 and get y = -4. If we now insert this into the starting equation -x + y + z = 0, the result is x = -1.

Tips for solving systems of equations

Here are a few more tips and comments:

  • First practice solving systems of equations with 2 unknowns before you take more.
  • It is quite natural that you will have some problems in the beginning and do not see the errors at first. You then either have to search thoroughly again or do the math all over again.
  • Try to make your life as easy as possible and first take a look at the system in order to be the first to eliminate a variable that can be eliminated as easily as possible.
  • Solve our exercises on the next page to get security.


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