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Welcome to our Math lesson on Solving Systems of Linear Equations, this is the third lesson of our suite of math lessons covering the topic of Systems of Linear Equations. Methods for Solving Them., you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.
There are three methods for solving a system of linear equations, where two of which are numerical methods and the third one is graph - based. We will explain each of them in the following part of this tutorial.
This is the easiest method for solving linear equations if the proper conditions are met. It consists of the elimination of one of the variables through the addition of the two equations to each other. Creating the proper conditions means multiplying one or both equations by suitable coefficients, to create possible solutions to cancel the variable we choose to eliminate from the resulting equation. Then, we continue with the solution because the new equation obtained, now has only one variable. After calculating the value of this variable, we substitute it in any of the original equations to find the value of the other variable (the one that was eliminated before).
Let's clarify this point through a couple of examples.
Solve the following system of linear equations by the elimination method.
We have two choices for eliminating one of the variables:
Let's choose the first option. Thus, multiplying the first equation by - 3 yields
Adding the two equations means adding the left sides and the right sides separately. This eliminates the variable x because -3x + 3x = 0 and yields for the rest
Now, we can substitute this value found for y in any of the original equations, for example in the first one. Thus,
Thus, the number pair (5, 3) is a solution set for this system of linear equations.
Proof:
Sometimes, it is necessary to multiply both equations by different coefficients to eliminate any variable, as we will see in the next example.
Solve the following system of equations using the elimination method.
First, we must write each equation in the form ax + by = -c. Thus,
Now, let's multiply the first equation by 3 and the second equation by 4 to eliminate y. Thus,
Adding the two equations yields
Let's substitute this value in the first equation to find y (we substitute the value of x in the first equation as it has smaller numbers, but you can substitute it in the second equation as well if you wish). Thus,
Proof:
This is the second method for solving a system of linear equations. It consists of choosing an equation from the system for expressing one of the variables in terms of the other variable and then, substituting the corresponding expression in the other equation. In this way, we obtain a first - order equation with one variable.
For example, in the system of equations
we can write the variable y in terms of the other variable x in the first equation, i.e. y = 2x - 5 and substitute the variable y in the second equation with this expression obtained from the first. In this way, we obtain
As for the value of y, we write
Proof:
and
Solve the following system of linear equations by substitution method.
It is better to multiply the first equation by 3 first, to eliminate the fraction. Doing this, yields
From the first equation, we obtain y = 3x - 18. We substitute y with this expression into the second equation and solve it for x. Hence,
Now, we can replace the variable x with 10 in any of the equations of the system - for example in the first - to obtain
Proof:
and
Remark! It is not necessary to make the proof after every system you solve. It is recommended however that you make a quick mental proof to be sure that you have correctly solved the system.
This method consists of plotting the graphs of each linear equation in the system and checking the coordinates of the meeting point, as they represent the solution set of the system. This is because at the meeting point both equations have the same values for the variable x and also for the other variable y.
We discussed some linear equation graphs in the initial part of this tutorial. We explained that we need to have only two points for each equation given in order to plot the graph of linear equations. We will continue in this way in this part of the tutorial as well.
Let's take an example for illustration. In the system of linear equations
we have to plot the graphs of each equation in the same figure and see where they meet. But first, we have to find two points from each graph that allow us to plot them. We have to choose the values of the x-variable and find the corresponding values of the y-variable. Thus, in the first equation let's take for example x = 0 and x = 2. Substituting them in the first equation we obtain
and
Therefore, points A(0, 5) and B(2, 9) are points that belong to the first graph.
Then, we follow the same procedure for the second equation of the system as well. We can take x = -5 and x = 4 and find the corresponding y-values. Thus,
and
Therefore, points C( - 5, 0) and D(4, 3) are points that belong to the first graph.
Now, we plot both graphs in the same coordinate system as shown below.
As you see, the intercept point is at ( - 2, 1). This means the solution set of this system of equations is x = -2 and y = 1.
We can confirm this graphical solution through one of the other methods discussed earlier; for example through the elimination method. Thus, multiplying by 3 the first equation of the system
to eliminate y, yields
Adding the two equations, yields
Hence, substituting this value in the second equation (for convenience) yields
These results are identical to those obtained through the graphing method.
Solve the following system of linear equations through the graphing method.
We can create tables to have the values arranged better. Thus,
Hence, the graphs of the two equations are as shown below:
vFrom the figure, it is clear that point M is the intercept of the two lines. It has the coordinates M(2, 5). This means the solution set of our system of equations is x = 2 and y = 5. We can prove this solution by replacing the variables with these values in the system. Thus, for the first equation, we haveand for the second equation, we have
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