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Solving Systems of Linear Equations
When you are given an equation such as , you are being asked to find a value of x that will make that equation true. When you solve that equation, you get x = 17. That means 17 is THE value that makes that equation true.

Now consider the equation . We are now looking for x and y values that will make the equation true. But should we use x = 1 and y = 5 or instead use x = 3 and y = 3? It is also possible to have fractional and decimal values for x and y. In fact, there are infinitely many solutions for this equation.

So how can you possibly know which values are correct?

Usually when you are given an equation with two variables, you are given another equation that has those same two variables. This is referred to as a system of equations. Below is an example.

With a system of equations, the task is still to find values of x and y that will work, but now those values of x and y have to make BOTH equations true. You may be able to guess at values until you find two that work. In the case of the system shown above, x = 4 and y = 2.

However, not all systems can be solved by the guess and check method. In this lesson you will explore two methods commonly used for solving systems of equations.

Addition Method
The addition method allows you to add the equations given to you in a system. Let’s look at how this works using the system from above.



The addition method says we can just add everything on the left hand side and add everything on the right hand side and keep the equal sign in between.



Now it is possible to solve the new equation and get x = 4. Once you know one of the variables, substitute it into either equation to find the other variable, in this case y = 2.

The goal of the addition method is to eliminate one of the variables. In the problem we just worked, adding the equations as they were given to us eliminated the y-variable. This will not always be the case. Sometimes you will have to manipulate one or both equations to eliminate one of the variables.

Let's Practice:
  1. Solve
When working with the addition method, we have our choice of which variable we want to eliminate. Usually, you will find that one is easier to get rid of than the other. In this case, eliminating the x will be easier than eliminating the y. Why?

To eliminate the x, we will need to multiply the top equation by -2. In order to eliminate the y, we would have to multiply both equations. This strategy will be used in later examples, but is not necessary here.

Why does multiplying the top equation by -2 work? Because that makes the coefficient of x = -2 and when it is added to the coefficient of x in the bottom equation, the result is 0. In other words, the x-variable can now be eliminated.

It is critical that when you multiply the top equation by -2, you are careful to multiply EVERY term by -2.
Now simplify the new equation to make the addition easier.
Now add the two equations. Remember that the goal was to eliminate the x variable. If that does not happen, you should be concerned.
Now solve the equation that is left.
Substituting back into one of the original equations gives
So the solution to this system of equation is (1, 2)

  1. Solve
In this situation, there is single operation that will eliminate one of the variables. Solving this system will require multiplying both equations by a constant to eliminate a variable. You can still choose which variable you will eliminate. Some students find it easier in a situation like this to eliminate the y variable since those coefficients already have opposite signs. However, if you choose to eliminate the x variable, your answer will still come out the same. Let’s proceed with eliminating the y.

If we multiply the top equation by 3 and the bottom equation by 5, then in the top equation the coefficient of y will be 15 and in the bottom equation it will be -15. When added, these will give 0. Again, be careful to multiply EVERY term in the top equation by 3 and EVERY term in the bottom equation by 5.
Simplify each equation before adding.
Now add the two equations. Remember the goal was to eliminate the y. That should happen in this step.
Solving the remaining equation give x = 6. If we substitute that value into one of the original equations, we get y = 3.
So the solution to this system of equation is (6, 3)

Substitution Method
Another common method for solving systems of equations is the substitution method. This is when you have (or can get) one of the equations solved in terms of one of the variables. The easiest case of working with the substitution method is shown in the example below. Other examples will show other instances when more work may be required.

More Practice:
  1. Solve
Since the second equation is solved in terms of y (meaning that you have y on one side by itself) we can substitute that into the top equation.
This will produce an equation that only has one variable and can be solved for that variable.
Now that we know x = 2, we can substitute that back into one of the original equations (the bottom one is the easiest) and come up with y = 4.

So the solution is (2, 4).

  1. Solve
Since neither equation is solved in terms of one of the variables, to use the substitution method we will have to manipulate of the equations. The second equation will be the easiest to solve for either x or y because there are no coefficients to work with. You can choose either variable to solve for. Since we had an equation solved in terms of y in the last example, let’s solve this one in terms of x. Remember this means we want to have x on one side by itself.
Now use and substitute that expression into the top equations.
Now you can simplify the new equation and solve it for y.
Knowing that y = 5, substitute that value into one of the original equations and find that x = 1. So the solution is (1, 5)

There is another method that is used for solving systems of linear equations. It involves the use of matrices. If you would like to see how a system is solved using matrices, click here.

Examples
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S Taylor

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