Solutions to a System of Equations
A solution to a system of linear equations is a specific coordinate pair (x, y) that satisfies all equations in the system. It is important that the solution satisfies every equation in the system, not just one; otherwise it doesn't represent a solution to the entire system.
There are several ways to find solutions to a system of linear equations, such as by graphing or using the substitution method. This lesson focuses on one method known as the addition method or the elimination method.
The Addition Method
The main goal of the addition method is to add the equations that make up the system so that one of the variable terms will cancel, resulting in an equation with only one variable. From there, we can solve for that variable, and use that value to substitute it back into other equations in the system and eventually solve for the remaining variable(s). This is why this method is also referred to as the elimination method; because the goal is to eliminate one of the variable terms through equation addition.
Addition Method: also called the elimination method, a strategy to solving a system of equations by adding equations in order to cancel variable terms.
In order for a term to cancel when we add equations, we just have a variable term in one equation, and its opposite in another equation. For example, if we have a 3x in one equation and a –3x in another equation, we can add the two equations together, and there will be no x term. Let's take a look at a concrete example:
We noticed that we had opposite y-terms in our system. In one equation, we had a positive 3y, and in the other, we had a negative 3y. When adding the equations together, our x-terms summed to 7x, and our constant terms summed to 14, but what happened to the y-terms? They disappeared, because the sum of a quantity and its opposite is zero. This eliminated an entire variable from our equation, allowing us to easily solve for x.
This doesn't represent our full solution to the system, however. Now that we know that 2 is our x-coordinate, we need to find the associated y-coordinate. To do so, choose any equation in the system - it does not matter which one we choose. We'll substitute 2 in for x and solve for y. This is shown below:
The solution to the system is:
Use the addition method when you notice like-terms with opposite coefficients between two equations. When you add the equations, the variable term will disappear from the equation, because the opposite coefficients canceled each other out.
Multiplying by -1
Using the Addition Method is ideal when we recognize a variable term in one equation, and its opposite in another equation. Unfortunately, this isn't always the case. Consider this system:
Adding the two equations as they are would not cancel any variable term. Sure, we notice that we have a –4 coefficient in front of y in one equation, and +8 coefficient in front of y in the other, but –4 and +8 are not opposites; they do not sum to zero.
Look at the x-terms in each equation, however. Wouldn't it be nice if one of them were negative? That way, we would be able to add the two equations and eliminate the x-variable! Let's take one of the equations, and multiply the entire equation by –1:
When we multiplied by –1, every term throughout the entire equation changed sign: positive terms became negative, and negative terms became positive. As a result, we have two terms that are opposites of each other, and we can use the addition method to eliminate a variable term. This is shown below:
Now that we have a value for y, we can once again use any equation in our system to solve for x through back-substitution:
The solution to the system is:
Multiply an equation by –1 when you notice two identical terms between two equations. Multiplying by –1 will turn one of the identical terms into a like-term with opposite coefficients, allowing you to add the equations to cancel a variable term.
Multiplying by a Scalar Value
Let's once again consider the previous system of equations:
We already know from our previous example that the solution is (5, 0.75). We found this by multiplying one of our equations by –1, so that the x-terms would cancel during addition. There is, however, another way to solve this system. Earlier, we noted that –4y and 8y would not eliminate y, even though one term was positive and one term was negative. What would make these terms eliminate through addition? If we could turn –4y into –8y, or if we could turn –8y into –4y, we would be in business. We can do just this by multiplying one of the equations by a scalar value. Let's take our first equation, and multiply it by 2:
We can use this equivalent equation and add it to the other equation in our system to eliminate the y-terms altogether, allowing us to solve for x.
With a solution for x, we can plug 5 in for x in any of the other equations in the system and find our solution for y.
Once again, we have determined that the solution is (5, 0.75)
Multiply an equation by a scalar value when you notice that a coefficient of a variable term is a multiple of its like-term in another equation. You can even multiply by negative scalar values in order to create equations with opposite like-terms. Doing so will allow you to add the equations in order to eliminate a variable term.