This section explores using an algebraic approach to solving a system of linear equations.
Substitution
In the previous section, we focused on solving systems of equations by graphing. In addition to being time consuming, graphing can be an awkward method to determine the exact solution when the solution has large numbers, fractions, or decimals. There are two symbolic methods for solving systems of linear equations, and in this section we will use one of them: substitution.
“The Interview” generated roughly million in online sales and rentals during its first four days of availability, Sony Pictures said on Sunday. Sony did not say how much of that total represented digital rentals versus sales. The studio said there were about two million transactions overall.
A few days later, Joey Devilla cleverly pointed out in his blog, that there is enough information given to find the amount of sales versus rentals.
First, we will define variables. We need two variables, because there are two unknown quantities: how many sales there were and how many rentals there were. Let be the number of rental transactions and let be the number of sales transactions.
If you are unsure how to write an equation from the background information, use the units to help you. The units of each term in an equation must match because we can only add like quantities. Both and are in transactions. The article says that the total number of transactions is million. So our first equation will add the total number of rental and sales transactions and set that equal to million. Our equation is: Without the units:
The price of each rental was . That means the problem has given us a rate of to work with. The rate unit suggests this should be multiplied by something measured in transactions. It makes sense to multiply by , and then the number of dollars generated from rentals was . Similarly, the price of each sale was , so the revenue from sales was . The total revenue was million, which we can represent with this equation: Without the units:
Here is our system of equations:
To solve the system, we will use the substitution method. The idea is to use one equation to find an expression that is equal to but, cleverly, does not use the variable “.” Then, substitute this for into the other equation. This leaves you with one equation that only has one variable.
The first equation from the system is an easy one to solve for :
This tells us that the expression is equal to , so we can substitute it for in the second equation:
Now we have an equation with only one variable, , which we will solve for:
At this point, we know that . This tells us that out of the million transactions, roughly were from online sales. Recall that we solved the first equation for , and found . Now we can back-substitute our value for into this equation to find a value for .
To check our answer, we will see if and make the original equations true:
In summary, there were roughly copies sold and roughly copies rented.
Elimination
We just learned how to solve a system of linear equations using substitution above. Now, we will learn a second symbolic method for solving systems of linear equations.
Let be the dollar amount she gives to her older grandchild, and be the dollar amount she gives to her younger grandchild. (As always, we start solving a word problem like this by defining the variables, including their units.) Since the total she has to give is , we can say that . And since she wants to give more to the older grandchild, we can say that . So we have the system of equations:
We could solve this system by substitution as we learned previously but there is an easier method. If we add together the left sides from the two equations, it should equal the sum of the right sides:
So we have: Note that the variable is eliminated. This happened because the and the perfectly cancel each other out when they are added. With only one variable left, it doesn’t take much to finish:
To finish solving this system of equations, we need the value of . For now, an easy way to find is to substitute in our value of into one of the original equations:
To check our work, substitute and into the original equations:
This confirms that our solution is correct. In summary, Alicia should give to her older grandchild, and to her younger grandchild.
This method for solving the system of equations in the example above worked because and add to zero. Once the -terms were eliminated we were able to solve for . This method is called the elimination method. Some people call it the addition method, because we added the corresponding sides from the two equations to eliminate a variable.
If neither variable can be immediately eliminated, we can still use this method but it will require that we first adjust one or both of the equations. Let’s look at an example where we need to adjust one of the equations.
To eliminate the terms, we will multiply each side of the first equation by so that we will have . We can call this process scaling the first equation by .
We now have an equivalent system of equations where the -terms can be eliminated:
So we have:
To solve for , we can substitute for into either of the original equations or the new one. We use the first original equation, :
Our solution is and . We will check this in both of the original equations:
The solution to this system is and the solution set is .
As we mentioned in the previous section, not every system of equations has exactly one solution. Here’s an example where that doesn’t happen.
What are the solutions to our system in this case? Graphically, it’s clear that there are infinitely many solutions, since there are infinitely many intersection points between the graphs of the two functions. Be careful, however: this doesn’t mean that all ordered pairs are a solution to our system. Our task is to write down all solutions. To do this, we can solve for one variable in terms of the other: . Now, for each value of , we can find a value of such that is a solution to our system. Since is up to us to choose, we call the free variable. We write our solution in set notation as follows:
This is read as “all the ordered pairs of the form such that is a real number”. This sounds very intimidating, but this gives us a recipe to produce solutions to our system of equations. You take any -value you want, and your corresponding -value is given by . For example, if we want to find a solution whose -value is 0, we plug in 0 and find that . Therefore, is a solution to our system. If you want to find a solution whose -value is , we find that the corresponding -value is , and find that another solution is . The point is that gives us all the information we need to generate each solution.
- There are two methods for solving systems of equations: substitution and elimination. The choice of which method to use depends on the problem and your personal preference.
- If solving a system of equations yields an equation which is always true, your system has infinitely many solutions, and you can find an expression for all of them.
- If solving a system of equations yields an equation which is always false, your system has no solutions.