7.2: Solving Linear Systems by Substitution
Learning Objectives
- Solve systems of equations with two variables by substituting for either variable.
- Manipulate standard form equations to isolate a single variable.
- Solve real-world problems using systems of equations.
- Solve mixture problems using systems of equations.
Introduction
In this lesson, we’ll learn to solve a system of two equations using the method of substitution.
Solving Linear Systems Using Substitution of Variable Expressions
Let’s look again at the problem about Peter and Nadia racing.
Peter and Nadia like to race each other. Peter can run at a speed of 5 feet per second and Nadia can run at a speed of 6 feet per second. To be a good sport, Nadia likes to give Peter a head start of 20 feet. How long does Nadia take to catch up with Peter? At what distance from the start does Nadia catch up with Peter?
In that example we came up with two equations:
Nadia’s equation:
Peter’s equation:
Each equation produced its own line on a graph, and to solve the system we found the point at which the lines intersected—the point where the values for and satisfied both relationships. When the values for and are equal, that means that Peter and Nadia are at the same place at the same time.
But there’s a faster way than graphing to solve this system of equations. Since we want the value of to be the same in both equations, we could just set the two right-hand sides of the equations equal to each other to solve for . That is, if and , and the two ’s are equal to each other, then by the transitive property we have . We can solve this for :
Even if the equations weren’t so obvious, we could use simple algebraic manipulation to find an expression for one variable in terms of the other. If we rearrange Peter’s equation to isolate :
We can now substitute this expression for into Nadia’s equation to solve:
So we find that Nadia and Peter meet 20 seconds after they start racing, at a distance of 120 feet away.
The method we just used is called the Substitution Method. In this lesson you’ll learn several techniques for isolating variables in a system of equations, and for using those expressions to solve systems of equations that describe situations like this one.
Example 1
Let’s look at an example where the equations are written in standard form.
Solve the system
Again, we start by looking to isolate one variable in either equation. If you look at the second equation, you should see that the coefficient of is 1. So the easiest way to start is to use this equation to solve for .
Solve the second equation for :
Substitute this expression into the first equation:
Substitute back into our expression for :
As you can see, we end up with the same solution that we found when we graphed these functions back in Lesson 7.1. So long as you are careful with the algebra, the substitution method can be a very efficient way to solve systems.
Next, let’s look at a more complicated example. Here, the values of and we end up with aren’t whole numbers, so they would be difficult to read off a graph!
Example 2
Solve the system
Again, we start by looking to isolate one variable in either equation. In this case it doesn’t matter which equation we use—all the variables look about equally easy to solve for.
So let’s solve the first equation for :
Substitute this expression into the second equation:
Substitute into the expression we got for :
So our solution is . You can see how the graphical solution might have been difficult to read accurately off a graph!
Solving Real-World Problems Using Linear Systems
Simultaneous equations can help us solve many real-world problems. We may be considering a purchase—for example, trying to decide whether it’s cheaper to buy an item online where you pay shipping or at the store where you do not. Or you may wish to join a CD music club, but aren’t sure if you would really save any money by buying a new CD every month in that way. Or you might be considering two different phone contracts. Let’s look at an example of that now.
Example 3
Anne is trying to choose between two phone plans. The first plan, with Vendafone, costs $20 per month, with calls costing an additional 25 cents per minute. The second company, Sellnet, charges $40 per month, but calls cost only 8 cents per minute. Which should she choose?
You should see that Anne’s choice will depend upon how many minutes of calls she expects to use each month. We start by writing two equations for the cost in dollars in terms of the minutes used. Since the number of minutes is the independent variable, it will be our . Cost is dependent on minutes – the cost per month is the dependent variable and will be assigned .
For Vendafone:
For Sellnet:
By writing the equations in slope-intercept form , you can sketch a graph to visualize the situation:
The line for Vendafone has an intercept of 20 and a slope of 0.25. The Sellnet line has an intercept of 40 and a slope of 0.08 (which is roughly a third of the Vendafone line’s slope). In order to help Anne decide which to choose, we’ll find where the two lines cross, by solving the two equations as a system.
Since equation 1 gives us an expression for , we can substitute this expression directly into equation 2:
So if Anne uses 117.65 minutes a month (although she can’t really do exactly that, because phone plans only count whole numbers of minutes), the phone plans will cost the same. Now we need to look at the graph to see which plan is better if she uses more minutes than that, and which plan is better if she uses fewer. You can see that the Vendafone plan costs more when she uses more minutes, and the Sellnet plan costs more with fewer minutes.
So, if Anne will use 117 minutes or less every month she should choose Vendafone. If she plans on using 118 or more minutes she should choose Sellnet.
Mixture Problems
Systems of equations crop up frequently in problems that deal with mixtures of two things—chemicals in a solution, nuts and raisins, or even the change in your pocket! Let’s look at some examples of these.
Example 4
Janine empties her purse and finds that it contains only nickels (worth 5 cents each) and dimes (worth 10 cents each). If she has a total of 7 coins and they have a combined value of 45 cents, how many of each coin does she have?
Since we have 2 types of coins, let’s call the number of nickels and the number of dimes . We are given two key pieces of information to make our equations: the number of coins and their value.
We can quickly rearrange the first equation to isolate :
Janine has 3 nickels and 4 dimes.
Sometimes a question asks you to determine (from concentrations) how much of a particular substance to use. The substance in question could be something like coins as above, or it could be a chemical in solution, or even heat. In such a case, you need to know the amount of whatever substance is in each part. There are several common situations where to get one equation you simply add two given quantities, but to get the second equation you need to use a product. Three examples are below.
Type of mixture | First equation | Second equation |
---|---|---|
Coins (items with $ value) | total number of items | total value (item value no. of items) |
Chemical solutions | total solution volume | amount of solute (vol concentration) |
Density of two substances | total amount or volume of mix | total mass (volume density) |
For example, when considering mixing chemical solutions, we will most likely need to consider the total amount of solute in the individual parts and in the final mixture. (A solute is the chemical that is dissolved in a solution. An example of a solute is salt when added to water to make a brine.) To find the total amount, simply multiply the amount of the mixture by the fractional concentration. To illustrate, let’s look at an example where you are given amounts relative to the whole.
Example 5
A chemist needs to prepare 500 ml of copper-sulfate solution with a 15% concentration. She wishes to use a high concentration solution (60%) and dilute it with a low concentration solution (5%) in order to do this. How much of each solution should she use?
Solution
To set this problem up, we first need to define our variables. Our unknowns are the amount of concentrated solution and the amount of dilute solution . We will also convert the percentages (60%, 15% and 5%) into decimals (0.6, 0.15 and 0.05). The two pieces of critical information are the final volume (500 ml) and the final amount of solute (15% of ). Our equations will look like this:
Volume equation:
Solute equation:
To isolate a variable for substitution, we can see it’s easier to start with equation 1:
So the chemist should mix 91 ml of the 60% solution with 409 ml of the 5% solution.
Further Practice
For lots more practice solving linear systems, check out this web page: http://www.algebra.com/algebra/homework/coordinate/practice-linear-system.epl
After clicking to see the solution to a problem, you can click the back button and then click Try Another Practice Linear System to see another problem.
Review Questions
- Solve the system:
- Solve the system:
- Solve the system:
- Solve the system:
- Solve the system:
- Solve the system:
- Solve the system:
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- How many pounds of each should I buy?
- If I suddenly realize I need to set aside $5 to buy chips, can I still buy 5 pounds of nuts with the remaining $10?
- What’s the greatest amount of nuts I can buy?
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- How many liters of each should be mixed to give the acid needed for the experiment?
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- If the whole test is worth 100 points and has 35 questions, how many of the questions are multiple-choice and how many are short-answer?
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- How could you have set up part b differently?