- Solve a linear system of equations using elimination by addition.
- Solve a linear system of equations using elimination by subtraction.
- Solve a linear system of equations by multiplication and then addition or subtraction.
- Compare methods for solving linear systems.
- Solve real-world problems using linear systems by any method.
In this lesson, we’ll see how to use simple addition and subtraction to simplify our system of equations to a single equation involving a single variable. Because we go from two unknowns ( and ) to a single unknown (either or ), this method is often referred to by solving by elimination. We eliminate one variable in order to make our equations solvable! To illustrate this idea, let’s look at the simple example of buying apples and bananas.
If one apple plus one banana costs $1.25 and one apple plus 2 bananas costs $2.00, how much does one banana cost? One apple?
It shouldn’t take too long to discover that each banana costs $0.75. After all, the second purchase just contains 1 more banana than the first, and costs $0.75 more, so that one banana must cost $0.75.
Here’s what we get when we describe this situation with algebra:
Now we can subtract the number of apples and bananas in the first equation from the number in the second equation, and also subtract the cost in the first equation from the cost in the second equation, to get the difference in cost that corresponds to the difference in items purchased.
That gives us the cost of one banana. To find out how much one apple costs, we subtract $0.75 from the total cost of one apple and one banana.
So an apple costs 50 cents.
To solve systems using addition and subtraction, we’ll be using exactly this idea – by looking at the sum or difference of the two equations we can determine a value for one of the unknowns.
Solving Linear Systems Using Addition of Equations
Often considered the easiest and most powerful method of solving systems of equations, the addition (or elimination) method lets us combine two equations in such a way that the resulting equation has only one variable. We can then use simple algebra to solve for that variable. Then, if we need to, we can substitute the value we get for that variable back into either one of the original equations to solve for the other variable.
Solve this system by addition:
We will add everything on the left of the equals sign from both equations, and this will be equal to the sum of everything on the right:
A simpler way to visualize this is to keep the equations as they appear above, and to add them together vertically, going down the columns. However, just like when you add units, tens and hundreds, you MUST be sure to keep the s and s in their own columns. You may also wish to use terms like as a placeholder!
Again we get , or . To find a value for , we simply substitute our value for back in.
Substitute into the second equation:
The reason this method worked is that the coefficients of the two equations were opposites of each other: 2 and -2. Because they were opposites, they canceled each other out when we added the two equations together, so our final equation had no term in it and we could just solve it for .
In a little while we’ll see how to use the addition method when the coefficients are not opposites, but for now let’s look at another example where they are.
Andrew is paddling his canoe down a fast-moving river. Paddling downstream he travels at 7 miles per hour, relative to the river bank. Paddling upstream, he moves slower, traveling at 1.5 miles per hour. If he paddles equally hard in both directions, how fast is the current? How fast would Andrew travel in calm water?
First we convert our problem into equations. We have two unknowns to solve for, so we’ll call the speed that Andrew paddles at , and the speed of the river . When traveling downstream, Andrew speed is boosted by the river current, so his total speed is his paddling speed plus the speed of the river . Traveling upstream, the river is working against him, so his total speed is his paddling speed minus the speed of the river .
Next we’ll eliminate one of the variables. If you look at the two equations, you can see that the coefficient of is +1 in the first equation and -1 in the second. Clearly , so this is the variable we will eliminate. To do this we simply add equation 1 to equation 2. We must be careful to collect like terms, and make sure that everything on the left of the equals sign stays on the left, and everything on the right stays on the right:
Or, using the column method we used in example 2:
Again we get , or . To find a corresponding value for , we plug our value for into either equation and isolate our unknown. In this example, we’ll plug it into the first equation:
Andrew paddles at 4.25 miles per hour. The river moves at 2.75 miles per hour.
Solving Linear Systems Using Subtraction of Equations
Another, very similar method for solving systems is subtraction. When the or coefficients in both equations are the same (including the sign) instead of being opposites, you can subtract one equation from the other.
If you look again at Example 3, you can see that the coefficient for in both equations is +1. Instead of adding the two equations together to get rid of the s, you could have subtracted to get rid of the s:
So again we get , and we can plug that back in to determine .
The method of subtraction is just as straightforward as addition, so long as you remember the following:
- Always put the equation you are subtracting in parentheses, and distribute the negative.
- Don’t forget to subtract the numbers on the right-hand side.
- Always remember that subtracting a negative is the same as adding a positive.
Peter examines the coins in the fountain at the mall. He counts 107 coins, all of which are either pennies or nickels. The total value of the coins is $3.47. How many of each coin did he see?
We have 2 types of coins, so let’s call the number of pennies and the number of nickels . The total value of all the pennies is just , since they are worth each. The total value of the nickels is . We are given two key pieces of information to make our equations: the number of coins and their value in cents.
We’ll jump straight to subtracting the two equations:
Substituting this value back into the first equation:
So Peter saw 47 pennies (worth 47 cents) and 60 nickels (worth $3.00) making a total of $3.47.
Solving Linear Systems Using Multiplication
So far, we’ve seen that the elimination method works well when the coefficient of one variable happens to be the same (or opposite) in the two equations. But what if the two equations don’t have any coefficients the same?
It turns out that we can still use the elimination method; we just have to make one of the coefficients match. We can accomplish this by multiplying one or both of the equations by a constant.
Here’s a quick review of how to do that. Consider the following questions:
- If 10 apples cost $5, how much would 30 apples cost?
- If 3 bananas plus 2 carrots cost $4, how mush would 6 bananas plus 4 carrots cost?
If you look at the first equation, it should be obvious that each apple costs $0.50. So 30 apples should cost $15.00.
The second equation is trickier; it isn’t obvious what the individual price for either bananas or carrots is. Yet we know that the answer to question 2 is $8.00. How?
If we look again at question 1, we see that we can write an equation: ( being the cost of 1 apple). So to find the cost of 30 apples, we could solve for and then multiply by 30—but we could also just multiply both sides of the equation by 3. We would get , and that tells us that 30 apples cost $15.
And we can do the same thing with the second question. The equation for this situation is , and we can see that we need to solve for , which is simply 2 times ! So algebraically, we are simply multiplying the entire equation by 2:
So when we multiply an equation, all we are doing is multiplying every term in the equation by a fixed amount.
Solving a Linear System by Multiplying One Equation
If we can multiply every term in an equation by a fixed number (a scalar), that means we can use the addition method on a whole new set of linear systems. We can manipulate the equations in a system to ensure that the coefficients of one of the variables match.
This is easiest to do when the coefficient as a variable in one equation is a multiple of the coefficient in the other equation.
Solve the system:
You can easily see that if we multiply the second equation by 2, the coefficients of will be +4 and -4, allowing us to solve the system by addition:
2 times equation 2:
Now simply substitute this value for back into equation 1:
Anne is rowing her boat along a river. Rowing downstream, it takes her 2 minutes to cover 400 yards. Rowing upstream, it takes her 8 minutes to travel the same 400 yards. If she was rowing equally hard in both directions, calculate, in yards per minute, the speed of the river and the speed Anne would travel in calm water.
Step one: first we convert our problem into equations. We know that distance traveled is equal to speed time. We have two unknowns, so we’ll call the speed of the river , and the speed that Anne rows at . When traveling downstream, her total speed is her rowing speed plus the speed of the river, or . Going upstream, her speed is hindered by the speed of the river, so her speed upstream is .
Distributing gives us the following system:
Right now, we can’t use the method of elimination because none of the coefficients match. But if we multiplied the top equation by 4, the coefficients of would be +8 and -8. Let’s do that:
Now we divide by 16 to obtain .
Substitute this value back into the first equation:
Anne rows at 125 yards per minute, and the river flows at 75 yards per minute.
Solving a Linear System by Multiplying Both Equations
So what do we do if none of the coefficients match and none of them are simple multiples of each other? We do the same thing we do when we’re adding fractions whose denominators aren’t simple multiples of each other. Remember that when we add fractions, we have to find a lowest common denominator—that is, the lowest common multiple of the two denominators—and sometimes we have to rewrite not just one, but both fractions to get them to have a common denominator. Similarly, sometimes we have to multiply both equations by different constants in order to get one of the coefficients to match.
Andrew and Anne both use the I-Haul truck rental company to move their belongings from home to the dorm rooms on the University of Chicago campus. I-Haul has a charge per day and an additional charge per mile. Andrew travels from San Diego, California, a distance of 2060 miles in five days. Anne travels 880 miles from Norfolk, Virginia, and it takes her three days. If Anne pays $840 and Andrew pays $1845, what does I-Haul charge
a) per day?
b) per mile traveled?
First, we’ll set up our equations. Again we have 2 unknowns: the daily rate (we’ll call this ), and the per-mile rate (we’ll call this ).
We can’t just multiply a single equation by an integer number in order to arrive at matching coefficients. But if we look at the coefficients of (as they are easier to deal with than the coefficients of ), we see that they both have a common multiple of 15 (in fact 15 is the lowest common multiple). So we can multiply both equations.
Multiply the top equation by 5:
Multiply the lower equation by 3:
Substitute this back into the top equation:
I-Haul charges $60 per day plus $0.75 per mile.
Comparing Methods for Solving Linear Systems
Now that we’ve covered the major methods for solving linear equations, let’s review them. For simplicity, we’ll look at them in table form. This should help you decide which method would be best for a given situation.
Best used when you...
...don’t need an accurate answer.
Often easier to see number and quality of intersections on a graph. With a graphing calculator, it can be the fastest method since you don’t have to do any computation.
Can lead to imprecise answers with non-integer solutions.
...have an explicit equation for one variable (e.g. )
Works on all systems. Reduces the system to one variable, making it easier to solve.
You are not often given explicit functions in systems problems, so you may have to do extra work to get one of the equations into that form.
Elimination by Addition or Subtraction
...have matching coefficients for one variable in both equations.
Easy to combine equations to eliminate one variable. Quick to solve.
It is not very likely that a given system will have matching coefficients.
Elimination by Multiplication and then Addition and Subtraction
...do not have any variables defined explicitly or any matching coefficients.
Works on all systems. Makes it possible to combine equations to eliminate one variable.
Often more algebraic manipulation is needed to prepare the equations.
The table above is only a guide. You might prefer to use the graphical method for every system in order to better understand what is happening, or you might prefer to use the multiplication method even when a substitution would work just as well.
Two angles are complementary when the sum of their angles is . Angles and are complementary angles, and twice the measure of angle is more than three times the measure of angle . Find the measure of each angle.
First we write out our 2 equations. We will use to be the measure of angle and to be the measure of angle . We get the following system:
First, we’ll solve this system with the graphical method. For this, we need to convert the two equations to form:
The first line has a slope of -1 and a intercept of 90, and the second line has a slope of and a intercept of -3. The graph looks like this:
In the graph, it appears that the lines cross at around , but it is difficult to tell exactly! Graphing by hand is not the best method in this case!
Next, we’ll try solving by substitution. Let’s look again at the system:
We’ve already seen that we can start by solving either equation for , so let’s start with the first one:
Substitute into the second equation:
Substitute back into our expression for :
Angle measures ; angle measures .
Finally, we’ll try solving by elimination (with multiplication):
Rearrange equation one to standard form:
Multiply equation two by 2:
Substitute this value into the very first equation:
Angle measures ; angle measures .
Even though this system looked ideal for substitution, the method of multiplication worked well too. Once the equations were rearranged properly, the solution was quick to find. You’ll need to decide yourself which method to use in each case you see from now on. Try to master all the techniques, and recognize which one will be most efficient for each system you are asked to solve.
The following Khan Academy video contains three examples of solving systems of equations using addition and subtraction as well as multiplication (which is the next topic): http://www.youtube.com/watch?v=nok99JOhcjo (9:57). (Note that the narrator is not always careful about showing his work, and you should try to be neater in your mathematical writing.)
For even more practice, we have this video. One common type of problem involving systems of equations (especially on standardized tests) is “age problems." In the following video the narrator shows two examples of age problems, one involving a single person and one involving two people. Khan Academy Age Problems (7:13)
- Solve the system:
- Solve the system:
- Solve the system:
- Nadia and Peter visit the candy store. Nadia buys three candy bars and four fruit roll-ups for $2.84. Peter also buys three candy bars, but can only afford one additional fruit roll-up. His purchase costs $1.79. What is the cost of a candy bar and a fruit roll-up individually?
- A small plane flies from Los Angeles to Denver with a tail wind (the wind blows in the same direction as the plane) and an air-traffic controller reads its ground-speed (speed measured relative to the ground) at 275 miles per hour. Another, identical plane, moving in the opposite direction has a ground-speed of 227 miles per hour. Assuming both planes are flying with identical air-speeds, calculate the speed of the wind.
- An airport taxi firm charges a pick-up fee, plus an additional per-mile fee for any rides taken. If a 12-mile journey costs $14.29 and a 17-mile journey costs $19.91, calculate:
- the pick-up fee
- the per-mile rate
- the cost of a seven mile trip
- Calls from a call-box are charged per minute at one rate for the first five minutes, then a different rate for each additional minute. If a 7-minute call costs $4.25 and a 12-minute call costs $5.50, find each rate.
- A plumber and a builder were employed to fit a new bath, each working a different number of hours. The plumber earns $35 per hour, and the builder earns $28 per hour. Together they were paid $330.75, but the plumber earned $106.75 more than the builder. How many hours did each work?
- Paul has a part time job selling computers at a local electronics store. He earns a fixed hourly wage, but can earn a bonus by selling warranties for the computers he sells. He works 20 hours per week. In his first week, he sold eight warranties and earned $220. In his second week, he managed to sell 13 warranties and earned $280. What is Paul’s hourly rate, and how much extra does he get for selling each warranty?
Solve the following systems using multiplication.
Solve the following systems using any method.
- Supplementary angles are two angles whose sum is . Angles and are supplementary angles. The measure of Angle is less than twice the measure of Angle . Find the measure of each angle.
- A farmer has fertilizer in 5% and 15% solutions. How much of each type should he mix to obtain 100 liters of fertilizer in a 12% solution?
- A 150-yard pipe is cut to provide drainage for two fields. If the length of one piece is three yards less that twice the length of the second piece, what are the lengths of the two pieces?
- Mr. Stein invested a total of $100,000 in two companies for a year. Company A’s stock showed a 13% annual gain, while Company B showed a 3% loss for the year. Mr. Stein made an 8% return on his investment over the year. How much money did he invest in each company?
- A baker sells plain cakes for $7 and decorated cakes for $11. On a busy Saturday the baker started with 120 cakes, and sold all but three. His takings for the day were $991. How many plain cakes did he sell that day, and how many were decorated before they were sold?
- Twice John’s age plus five times Claire’s age is 204. Nine times John’s age minus three times Claire’s age is also 204. How old are John and Claire?