Idiot's Guides: Algebra I (2015)
Part III. Variations on the Line
Chapter 10. Systems of Equations and Inequalities
In This Chapter
· Solving systems of linear equations by graphing
· Solving systems of linear equations by substitution and linear combinations
· Solving problems using linear systems
· Solving systems of inequalities by graphing
Imagine you’re standing in line to buy lunch. You want a hot dog and a soda, but you don’t see the prices listed anywhere. There are two people in front of you. The first one orders 3 hot dogs and 2 sodas and is charged $8. The person just in front of you orders 4 hot dogs and 2 sodas and pays $10. How much will you pay for your hot dog and soda?
Until now, we’ve talked about solving equations with one variable and graphing equations with two variables. That general rule holds because there’s no way to be certain of a solution when you have one equation with two variables. If the only information you had was the order of the first person in line, 3 hot dogs and 2 sodas for $8, or 3x + 2y = 8, you could find more than one set of values for x and y that might work. You’d have no way of knowing which combination is correct. But when you add that second order to your information, you have a chance to come up with one right answer.
In this chapter, we’ll look at algebraic techniques for finding that one right answer. Guess and test can sometimes be helpful, but it’s time consuming, so we’ll look at three strategies for solving problems in which you’re looking for two variables and you have two pieces of information, or two equations, to work with. The first is a natural for two variable equations, because it’s a graphing method. The other two methods are based on the techniques you use to solve one equation.
Solving Linear Systems
Two linear equations with the same two variables form a linear system of equations. The key to solving for the variables is to use both equations at the same time, which is why systems are sometimes called simultaneous equations. There are three methods for finding the solution of a system: graphing, substitution, and combination.
A linear system is a set of two linear equations, each involving the same two variables. The solution of the system is the one pair of values that satisfy both equations.
Solving by Graphing
The graph of an equation is a picture of all the points that represent pairs of numbers that balance, or solve, the equation. If you graph two equations on the same set of axes, either the lines will be parallel, or they will turn out to be the same line, or they will cross. If they’re parallel, there is no point on both lines, so there’s no pair of numbers that solve both equations. The system has no solution if the lines are parallel. If they are the same line, they have infinitely many points in common so the system has infinitely many solutions. If the lines intersect, that one point where they cross represents the one pair of numbers that satisfies both equations. The one point common to both lines is the solution to the system.
Remember the situation in the hot dog line? You can write an equation to describe the transaction of each of the customers ahead of you. Let x stand for the number of hot dogs a customer buys and let y stand for the number of sodas. The customer who orders 3 hot dogs and 2 sodas and is charged $8 gives you the equation 3x + 2y = 8. The person who orders 4 hot dogs and 2 sodas and pays $10 gives you the equation 4x + 2y = 10. To figure out what your one hot dog and one soda will cost, you need to solve the system
3x + 2y = 8
4x + 2y = 10
To find the solution by graphing, first put each equation in slope-intercept form.
Graph each equation. The first has a y-intercept of 4 and a slope of . The second equation has a y-intercept of 5 and a slope of -2. Focus on the first quadrant for this problem, because the number of hot dogs and the number of sodas cannot be negative numbers.
The point at which the lines cross represents the solution of the systems. In this case, the point of intersection is (2, 1), which means x = 2 and y = 1 is the solution of the system. That means hot dogs cost $2 and sodas cost $1. If you buy one of each, you’ll pay $3.
Solve each system of equations by graphing both equations and locating the point of intersection. Check your answers in both equations. If the lines are parallel, say that the system has no solution.
1. y = 2x + 3
y = 3x − 1
3. x + y = 9
x − y = 1
4. 3x − y = 5
6x − 2y = 8
5. 2x − 5y = −44
y = 6 − x
6. y = 3x − 7
2x − 3y = 14
8. 2x + y = −8
3x − 5y = 14
9. x + y = 6
y = −x − 4
10. 3x − 2y = 5
x + 4y = − 3
The graphing method can often be quick and simple, but it does have its drawbacks. If the solution involves extremely large or extremely small numbers, it’s hard to find a way to scale the axes that will let you see the intersection on the page and still have enough detail to see exactly what the point of intersection is. If your solution is represented by the point (4672, 5928), graphing may not be your best method.
Graphing is also not ideal for solutions that are not whole numbers. If your solution is (3.082, -7.121), looking at a graph will probably make you think it’s (3, -7), which might be good enough in some situations, but it’s important to have other methods available.
Don’t be afraid to change the scale on the axes when you draw a graph. One box doesn’t have to be 1 unit. You can count by 2s or 5s or 10s or whatever number is convenient if your problem involves large numbers, and you can make each box worth a half or a tenth, if your numbers are small. There’s no rule that says you have to use the same scale on the vertical and horizontal axes, but if they’re different, you need to be very careful when counting slope.
Solving by Substitution
You would not be able to find a unique solution to a single equation with two variables, like 3x − 4y = 23, but if I tell you that y = −5, you can replace y with –5, and then you’ll be able to solve.
3x − 4y = 23
3x − 4·(−5) = 23
3x + 20 = 23
3x = 3
x = 1
The substitution method for solving a system works on a similar idea. In this method, you use one equation to tell you what one of the variables equals, and then you substitute that into the other equation.
2x − 5y = −44
y = 6 − x
To solve this system by the substitution method, you can use the bottom equation to tell you what y equals. It equals 6 − x. Substitute 6 − x for y in the top equation.
2x − 5y = −44
2x − 5(6 − x) = −44
2x − 30 + 5x = −44
7x − 30 = −44
Making that substitution and simplifying gives you an equation you can solve easily.
7x – 30 = –44
7x = –14
x = –2
Now you know that x = −2, but you’re not done. You still need to find out the value of y, so substitute once more. This time, choose either one of the original equations and substitute -2 for x. Let’s use y = 6 − x and replace x with -2.
y = 6 – x
y = 6–(– 2)
y = 6 + 2
y = 8
Now we have a solution of the system:
x = −2 and y = 8
That example conveniently had one equation that directly told you what y equaled. That’s great, but it doesn’t always happen. What should you do with a system like this one?
3x − 2y = 5
x + 4y = −3
Carefully look over both equations. Your job will be to pick one equation, either one, and isolate one variable, either variable, in that equation. Look for the easiest one. I think using the bottom equation and isolating x may be the way to go. You might very well choose a different equation and/or a different variable and that’s fine. It’s your choice.
Some people think you can only isolate y, but that’s not the case. It’s a mistaken impression that comes from the fact that you often see linear equations in y = mx + b form. If it’s easier to isolate x, that’s fine.
Taking the bottom equation, you can isolate x by just subtracting 4y from both sides.
x + 4y = −3
x = −3 − 4y
Now you’ll go to the top equation and replace x with −3 − 4y. Then you can solve for y.
3x − 2y = 5
3(−3 − 4y) − 2y = 5
−9 − 12y − 2y = 5
−9 − 14y = 5
−14y = 14
y = − 1
Don’t wander off before you finish the job. You have to find the value of x, too. Go back to either one of the original equations.
x + 4y = −3
x + 4(−1) = −3
x − 4 = −3
x = 1
The solution of the system is x = 1, y = − 1.
Solve each system by substitution. Check your answers in both equations.
11. x − y = −9
3x + 2y = 23
12. y = 2x − 7
4x + y = 8
13. 5x + y = 12
3x − 7y = 30
14. 2x − y = 1
7x − 3y = 1
15. b = 8 − 2a
3a + 2b = 15
16. 2x + 3y = 21
4x + y = 9
17. 4x = 3y + 14
2y = 19 − 3x
18. 5x − 2y = 29
3x + 4y = 33
19. 2x + 4y = 6
4x + 5y = 6
20. 2a + 3b = 5
3a − 8b = −30
The substitution method has some significant advantages over graphing, especially when large numbers or fractions are involved. However, there are situations in which substitution gets messy, too. There’s another method, also based on the techniques you use for solving equations, that will help solve many systems easily.
Solving by Linear Combinations
The method of linear combinations is sometimes called the elimination method, because it focuses on eliminating a variable. Truthfully, the substitution method eliminates a variable, too. The combination method does it by adding or subtracting, rather than substituting.
You know that when solving an equation, you can add the same amount to both sides and the equation will stay in balance. The linear combination method gets a little bit clever with that fact. Suppose you’re looking for two numbers that add to 59 and differ by 25. You call the larger number x and the smaller one y, and your problem can be represented by this system of equations.
x + y = 59
x − y = 25
You could take the top equation and add a constant to both sides. You could add 25 to both sides, but it wouldn’t really do much.
x + y = 59
x + y + 25 = 59 + 25
x + y + 25 = 84
You could add −x to both sides, but that would just rearrange things.
x + y = 59
x + y − x = 59 − x
y = 59 − x
You could even add x − y to both sides, but that really wouldn’t help.
x + y = 59
x + y + x − y = 59 + x − y
2x = 59 + x − y
So here comes the clever part. You’re allowed to add x − y to both sides, and you’re allowed to add 25 to both sides, but according to the second equation, x − y and 25 are the same amount. They’re equal. x − y = 25. So you’re going to add x − y to one side of the top equation and 25 to the other. It’s okay because x − y is just another name for 25.
x + y = 59
x − y = 25
x + y + x – y = 59 + 25
2x = 84
x = 42
Doing that makes the y’s disappear, leaving us with just one variable, so we can solve for x, then go back and find y.
x = 42
42 + y = 59
y = 59 − 42
y = 17
This method of solving a system is sometimes just called the addition method, but it takes its official name from the fact that we’re combining the two equations in a way that eliminates one variable.
If you combine the two equations and both variables disappear, and what’s left is true, like 0 = 0 or 42 = 42, the system has infinitely many solutions. If both variables disappear and what’s left is false, like 0 = 4 or 8 = −3, then the system has no solution.
You may wonder if it was just a coincidence that adding the two equations eliminated a variable. That doesn’t happen all the time, does it? The simple answer is no, it doesn’t, but you can make some adjustments in order to make it happen.
When you look at a system and you want to solve by linear combination, take a step at a time.
1. Simplify the two equations and arrange them so the x-terms, the y-terms, the constants, and the equal signs line up. This is not what you want.
3x + 4y = 25
x = 2y − 5
But you can rewrite it like this.
3x + 4y = 25
x − 2y = −5
2. Ask yourself if adding the equations together will eliminate a variable. If you add the like terms that are now lined up, and you still have two variables, you need to try something else. This is not good.
3. Check to see if subtracting the equations will eliminate a variable. Subtracting won’t do the job here either, but don’t give up.
4. Focus on one variable at a time, and ask yourself whether there is a common multiple for the two coefficients. In the example, 3x and x have coefficients of 3 and 1, which have a common multiple of 3. 4y and −2y could use 4 or -4. Don’t worry about signs just now. Let’s just say 4.
If you can find a convenient common multiple for one or both variables, you can still make the linear combination method work. You do it by multiplying the equations—one or both—by constants. You know you can multiply both sides of an equation by any non-zero constant without unbalancing the equation. With your equations, you can choose which variable to pay attention to, and that will determine what constant you multiply by.
If you focus on the x-terms, you can multiply the bottom equation by 3.
3x + 4y = 25
3(x − 2y) = (−5)·3
This will give you a system in which the x-terms have the same coefficient. Subtracting the equations will eliminate the x-terms but be careful to subtract all the way through.
If you choose instead to focus on the y-terms, you should multiply the bottom equation by 2. That will give you a system in which you can eliminate the y-terms by adding the equations.
Once you add, you can solve for x, and then for y.
The choice of which variable to eliminate is yours. You only have to eliminate one variable, and you can choose whichever one you think is easier.
When choosing which variable to eliminate, think about the signs. If the terms with a certain variable have opposite signs, you’ll be able to eliminate by adding. If they have the same sign, you’ll have to subtract, and errors are more common when subtracting. If you’d like to change the sign of a term, you can, but only by changing the signs of all terms in that equation. Multiply through by a negative number and all the signs will reverse.
Let’s work on solving this system by linear combinations.
3x + 4y = −1
2x + 5y = −3
If you want the x-terms to match, you’ll need to multiply the top equation by 2 and the bottom one by 3.
If you’d rather eliminate the y-terms, you’d need to multiply the top equation by 5 and the bottom one by 4.
Once you have a system in which one pair of variable terms have either exactly the same coefficients or exactly opposite coefficients, you can eliminate a variable by either adding or subtracting. If the coefficients are identical, subtract. If they have opposite signs, add. In this example, you’ll need to subtract.
Use linear combinations to solve each system. Check your answer in both equations.
21. x + y = 17
x − y = 3
22. x − 2y = 5
3x + 2y = 23
23. 2x + y = 7
8x + y = 19
24. 7x + 5y = 21
7x + 9y = 21
25. 5x + y = 12
3x − 7y = 30
26. 4x + 3y = 7
2x − 5y = 23
27. 5x − 2y = 29
3x + 4y = 33
28. 4x − 3y = 14
3x + 2y = 19
29. 3x − 2y = 1
2x + 3y = 18
30. 4x + 5y = 6
2x + 4y = 6
Applications of Systems
Systems of linear equations can be solved by graphing, by substitution, or by linear combination. The need to solve a system usually arises from a problem in which there is more than one unknown. Many of these problems are mixture problems. It may be that substances are being mixed, for a package of mixed nuts or candies, or the chemicals in a lab. It may be a collection of coins of different values, or investments at different interest rates.
To solve any sort of mixture problem that needs two variables, you can follow the same basic pattern. Start by defining what unknown quantity each variable stands for. Then ask yourself, and answer, a few questions. How much, or how many, of each substance do we have? How many pounds of nuts, or how many liters of a chemical, or how many of that type of coin? Then, what is the price per pound, or the concentration of the active ingredient, or the value of the coin?
The total value of the substance can usually be found by multiplying how much you have by the unit value. If you have 3 pounds of nuts that sell for $8 a pound, they are worth 3 × $8 = $24. Seven dimes, coins worth 10 cents each, are worth 70 cents. If a chemical is 80% acid and you have x liters of it, 80%·x = 0.8x liters are actually acid. It may be helpful to organize the information in a chart like this one.
Let’s look at an example. A collection of nickels and dimes, 16 coins in all, is worth $1.25. How many nickels and how many dimes does the collection contain?
Start by defining variables. Let’s use n for the number of nickels and d for the number of dimes. Your chart would look something like this. The last row will tell you about the total collection.
You can get your two equations from the table. One equation is produced by adding down the how much column. n + d = 16. The other equation comes from adding down the total value column. 0.05n + 0.10d = 1.25 To solve the problem, you need to solve the system below.
n + d = 16
0.05n + 0.10d = 1.25
If you multiply the bottom equation by 10, you’ll be able to eliminate the d by subtracting.
Then you can solve the system.
Your solution says there are 7 nickels, worth a total of 35 cents, and 9 dimes, worth a total of 90 cents. Sure enough, 35 cents plus 90 cents is $1.25.
Another common type of situation that leads to a system is the distance problem, which talks about two vehicles moving at different speeds. The basic rule for this sort of problem is distance equals rate times time. The total distance covered equals the rate of speed multiplied by the time spent traveling at that speed.
Suppose you need to get to a town 138 miles from your home. You start out on foot, walking at 4 miles per hour, and you walk until you get to the bus stop. You get on the bus, and travel at 28 miles per hour, until you reach your destination, exactly 6 hours after you left home. How long did you walk?
Let’s define variables. Let x represent the number of hours you walked. What’s your second variable? You know your walking speed and the speed of the bus, so it must be the time on the bus. Let y be the number of hours you spent on the bus. You can use a chart here, too.
You can generate one equation by adding down the time column. x + y = 6. Then you can get a second equation by adding down the distance column. 4x + 28y = 138. To solve the problem, solve the system.
x + y = 6
4x + 28y = 138
Multiply the top equation by 4 to eliminate the x’s or 28 to eliminate the y’s. Let’s multiply by 4, and then subtract.
The solution is and , which means you spent 1 hour and 15 minutes walking and 4 hours and 45 minutes on the bus.
Solve each problem using a system of equations. Use whatever method seems most efficient.
31. A collection of 20 coins, all dimes and quarters, totals $3.05. How many dimes and how many quarters are in the collection?
32. A stamp collector bought 30 stamps. Some stamps cost 14 cents and some cost 17 cents. The total price was $4.56. How many of each stamp did he buy?
33. A coffee merchant wants to mix coffee worth $6.00 a pound with coffee worth $5.10 a pound, to create a blend that she can sell for $5.40 a pound. She wants to produce a total of 60 pounds of the new blend. How many pounds of each type of coffee should she use?
34. I had $6,000 to invest. I invested part of it in a fund that pays 4% interest and the rest in a fund that pays 7% interest. If the total interest from the two accounts was $342, how much did I invest at each rate?
35. You need 48 ml of a 15% salt solution. You have a 5% salt solution and a 20% salt solution. How much of each should you combine to make 48 ml of a 15% solution?
36. A driver began a trip of 352 miles traveling at 50 miles per hour. After a period of time, she had to slow to 20 miles per hour because of construction, and completed the trip at that speed. If her total travel time was 8 hours, how long did she travel at each speed?
Define your variables, then translate the problem into two equations. Solve and check.
37. A group of 8 adults and 10 children paid a total of $115 for admission to an amusement park. A group of 3 adults and 4 children paid a total of $44 for admission to the same park. Find the price of admission for an adult, and the admission price for a child.
38. A number of nickels and dimes have a total value of $1.65. If the number of nickels and dimes were interchanged, the total value would be $1.35. How many nickels and how many dimes are there in the original collection?
39. An order of 3 pens and 5 pencils costs $3.90. In the same shop, an order
of 5 pens and 2 pencils costs $4.60. find the price of 1 pen and the price of 1 pencil.
40. The attendance at a carnival was 250 people, some adults, some children. The total collected was $735. The adult admission price was $3.50 and the child’s admission was $2.50. How many adults and how many children attended the carnival?
We often find ourselves in situations where we’re searching for more than one piece of information. Being able to solve a system of equations is an important tool for those problems. If you can represent the quantities you’re looking for by variables and you have enough information to write two equations involving those variables, you can solve to find them. But life doesn’t always cooperate by giving you two nice tidy equations. Sometimes our information comes in the form of inequalities.
Systems of Inequalities
Imagine that you’re in charge of organizing a party. You have a budget for food and you know how much it will cost to feed an adult and how much it will cost to find a child. The number of people you can invite is limited by the size of your home, but there is also a minimum number, because certain people must be invited.
This situation sounds as though it might be represented by a system of equations. Let x represent the number of adults you invite and y represent the number of children. When you start to write about the situation, however, you may realize that you don’t have to spend every penny of your budget. It would be okay to spend less than what you budgeted, as long as all your needs were covered. And there may not be a solution of the form “invite this many adults and that many children.” Instead, you’ll probably end up with a range of possibilities. You’re dealing with an inequality situation.
The choice between a system of equations and a system of inequalities is not always clear, but if you’re interested in all the possibilities, you probably want a system of inequalities. If you just want to know about the dividing line between more and less, or acceptable and unacceptable, you can probably use a system of equations.
Writing a set of inequalities to describe a situation is very much the same as writing equations, but only the graphing method lets you really see your solution, so that’s how systems of inequalities are tackled. The other difference between systems of equations and systems of inequalities is that you may find that you have more than two inequalities. Each inequality sets a boundary on the acceptable values in the solution set.
Let’s put some numbers into the party situation and see what it looks like. Let’s say you know it costs $5 for food for an adult and $3 for a child and your budget is $100. You must invite the 6 people in your immediate family at a minimum, and you don’t think your home can hold more than 25 people, so that’s your maximum.
Using x for the number of adults and y for the number of children, you would have three inequalities. (The last two could be combined into a compound inequality.)
5x + 3y ≤ 100
x + y ≥ 6
x + y ≤ 25
When graphing a system of inequalities, especially one with more than two equations, don’t rush to shade entire half-planes. You can wind up with so much shading that you can’t see where the overlap is. Instead, try putting a narrow fringe of shading on the correct side of the line. After you’ve done that for all your inequalities, look for the region that has fringe on all sides.
If you graph all those inequalities on one axis, you’ll get the following graph. You only need to look at Quadrant I because you won’t be inviting a negative number of people.
All that shading can be hard to understand, so it’s worthwhile to focus on the portion that is shaded by all three inequalities and mark it clearly.
Any point in that shaded region represents a number of adults and a number of children that you could invite while staying within the limitations you have.
Graph each system of inequalities and clearly mark the area that represents the solution.
41. y ≤ 2x + 1
y ≥ 4 − x
42. y < 7 − 3x
2x + y > 8
43. 3x − y < 1
y ≥ 2 − 3x
44. 5x + 2y ≥ 12
3x + 2y < 12
45. 3x + 4 ≥ 4y
x + 2y < 12
Use the graph of a system of inequalities to show the solution set for each problem.
46. The sum of two numbers is at least 40 and their difference is no more than 8. What are the numbers?
47. Find two numbers whose difference is no more than 12 and whose sum is at least 20.
48. A small airline uses planes that can seat no more than 80 passengers, some first class, some coach. For a certain flight, they charge $100 for a coach ticket and $300 for a first class seat. If the airline wants to collect at least $7500 in fares for the flight, how many first class and how many coach seats should they provide?
49. Gillian creates metal sculptures from nuts and bolts, and has decided to produce some of her more popular creations in quantity for sale to gift shops. Her smallest sculpture, Gull, uses 15 nuts and 12 bolts. Her largest piece, Giant, is built from 20 nuts and 30 bolts. Gillian has 400 nuts and 300 bolts available for sculpting. How many of each sculpture can Gillian produce?
50. A small bakery sells cookies and cakes, and bakes large batches of each, using flour and sugar in each recipe. Each batch of cookies uses 2 pounds of flour and 3 cups of sugar. Each batch of cakes uses 9 pounds of flour and 4 cups of sugar. On any day, there are 25 pounds of flour and 36 cups of sugar available. How many batches of cookies and how many batches of cakes can they produce in a day?
The Least You Need to Know
· A pair of linear equations using the same two variables that are solved simultaneously are called a linear system.
· You can solve a system by graphing both equations and finding the point of intersection.
· You can solve a system by isolating a variable in one equation and using the result to substitute for that variable in the other equation.
· You can solve a system by combining the equations, or multiples of the equations, by addition or subtraction to eliminate one variable.
· Once the value of one variable has been found, you can substitute that variable into one of the equations to solve for the other variable.