# Systems of Linear Equations in Two Variables

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1 Chapter 6 Systems of Linear Equations in Two Variables Up to this point, when solving equations, we have always solved one equation for an answer. However, in the previous chapter, we saw that the equation of a line Ax + By + C = 0 which contains two variables, x and y, does not have one answer. In fact, it has an innite number of answers and the plot of all these answers forms a line in the 2-dimensional Cartesian Coordinate System. There is a property in mathematics that states that if you want to solve an equation that has multiple variables in it for one specic answer, then in order to even have a chance to solve for the one answer, the number of equations involving the variables must be the same as the number of variables. So in the case of linear equations in two variables, if we wish to get just one answer, then we must have two dierent equations. These two equations are what we call a system of equations. Denition System of equations In general, a system of equations means a collection of equations and the solution to the system of equations is the set of values that solves all the equations in the system at the same time. In this chapter we discuss some techniques for solving systems of linear equations in two variables. 6.1 Graphing Method View the Video Tutorial for this section here. Although the graphing method is the least practical in getting the exact solution, it is the best method for illustrating the conceptual meaning of the solution to a system of equations. Proposition Graphing Method for solving a System of Linear Equations in Two Variables. Graph the solutions of each equation. Since they are linear, they will graph as lines. The point where the two lines intersect is the point whose x and y coordinates will satisfy both equations and hence will be the solution to the system of equations. Example Solve the system of equations by using the graphing method. 1

2 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 2 1) x 2y = 8 3x 4y = 6 2) 2x 3y = 12 4x + 6y = 12 3) 15x + 18y = 6 5x 6y = 2 Solutions: 1) We know that both equations will graph as lines. We have two methods for graphing a line and the method chosen will depend on the equation. In this case, for the line x 2y = 8 It is easy to nd the x and y intercepts and so graph the line using those two points. x-intercept y-intercept x 2(0) = 8 x = 8 x = 8 (8,0) (0) 2y = ( 2y) = 1 2 ( 8) y = 4 (0,4) For the second equation, if we solve for the x and y intercepts, we get x-intercept 3x 4(0) = 6 3x = 6 x = 2 (-2,0) y-intercept 3(0) 4y = 6 4y = 6 y = 6 4 = 3 2 (0, 3 2 ) Here we run into one of the weaknesses of the graphing technique. Since we are reading the solution, the intersection point, from the graph, we need to be sure that the graph is accurate. Although we could label the tick marks in fractional values, it could become unwieldy if the numbers get large. Notice that to plot the point (8,0) would require 16 tick marks along the x-axis if going by 1 2. So, in general, we try to avoid fractional coordinates. So we can try to nd a second point with integer coecients by trial and error or since we have one nice point, (-2,0), we can use the slope to nd another point. To nd the slope, we will use the slope-intercept form. So we rst solve for y. 3x 4y = 6 4y = 3x 6 y = 3 4 x y = 3 4 x The slope of the line is m = 3 4 and since we know (-2,0) is on the line, we can start from there and go up 3 and to the right 4, to get another point, (2,3), see graph below.

3 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 3 Now, if we graph the two line together, we get the following graph. From the graph, it would appear that the point (2,3) is the solution to the system. We can check this by plugging in 2 for x and 3 for y into the two equations to see if it satises both equations. x 2y = 8 (2) 2(3) = 8 3x 4y = 6 3(2) 4(3) = 6 So (2,3) is a solution to both equations and hence the solution to the system. 2) In this problem, graphing the lines by nding two points on each line leads to two lines that look parallel. The problem is that when you are looking at a graph, you are only seeing the graph for a xed range of x and y values and so it is possible that lines intersect at x and y values that are not shown on the graph. So we need an analytic way to ensure that the lines are indeed parallel. Recall that two lines are parallel if they have the same slopes. So we rst put both equations in slope-intercept form 2x 3y = 12 2x 2x 3y = 2x + 12 y = 1 ( 2x + 12) 3 y = 2 3 x 4 4x + 6y = 12 +4x +4x 6y = 4x + 12 y = 1 (4x + 12) 6 y = 2 3 x + 2

4 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 4 So both lines have the same slope m = 2 3, which means that the two lines are parallel. Since parallel lines never intersect, there is no solution to the system of equations. 3) In this problem, we again can graph the lines by nding two points on each line, but it will be more instructive to put both equations in slope-intercept form. 15x + 18y = 6 +15x +15x 18y = 15x 6 y = 1 (15x 6) 18 y = x 6 18 y = 5 6 x 1 3 5x 6y = 2 5x 5x 6y = 5x + 2 y = 1 ( 5x + 2) 6 y = 5 6 x 2 6 y = 5 6 x 1 3 Notice that, in slope intercept form, both equations are the same, which means that the two lines are actually on top of each other (or are the same line). In other words, the two lines intersect each other at every point along the line. The solution is every point on the line. Note that there are an innite number of solutions but that not every point is a solution. Only the points that satisfy the equation y = 5 6 x 1 3 are solutions to the system. This means that for any given x value, the y value is 5 6 x 1 3. For example, if x = 0 then y = 5 6 (0) 1 3 = 1 3. So ( 0, 5 6 (0) ( 3) 1 or 0, 1 3) is a solution. In general, we write the solution as the ordered pair ( x, 5 6 x 3) 1.

5 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES Substitution Method View the Video Tutorial for this section here. As mentioned before, one of the problems with the graphing method is that it is dependent on the accuracy of the graph. So if the solution to a system contained fractional values, the graphing method may not be able to get the exact answer. The substitution method is an analytical method that lets us nd the exact solution to a system of equations. The substitution method is based on the idea that if we say that A = B, then wherever we see A, we can replace it with B and likewise, wherever we see B we can replace it with A. We use the following example to describe the method. Example Solve x 2y = 8 (Equation 1) 3x 4y = 6 (Equation 2) Solution: x 2y +2y = 8 +2y x = 2y 8 x = (2y 8) x = 2y + 8 Step 1: We choose one of the equations and solve for one of the variables. Here we choose the Equation 1 and solve for x as that will be the easiest to do. Step 2: 3x 4y = 6 3( 2y + 8) 4y = 6 6y y = 6 Now, in the equation that was not used above, in this case Equation 2, we substitute in for x the expression that we found, 2y + 8. By making this substitution we end up with an equation that contains only one variable which can then be solved our normal way. Note that you must use the equation that was not solved for in the rst step, in this case we must use Equation 2. If you substitute into the equation that has already been solved, in this case Equation 1, the variables will always drop out and you will end up with a statement that is always true, such as 0 = 0, which will tell us nothing. 6y y = 6 10y = y = 30 y = 1 10 ( 30) y = 3 continuation of Step 2: Solve for y. x = 2y + 8 x = 2(3) + 8 x = 2 Step 3: Substitute the 3 for y into the equation that we solved originally for x in the rst step.

6 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 6 The solution is therefore (2, 3). Example Solve the following using the substitution method. 1) 39x + 4y = 5 y = 21x + 5 2) 2x + 5y = x = 1 y 3) y + 2 = 3 4 x 8 = 3x 4y Solutions: 1) Since the second equation in this system has already been solved for y, we will substitute 21x + 5 for y into the rst equation and solve for x. Substitute 1 3 in for x in the second equation. 39x + 4y = 5 39x + 4( 21x + 5) = 5 39x 84x + 20 = 5 45x = x = 15 x = 1 45 ( 15) x = = 1 3 y = ( ) = = 2 So the solution is the point ( 1 3, 2). 2) For step 1, we solve the second equation for y. Again, it does not matter which equation you solve or for what variable. 2 5 x 2 5 x+y = x y +y y = x Step 2, we substitute this in for y in the rst equation. 2x + 5y = 20 2x + 5(1 2 x) 5 = 20 2x + 5 2x = 20 5 = 20 Since 5 20, this tells us that this equation has no solution which means that the overall system of equations has no solution. In other words, the two equations represent parallel lines and so do not intersect anywhere. Note that if you put both equations in slope-intercept form, y = 2 5 x + 4 y = 2 5 x + 1 they have the same slope but dierent y-intercepts, which means that the two lines are parallel. So again, we say that the system has no solution.

7 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 7 3) For this system, solving the rst equation for y would be easiest. Now we substitute this into the second equation. y + 2 = 3 4 x y = 3 4 x 2 8 = 3x 4y ( ) 3 8 = 3x 4 4 x 2 8 = 3x 3x = 8 Since this equation is always true, what this tells us is that the two equations represent the same line. 1 You can prove that they are the same line by putting both equations in slope-intercept form and seeing that you get the same equation in that form. So any point on the line, y = 3 4x 2, will be a solution to the system. As an ordered pair, this means that for any given x value, the y-coordinate value is 3 4x 2, we write the solutions as y }} 3 (x, 4 x 2) 1 This is why it is important that you make sure that you substitute into the correct equation. If you make the substitution into the wrong equation, you will always get a solution similar to this one (all variables drop out and you get a true statement) and think the two equations represent the same line. For example, in the rst example problem, if you substituted the expression for y into the second equation, you would end up with all the variables dropping out and think the two equations represented the same line, when we know they do not.

8 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES Elimination Method View the Video Tutorial for this section here. This video discusses how to enter answers for Dependent Systems. The elimination method is based on the concept that if you have two pairs of quantities that are equivalent to each other, A = B C = D and you add the left side together and the right side together, you still end up with the left and right side equaling each other, A + C = B + D The way we use this idea in solving a system of equations is to rewrite the equations such that when we add the two equations, all the terms containing one of the variables cancel each other out. We describe the method with the following example. Example Solve x 2y = 8 (Equation 1) 3x 4y = 6 (Equation 2) Solution: 3( x 2y) = 3( 8) 3x 4y = 6 3x 6y = 24 3x 4y = 6 Step 1: Rewrite one of the equations so that the terms containing the variable we wish to eliminate are exact opposites. Here we will get rid of the terms with the variable x. To do this, we multiply both sides of Equation 1 by 3. Since we multiplied both sides by 3, the new and old equation are equivalent. 3x 6y = x 4y = 6 6y + ( 4)y = 24 + ( 6) 10y = 30 y = 3 Step 2: Add the two equations together. If we did step 1 correctly, one of the variables should drop out in the resulting equation. Solve for the remaining variable. Here, the terms with x in them cancel each other out and so we solve for y. x 2y = 8 x 2(3) = 8 x 6 = 8 x = 2 x = 2 Step 3: Substitute the value found in step 2 into either of the equations and solve for the other variable. In this case, we substitute 3 in for y into equation 1. The solution is the point (2, 3). Before doing more examples, we need to discuss the idea of the least common multiple.

9 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 9 Denition Least Common Multiple The least common multiple of two whole numbers, a and b, is the smallest number that both a and b divide into. Let L be the least common multiple of a and b. Then L a L b = m = n where m and n are whole numbers. Note that if we solve the above for L, we get L = a m L = b n which means that the least common multiple of a and b is the smallest number that you can get by multiplying a and b by some other whole numbers. Example Find the least common multiple of the following pairs of numbers. 1) 4 and 2 2) 3 and 5 3) 6 and 8 Solutions: 1) The following are the steps to nd the least common factor. 4 = = 2 Step 1: Prime factor both numbers. Write both numbers as a product of prime numbers. Least common factor is 2 2 or 4. Step 2: The least common multiple must contain all the factors of both numbers. If they share a common factor, use the one that has more. Since both 4 and 2 share the common factor 2, we use the two 2's from 4. Note that 4 is the smallest number that you can get by multiplying some number to 4 and 2. In this case, 4 1 = 4 and 2 2 = 4. 2) For 3 and 5, since they are both prime numbers, there is nothing to do for step 1. For step 2, we just multiply both 3 and 5. So, the least common multiple is 3 5 = 15. 3) 6 = = 2 3 Step 1: Prime factor both numbers. Write both numbers as a product of prime numbers.

10 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 10 Least common factor is = 24 Step 2: The least common multiple must contain all the factors of both numbers. If they share a common factor, use the one that has more. Since both 6 and 8 share the common factor 2, we use the three 2's from 8. The reason for the above discussion of the least common multiple is that in example 6.3.1, we saw that in order to get rid of the x terms, they had to be exact opposites of each other and the only way to change the terms is to multiply the entire equation by some number. So the goal is to change both terms to the least common multiple with opposite signs. In example 6.3.1, equation 1 had x and equation 2 had 3x. Ignoring the negative sign for a moment, we are dealing with the numbers 1 and 3. The least common multiple is 3 and so if we multiply x by 3, the x terms will be exactly opposite to each other. Example Solve using the elimination method. 1) 2x 3y = 1 5x 2y = 1 2) 18x 6y = 15 9x + 3y = 4 3) 3x + 6y = 18 y = 1 2 x + 3 Solutions: 1) We follow the steps of example (2x 3y) = 2(1) 3(5x 2y) = 3(1) 4x + 6y = 2 15x 6y = 3 Step 1: We will eliminate the y terms. The rst equation has a 3y while the second equation has 2y. Ignoring the signs for a moment, the least common multiple is 6. So if we multiply the rst equation by -2 and the second by 3, we will end up with the y terms being exact opposites. 4x + 6y = x 6y = 3 11x = 1 Step 2: Add the two equations and solve for x. x = ( 1 11 ) 3y = 1 3y = (1) y = 9 11 y = 1 3 ( ) 9 = Step 3: Substitute the x value into the rst equation and solve for y. 2) So the solution is the point ( 1 11, 3 11).

11 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 11 18x 6y = 15 2( 9x + 3y) = 2( 4) 18x 6y = 15 18x + 6y = 8 Step 1: Here we will eliminate the y terms. Least common multiple is 6, so we multiply the second equation by 2. 18x 6y = x + 6y = 8 0 = 23 Step 2: Add the two equations. No solutions. Since 0 is never equal to -23, that means that the system of equations has no solutions. In other words the lines do not intersect. They are parallel lines. 3) 3x + 6y = 18 y = 1 2 x + 3 3x + 6y = x + y = 3 First, we will rearrange the second equation so that the x and y terms line up in both equations. 3x + 6y = 18 6 ( 1 2 x + y) = 6(3) 3x + 6y = 18 3x 6y = 18 We will get rid of the y terms. Since the least common multiple of the y terms is 6, we multiply the second equation by -6. 3x + 6y = x 6y = 18 0 = 0 Add the two equations together. Since all the variables drop out and we end up with a statement that is always true, this means that the two equations represent the same line. (x, 12 x + 3 ) So the solutions are any points on the line y = 1 2 x + 3. As an ordered pair, this means that for any value x, the y value will be 1 2 x + 3.

12 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES Applications View the Video Tutorial for this section here Distance/Rate in Moving Medium Recall that in a previous chapter we used the fact that if an object travels at a constant speed (rate), r, for some time, t, then the distance traveled, d, is the product of the speed and time, d = rt What we want to study now is what happens if the object is moving at a constant speed in a medium that is also moving. For example a boat moving in a owing river or an airplane ying in air that is also moving. In such cases, we have to understand the relationship between the speed of the object and the speed of the medium. To understand this relationship, we give the following illustration. Example Suppose there are three people, a newlywed couple on a train and the bride's ex standing at the station. The train is moving at 20 feet/sec. away from the station (and the ex) while the bride on the train walks towards the front of the train at 3 feet/sec. and the groom stays at the end of the train, see gure below. After 5 seconds, we have the following: To the groom, the bride would appear to be walking 3 feet/sec. away from him. To the ex, the groom would appear to be moving at 20 feet/sec. away from him, since the train is moving at 20 feet/sec. However, since the bride is walking on the train as the train moves, the bride would appear to be moving away from the ex faster than the groom, who is standing still on the train. To the ex, it would appear that the bride is moving away from him at the speed of the train plus her walking speed, = 23 feet/sec. We can verify this as follows. Since d = rt, if we solve for the rate, r, we get r = d t. Now after 5 sec. (t = 5), the distance between the ex and the groom is 100 ft., while the distance from the ex to the bride would be 115 ft. The speed of the groom, as viewed by the ex, is r = = 20 ft/sec. which is the speed of the train. This makes sense, since the groom is not moving on the train. The speed of the bride, as viewed by the ex, is r = = 23 ft/sec. After 10 seconds, we have a similar situation. The bride will be 30 feet away from the groom and so would appear to the groom as walking at a speed of r = = 3 ft/sec. To the ex, the bride would be 230 feet away and so would appear to be moving at a speed of r = = 23 ft/sec. or the speed of the train plus her walking speed.

13 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 13 What this illustrates is that the measure of speed is relative to where you are when observing a moving object. Proposition Relative Speed of Objects in a Moving Medium Let r s be the speed of the object in medium that is not moving (e.g. the speed of a boat in standing water.) Let r m be the speed of the moving medium (e.g. the speed of the river current.) Then to the observer who is standing still (not in the moving medium), if the object and medium are moving in the same direction(e.g. a boat going down river with the current), the speed of the object to the observer would be the sum r s + r m Whereas, if the object is moving in the opposite direction of the moving medium (e.g. a boat going up river against the current), the speed of the object to the observer would be the dierence r s r m if r s > r m r m r s if r s < r m We use which ever dierence gives us a positive value, since speed must be a positive number. Example A boat takes 7 hours to go up river 42 miles against the river current, while it takes 3 hours to cover the same distance going down river with the current. Find the speed of the boat in standing water and the speed of the river current. Solution: Let b be the speed of the boat in standing water and c the speed of the river current. Then according to Proposition above, we have that the overall speed going up river is b c and going down river is b + c, to someone watching the boat from the bank of the river. Note that we assume that b > c, since if the current is faster than the boat, then the boat could not go up river against the current. Now we build our distance, rate, and time table. rate time = distance up river b c 7 7(b c) down river b + c 3 3(b + c) The dierence between this problem and the distance/rate problems we did when talking about linear equations in one variable is that for those problems, we looked at the relationship between the two distances to create one equation, while here we have two unknowns, b and c, and so we need two equations instead of one. We know that the distance traveled up river in 7 hours is 42 miles and like wise the 42 miles was

14 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 14 covered going down river in 3 hours. 7(b c) = 42 3(b + c) = 42 An easy way to solve this system of equations is to rst get rid of the 7 and 3 in the two equations by dividing them out and then use the elimination method. 1 7 (7)(b c) = 1 7 (42) 1 3 (3)(b + c) = 1 3 (42) b c = 6 b + c = 14 b c = 6 + b + c = 14 2b = 20 b = 10 b + c = 14 (10) + c = 14 c = 4 So the speed of the boat is 10 mph while the speed of the river current is 4 mph Miscellaneous Problems The key to working out word problems is to rst restrain yourself from trying to get equations right away. Here is a general guideline for how to approach word problems. General Approach to Solving Word Problems: Keep in mind that you will have to read the problem at least two times. Step 1: On the rst read, ignore all the numbers and just focus on what the problem is talking about, e.g. is it talking about a car going from point A to B and back, two planes going in opposite directions, two dierent savings accounts, etc... Step 2: On the second read, write down on paper all the information given in the problem about what you identied in step 1. Just write the information down in words as done in the problem. The information should be organized so that you can, at a glance, determine what information belongs to what. A table to organize the information is often helpful. Step 3: Determine what the variable(s) should represent. Then convert the information found in step 2, into mathematical expressions using the variable(s) we just identied. Step 4: Now we step back and look at the information written out and try to nd the relationship between the quantities identied in step 1. This is the step at which we want to create equations. Step 5: Once the equation(s) are made, we solve them. Example How much 10% saline solution should be mixed with a 40% saline solution to get 50 ounces of a 28% saline solution? Find the amount of each saline solution used. Solution: Note that we have done problems like this before when discussing mixture problems. Here, we show a dierent approach to solving the problem using the idea of systems of equations.

15 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 15 Step 1: The problem involves mixing two dierent types of liquid to get a third type. So we are talking about a 10%, 40%, and 28% saline solutions. Step 2 and 3: Here we mix steps 2 and 3 a little because mixture problems are fairly straightforward, in that the table created for them always follows the same format. Since we do not know how much of either saline solution is mixed in, we let x = total amount of the 10% saline solution used y = total amount of the 40% saline solution used and make our table, as we did when we talked about mixture problems. total amount of solution in each bottle. actual amount of salt in each bottle. 10% 40% 28% x + y = x + 0.4y = 0.28(50) Step 4: So we end up with the system of equations, x + y = x + 0.4y = 0.28(50) Step 5: Here we use the substitution method. x + y = 50 y = 50 x Solve the rst equation for y. 0.1x + 0.4(50 x) = 14 Substitute what we found for y above into the second equation. 0.1x x = x + 20 = x = 6 x = = 20 Solve for x. y = 50 (20) = 30 Substitute the x value into the equation above where we solved for y. So we need to use 20 ounces of the 10% saline solution and 30 ounces of the 40% saline solution. Example A total of \$25,000 is to be distributed into two savings accounts which give a simple interest of 2% and 5%, respectively. How much should be invested into each account so that the total interest earned at the end of the year is \$1300? Solution: Step 1: The problem involves two savings accounts, a 2% interest account and a 5% interest account.

16 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 16 Step 2 & 3: Let x = amount invested at 2% y = amount invested at 5% This problem is very similar to the previous example. The only dierence is that instead of talking about the total liquid and the amount of salt in each bottle, we instead deal with total dollar amounts invested and the interest earned in each account. We can make a table similar to that used for the previous mixture problem. 2% 5% \$ amount in each account. x + y = 25,000 interest earned in each account 0.02x y = 1300 Step 4: So the system of equations that we need to solve is x + y = x y = 1300 Step 5: We use the substitution method to solve the system where we solve the rst equation for y and substitute that into the second equation to solve for x. y = x 0.02x (25000 x) = x x = x = 50 x = = Before substituting this into the rst equation to nd y, we note that the above is saying that we need to invest -\$ into the 2% account. The negative does not make any sense in the context of this problem and the way we set up the variables. This means that although there is a mathematical answer, it does not make any physical sense. So it is not possible to earn \$1300 in interest under the given conditions (note that this makes sense, since even if you invested everything into the 5% account, you only get 0.05(25000) = 1250.) Example Randy needs to buy a new printer to print out iers for a new math class. An inkjet printer will cost \$69.95 for the printer and 10 cents per page to print, while a laser printer will cost \$ for the printer and cost 5 cents per page to print. Determine at what number of pages printed, the total cost will be the same in using either printer. Solution: Step 1: We are talking about two printers, an inkjet and laser. Step 2: inkjet laster printer cost \$69.95 \$ cost per page to print \$0.1 \$0.05 We want the total cost of using either printer to be the same. Step 3: Let x be the number of pages printed. Then the total cost to use each printer is the sum of the cost of the printer and the cost to print x pages. We let C I be the total cost of using the inkjet printer and C L the total cost of using the laser printer. Total cost = Cost to print x pages + Cost of printer C I = 0.1x C L = 0.05x

17 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 17 Since we want to know when the total cost is the same, we also have the equation C I = C L So here, we have three equations and three unknowns. Step 4: We use the substitution method, starting with the equation C I = C L 0.1x = 0.05x x x x = 50 x = = 1000 So if 1000 pages are printed, then the cost will be the same for either the inkjet or laser printer. Another way to approach this problem would have been to let the variable y represent the total cost, whether we are talking about the inkjet or the laser, just as x represents the number of pages printed in either printer. In this way, the printer we are talking about is simply identied by the equation rather than by the variables. y = 0.1x the inkjet printer y = 0.05x the laser printer So we get two lines and we seek the point at which the two lines intersect each other. If we use the substitution method (set the y's equal to each other), we end up with the same equation as above.

18 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES Practice Problems Be sure to write your answer as a sentence at the end. 1) The ACME Anvil Company sells two types of anvils. The base model is sold for \$30 while the deluxe model is sold for \$50. In conducting a market study, ACME found that their best customer, Mr. Wile E. Coyote, purchased a total of 350 anvils in the previous year which generated a total revenue of \$12,700 for the company. Determine how many of the deluxe model Mr. Coyote purchased.

19 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES Systems of Linear Inequalities in Two Variables View the Video Tutorial for this section here. In the previous chapter we discussed how to solve a linear inequality in two variables. To solve a system of inequalities, we add just one more step at the end. We summarize the steps as follows: Steps to Solve a System of Linear Inequalities in Two Variables Graph the solution of each inequality on the same graph. Recall that this involves graphing the associated line which divides the cartesian plane in two. Then you shade in the half-plane that solves the inequality. The points at which the lines intersect are a key feature and so should be found analytically using the techniques we discussed in the previous sections of this chapter. Once you have shaded the solution to each inequality, since the solution of the system are the points that solve all the inequalities, the region where all the shadings from the rst step overlap will be the solution to the system. We illustrate the method with the following example. Example Solve 2x + y < 4 3x + y 1 Solution: Solve each inequality. 2x + y < 4 3x + y 1 y < 2x 4 y 3x + 1 Graphing them together, we get

20 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 20 The solution is where the two shaded regions overlap each other. An important feature of the graphs is where the two lines intersect, as that forms a vertex (corner) of the region that represents the solution. We nd the intersection by converting the original system of inequalities into a system of equations. 2x + y = 4 3x + y = 1 We will solve this system using the elimination method by multiplying the rst equation by -1. Substituting this into the rst equation, we get 2x y = 4 + 3x + y = 1 5x = 5 x = 1 2( 1) + y = 4 y = 2 So the point at which the two lines intersect is ( 1, 2). Note that this tells us that the region containing the solution is to the left of x = 1. The graph of the nal answer is Note that the points on the line y = 2x 4 are not included in the solution and so the point where the two lines intersect is also not included in the solution, which is why we use a hallow circle to represent that point. Example Graph the regions described by the following systems of linear inequalities. Be sure to nd and label the vertices (the corner points of the region.)

21 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 21 1) 5x 2y > 20 y < 5x 15 2) x 4 y > 2 4x + 7y 26 Solutions: 1) The rst step is to graph the regions dened by each inequality. We do this by rst graphing the lines associated with each inequality. In the rst equation, we note that the x and y intercepts are easy to nd and they are integer values. x-intercept y-intercept 5x 2(0) = 20 5(0) 2y = 20 5x = 20 2y = 20 x = 4 y = 10 (4, 0) (0, 10) So we can graph the line using the intercepts. Since the y-intercept is (0, 10), we know that the point (0,0) is above the line and so will use that as our test point. Plugging this into the rst inequality, we get 5(0) 2(0) > 20 0 > 20 which is a false statement. That means that the solutions lie on the other side (below) the line 5x 2y = 20, (see the gure below.) As for the second inequality, the associated line is y = 5x 15 which means it has a slope of m = 5 and y-intercept (0, 15). Using this we can graph the line and since the inequality says that y must be less than 5x 15, this means we want the region below the line y = 5x 15 (see the gure above.) Therefore, the solution is the overlapping region. We nd the point of intersection of the two lines (the vertex of the region) by solving the system of equations 5x 2y = 20 y = 5x 15 Using the substitution method to substitute 5x 15 in for y in the rst equation, we get 5x 2(5x 15) = 20 5x 10x + 30 = 20 5x + 30 = 20 5x = 10 x = 2

22 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 22 To nd y, we substitute the 2 for x in the second equation, y = 5(2) 15 y = 5 So the lines intersect at (2, 5). This point is not included in the region and so we use a hallow circle. The graph of the nal solution is 2) The method for solving the system of inequalities is the same whether there are two, three, or more inequalities. The rst step is to graph the solution of each inequality. The line associated with x 4 is x = 4, which is a vertical line. We use a solid line and shade to the right of the vertical line, since x is greater than or equal to -4. The line associated with y > 2 is y = 2, which is horizontal line. We use a dashed line and shade everything above the line since y must be greater than 2. For the third inequality, we will rst solve it for y, to put it in slope-intercept form. 4x + 7y 26 7y 4x + 26 y 4 7 x So the associated line is y = 4 7 x Since the intercepts have fractional values, we seek an easier point to graph. If we use x = 3, y = 4 26 (3) = 7 = 14 7 = 2 Which means the point (3, 2) is on the line and using the slope m = 4 7, we can nd another point by going up 4 and to the left by 7, which gives us the point ( 4, 6) (see the gure below.)

23 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 23 The inequality y 4 7 x denes the region below the line y = 4 7 x Note that the three lines create a triangle and the region inside the triangle is where all the solutions overlap each other. So we need to nd the vertices (the corners of the triangle, point A, B, and C on the graph below.) Point A is where the lines x = 4 and y = 2 intersect, which is ( 4, 2). Point B is where the lines y = 2 and y = 4 7 x intersect. Using substitution with the variable y, we get 2 = 4 7 x + 26 ( 7 7(2) = x + 26 ) 7 14 = 4x = 4x 3 = x So point B is (3, 2). Point C is where the lines x = 4 and y = 4 7 x variable x, we get intersect. Using substitution with the y = 4 7 ( 4) = = 6 So point C is (3, 6). Note that looking at the graph above, it looked as though we could read the coordinates of A, B, and C from the graph. However, we need to check analytically (using the equations) to be sure of the exact coordinates as the graph may not be accurate. Note that points A and B are not included because points on the line y = 2 are not in the solution and so we use hallow circles to represent them. Point C is included because points on the line

24 CHAPTER 6. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 24 x = 4 and y = 4 7 x graph of the region is are in the solution and so we use a lled circle to represent C. So the

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