1 . Systems of Linear Equations Question : What is a system of linear equations? Question : Where do systems of equations come from? In Chapter, we looked at several applications of linear functions. One set of linear functions was the cost and revenue functions for a dairy. These functions can be analyzed to determine when the dairy farm is making money and when it is losing money. The break-even point is the point at which a farm s costs are equal to its revenue. In this section we ll learn how to find this point graphically. We also looked at supply and demand functions for milk. These functions model the consumer s and supplier s behavior with respect to the quantity and price of milk in a market. In economics, we are interested in knowing how the quantity of milk sold is related to the price of milk. The equilibrium point describes the price of a commodity, such as milk, when the quantity demanded by consumers matches the supply that manufacturers are willing to provide. In this section we ll locate the equilibrium point graphically. For both of these applications, we need to find a point of intersection of two lines. Many other applications require the same mathematical process. At the end of this section we ll build an application from the ground up. In this application, we ll learn how different types of gasoline containing ethanol can be blended together to yield a mixture with a particular volume and level of ethanol.
2 Question : What is a system of linear equations? In many applications there is more than one equation that governs the solution to the problem. For instance, suppose we want to process 00,000 gallons of raw milk into whole, %, and % milk. Each type of milk sells for $.49, $4.9 and $4.59 per gallon respectively. The dairy needs $44,000 in revenue from the sale of the milk. If Q, Q, and Q are the amounts of whole, % and % milk processed in gallons, the solution to this problem must satisfy Q Q Q 00,000 gallons of raw milk.49q 4.9Q 4.59Q 44,000 dollars of revenue By using the term satisfy, we mean that when values are substituted into the variables Q, Q, and Q, the left hand side of each of the equations is equal to the right hand side of the same equation. When the two sides of the equation yield the same values, we say that the values result in a true statement. If the values substituted into the equation result in different values on each side of the equation, we say that the values result in a false statement. There are two equations for this problem since the dairy wants to use all of the raw milk and have $44,000 in revenue from the sales of the processed milk. Any combination that allows the dairy to meet both of these requirements is a reasonable solution to this application. To check whether there exists a combination of milk products that works, we need to substitute values for the variables into each equation. Example Do Specific Combinations Solve Both Equations? For each part below, determine if the combination of milk is a solution to the equations Q Q Q 00, Q 4.9Q 4.59Q 44, 000
3 where Q, Q, and Q are the amounts of whole, % and % milk processed in gallons. a. 50,000 gallons of whole milk, 50,000 gallons of % milk and no % milk. Solution For this combination of milk, we have Q equal to 50,000, Q equal to 50,000, and Q equal to 0. If we substitute these values into the first equation, we get 50,000 50, ,000 TRUE Since the left hand side of the equation is equal to 00,000, this combination of milk results in a true statement and combines to give the proper amount of milk. However, when we put this combination of milk products in the second equation, we get?.49 50, , ,000 84,000 44,000 FALSE The left hand side of the equation simplifies to 84,000 to yield a false statement. This combination of milk products does not solve the second equation. This combination solves the first equation so the amounts total to 00,000 gallons of milk, but does not solve the second equation. The amount of revenue from the sales does not equal $44,000. b. 0,000 gallons of whole milk, 60,000 gallons of % milk and 0,000 gallons of % milk. Solution Put the values Q 0,000, Q 60,000, and Q 0,000 into each equation:
4 0,000 60,000 0,00000,000? 00,000 00,000 TRUE.49 0, , ,000 44,000? 44,000 44,000 TRUE yields two true statements. Since this combination makes each equation true, we know that it gives both the proper amount of milk and the proper revenue. For this set of equations, each variable was raised to the first power so the equations are linear in each variable. When the combination of values solves each of the linear equations, the values for the variables are a solution to the system of linear equations formed by the two equations. Let s generalize these ideas to help us understand exactly what a system of linear equations is. A system of linear equations consists of several equations containing variables and constants. Recall that variables are letters or symbols that represent unknown quantities. In any of the linear equations, the variables can vary. Constants are values that do not change in an equation. Constants may be numbers like or -5. Or a constant may be a letter that represents a fixed but unknown number. As illustrated in Example, the equations in a system of linear equations can have any number of variables. It is best to name them in a systematic way. Since we were referring to quantities of milk, the letter Q was useful. There were three variables so subscripts distinguish each type of milk and yield Q, Q, and Q. For business problems we need to allow for any number of variables representing a variety of quantities. We could use variables without subscripts like x and y, but we would run out of names quickly for most realistic applications. If we use the letter x, n variables are represented by x, x,, xn. Letters representing constants can be handled in a similar manner. We can represent n constants using the letter a as a, a,, an. The 4
5 single constant on the right side of a linear equation is represented by the letter k. With these constants and variables, a definition for a linear equation can be written. Keep in mind that there is nothing special about the letters used and any letter could be used in place of a, k or x. However, since a, a,, an and k are constants, these letters represent fixed values. This makes them different from the variables x, x,, xn which represent values that can vary in the equation. A linear equation in n variables x, x,, xn is any equation that can be written in the form ax ax ax k n n where a, a,, an and k are constants representing fixed values. Two special cases must be noted. Many simple problems involve only two variables. We could write the linear equation as ax ax k, but often the letters a, b, x, and y are used. An equivalent linear equation in two variables is axby k. Similarly, for three variables you can write ax ax ax k or axbycz k. For more than three variables we run out of letters at the end of the alphabet and resort to subscripts. The key concept is that each term in the equation has first degree. Another way to say this is that the variable in each term is raised to the first power. As long as each term is a first degree term, the equation is called a linear equation. If we collect several linear equations in the same variables together, we get a system of linear equations. 5
6 A system of linear equations in n variables is a finite number of equations where each equation can be written in the form ax ax ax k n n A solution to the system is a collection of values for the variables that makes all of the equations true. Example Do Certain Values Solve the System of Linear Equations? The equations y x 5 y x form a system of linear equations since the collection of equations can be written as x y5 x y Do the values x 7 and y solve the system? Solution To see if the values solve the system, we need to put the values into each equation and make sure both equations are true.?? TRUE TRUE 6
7 Since both equations are true, the values x 7 and y solve the system of linear equations. It is convenient to think of this solution in Example as an ordered pair xy, and to write the solution as 7,. There is a good reason for doing this. The equations in Example, y x5 y x both have graphs that are lines. These lines intersect at the solution to the system of linear equations. The first equation corresponds to a line with a slope of and a vertical intercept of -5. The second line has a slope of - and a vertical intercept of. Figure The two lines corresponding to the system of linear equations intersect at the solution (7, ). Graphing each equation is a simple way of estimating the solution to a system of equations. If the system of equations contains only two variables, the solution to the system is the point of intersection of all lines in the system if such a point exists. 7
8 Example Find the Point of Intersection Find the solution to the system of equations x0y 00 50x0y 40.9 by finding where the graphs of the equations intersect. Solution To graph each equation, solve it for y. For the first equation, 0y x 00 y 0 x0 Divide both sides by 0 Add x to both sides To solve the second equation in the system for y, 0y 50x40.9 y 50 0 x.9 Divide both sides by 0 Add 50x to both sides The system can now be written as y y x0 x.9 Each of these equations is in the slope-intercept form for a line, y mx b. In this form, m is the slope of the line and b is the vertical intercept of the line. 8
9 Figure A graph of the system in Example. It is a good idea to make sure the lines are graphed properly. The blue line has a vertical intercept of 0 and a slope that is small and positive, The red line has a vertical intercept of.9 and a slope that is larger, Because the slope of the red line is larger than the slope of the blue line, the red line is steeper than the blue line. Based on the graph in Figure, the point of intersection appears to be about 0, 9. However this can t be correct since x 0 and y 9 do not satisfy the two equations in the system ? FALSE ? FALSE The best we can say, based on the graph, is that the solution is close to (-0, 9). We ll need to use a graphing calculator or algebraic techniques to be more precise. However, keep in mind that a graphing calculator 9
10 will give a better answer, but is still an approximation. Algebraic techniques give the exact answer as long as numbers are not rounded in the process of finding the point of intersection. By substituting these values into the system of equations, we can determine that these values are the exact solution:?? TRUE TRUE When utilizing technology to find a solution, keep in mind that the solution is only an estimate unless the solution satisfies each equation in the system.. Each of the linear systems of equations we have considered has a single point of intersection. You can easily imagine a situation where two lines do not intersect. A system of equations for which the graphs do not intersect has no solutions. Example 4 Find the Solution to the System of Equations Solve the system of equations x y 7 0.5x0.75y 6.75 by graphing both equations on the same graph. Solution To graph each equation, we need to solve each equation for y. x y 7 yx7 y x 7 y x9 Subtract x from both sides Divide each side by - Simplify 0
11 0.5x0.75y y 0.5x y 0.5x y x9 Add 0.5x to both sides Divide each side by 0.75 Simplify When each equation is solved for y, we get an equivalent system of equations, y y x9 x9 Figure A system of equations whose graphs are parallel. Examining this system, we see that the slope of each line is and the y intercepts are different. The lines in the system, graphed in Figure, are parallel and so the system has no solution.
12 Example 5 Find the Solution to the System of Equations Graph the system of linear equations x y x0 y 5 to find the solution of the system. Solution The first equation is already solved for y, but the second is not. Start by solving the second equation for y: x y 5 y x y x5 5 y x0 Subtract x from both sides Multiply both sides by Simplify This equation is identical to the first equation so its graph is the same as the first equation. In other words, the two equations are different descriptions of the same line. Figure 4 The graphs of the two equations in Example 5 coincide everywhere along the line.
13 Any point on the line is a solution to the system of equations. For instance, the ordered pair (-, 6) is a solution,?? TRUE 5 5 TRUE as well as the ordered pair (,4):?? TRUE 55 TRUE We can symbolize this fact by writing the solution as any ordered pair x, y such that y x 0. Since both equations represent the same line, we could also say that the solution is any ordered pair x, y such that x y. 5
14 Question : Where do systems of equations come from? In Examples 4 and 5 of section., we found the total cost CQand ( ) the total revenue RQ ( ) for a dairy as a function of the number of cows, Q. The total cost function, CQ ( ) 890.8Q 68688, and the revenue function, RQ ( ) 547Q, are graphed in Figure 5. Figure 5 Revenue and cost functions for the dairy. As the number of cows at the dairy is increased, the cost to maintain the dairy and the revenue from dairy products both increase. In the window shown in Figure 5, the two graphs do not intersect. But the revenue function has a slope of 547 and the cost function has a slope of Since the revenue function is steeper than the cost function they will eventually intersect. We can form a system of equations whose solution corresponds to the point of intersection by replacing the function notation with a variable. Let Y represent the amount of money in dollars and use Y to replace both RQ ( ) and CQ. ( ) This is reasonable since both functions output amounts of money in dollars. The system Y Y 890.8Q Q 4
15 allows us to see the relationship between the revenue and cost for a business. By graphing this system, we can interpret what the point of intersection means in the context of the dairy and other types of businesses. Example 6 Find the Point of Intersection for the Dairy Graph the system of equations Y Y 890.8Q Q to find the solution to the system. Solution Each equation in the system is solved for the dependent variable Y and can be graphed directly. Figure 6 The system of equations representing the cost (blue) and revenue (red) of a dairy. To find an estimate of the location of the point of intersection, we can use a graphing calculator or other type of technology. 5
16 Figure 7- Screenshot on a graphing calculator of the system of equations in Example 6. The point of intersection is at Q and Y This tells us that at approximately dairy cows, the total cost of maintaining the herd is about $7,8.66 and the revenue generated by the herd is about $7,8.66. At this point, the total cost and revenue are exactly the same. A break-even quantity of approximately dairy cows leads to an interesting point. Is it possible to have of a dairy cow? The quantity of cows Q must take on non-negative integer values. The dairy owner will have either 04 cows or 05 cows. Let s examine the costs and revenue at each of these herd sizes. At a herd size of 04, R(04) ,888 C(04) ,.0 the costs are slightly higher than the revenue by about 69,.0 68,888 or 44. dollars. This indicates that the dairy would be losing money. In terms of profit, the dairy has a negative profit since the costs are greater than the revenue. At a herd size of 05, R(05) , 45 C(05) , the revenue is slightly higher than the cost by about 7,45 7, or dollars. The dairy is making money since the profit is $ at this herd size. 6
17 Neither herd size results in the revenue equal to the cost. For realistic herd sizes, the revenue and costs are not exactly equal. However, a herd size of 05 is closer to the point of intersection. The point at which a business s costs and revenue are equal is called the break-even point. As demonstrated in Example 6, this point can be found by solving a system of equations consisting of a revenue equation and a cost equation. In specifying a break-even point, we may find that the point of intersection is at an unrealistic value due to the nature of the items being produced. A complete analysis should examine the break-even point as well as realistic values for the variable near the break-even point. We can also examine where other graphs intersect to find business-related quantities. In Chapter, we introduced supply and demand functions. The supply function SQ ( ) represents the price at which suppliers would be willing to supply Q products. The demand function DQ ( ) represents the price at which consumers would be willing to buy Q products. In a market, the demand by consumers and the supply by manufacturers may or may not be in balance. The quantity demanded by consumers may be in perfect balance with the quantity supplied by suppliers. At this price and quantity, called the equilibrium point, consumers demand the exact quantity that manufacturers are willing to supply. On a graph of the supply and demand functions, the equilibrium point is the point of intersection of the two graphs. This ordered pair is labeled Q, P, where the subscript refers to equilibrium. e e 7
18 Figure 8 - At the equilibrium point Qe, P e the quantity demanded by consumers is matched by the quantity that manufacturers are willing to supply. At prices lower than the equilibrium price, the quantity demanded by consumers is larger than the quantity supplied by manufacturers. In Figure 9, the red dashed horizontal line corresponds to a price that is lower than the equilibrium price. When the quantity demanded by consumers is larger than the amount supplied by manufacturers, there is a shortage of the product. 8
19 Figure 9 - If the price is below the equilibrium price, the quantity demanded by consumers is greater than the quantity that manufacturers are willing to supply. This results in a shortage of the item on the market. When the quantity that manufacturers are willing to supply is greater than the quantity the consumer is willing to buy, there is a surplus of the item on the market. A surplus occurs when the price of the product is greater than the equilibrium price. Figure 0 - If the price is above the equilibrium price, the quantity supplied by firms is greater than the quantity that suppliers are willing to supply. This results in a surplus of the item on the market. 9
20 A surplus of an item will lead to lower prices since it costs businesses money to keep inventory on hand. The price will tend to lower until the price reaches the equilibrium price. At this point the quantity demanded by consumers and the quantity that manufacturers are willing to supply will be equal. Example 7 Find the Milk Equilibrium Point In earlier examples, we found the supply and demand functions pictured below for a small dairy. Figure - The demand function D(Q) and the supply function S(Q) for the dairy. Find the equilibrium point for the dairy. Solution The graph in Figure shows an equilibrium point at roughly (00, ). To find a better estimate of the equilibrium point, we need to utilize some sort of technology like a graphing calculator. Start by rewriting the functions DQ ( ) and SQ ( ) with a variable. Since the output from each of these functions is a price, we ll use P. The system of equations is 0
21 P 0.05Q7.75 P 95 Q The first equation relates the quantity of milk demanded by consumers to the price. As you would expect, higher quantities are demanded when the price is lower. The second equation relates the quantity of milk that dairy farms are willing to supply to the price. Higher quantities are supplied when the price is high. Figure A screen shot of the system of equations in Example 7. The equilibrium point is at (95, ). According to the estimate in Figure 7, at an average price per gallon of three dollars, 95,000 gallons of milk will be demanded by consumers and dairies will be willing to supply 95,000 gallons of milk. Since the demand and supply are in balance, this is the equilibrium point.