Chapter 2 Linear Systems in Two Variables and Inequalities with Applications Section 2.1

Section 2.1 Graphical and Numerical Solutions Definition of a System of Equations Solving a Linear System in Two Variables by Graphing Classification of Linear Systems Graphical Modeling with Linear Systems

Definition of a System of Equations In many real-world scenarios, sometimes we need to work with two or more equations to describe a particular situation. A system of equations is a set of two or more equations to be solved together, where the solution must satisfy all the equations simultaneously. The equations in the system can be linear or nonlinear. In this chapter, we examine linear systems consisting of two equations in two variables (“2 x 2” systems).

Example: x + y = 30 x – y = 10 Both equations, independently, would have an infinite number of solutions. x + y = 30: (10, 20), (15, 15), (0, 30), (20, 10), ( – 10, 40), … x – y = 10: (40, 30), (0, – 10), (20, 10), (30, 20), (15, 5), … Only one point satisfies both equations simultaneously. Therefore, we say that the point (20, 10) is the solution to the system of equations formed by these two linear equations.

Solving a Linear System in Two Variables by Graphing Graph each equation and see where the lines intersect. The intersection point is where the (x, y) coordinates will be exactly the same for both equations. Since the solution to the linear system will be a point that satisfies both equations, the intersection point will give us the solution to the system.

Using grid paper, solve the linear system graphically. Label both axes and show your scale and all tick marks. x + y = 30 x – y = 10 The point of intersection of the lines is (20, 10). We can check that this ordered pair is a solution to the linear system by substituting (20, 10) into the equations: x + y = = 30 True x – y = – 10 = 10True

Use the graphing calculator to solve the system graphically. y = –3x + 2 y – 5 = 2x 1. Input each equation into Y1 and Y2, respectively. (Make sure each y is isolated.) 2. Graph using an appropriate window. We are using [–10, 10, 1] by [–6, 10, 1]. (Window may vary.) 3. Use the Intersection feature. Keystrokes: 2nd TRACE(CALC) 5(Intersect) ENTER ENTER ENTER The solution to the system is: (–0.6, 3.8)

The average teen sends 3,339 texts per month. (Source: Jeremy and April sent 2,580 text messages last month. April sent twice as many texts as Jeremy. Let x = number of text messages that Jeremy sent, and y = number of text messages that April sent. Establish and solve a linear system to find how many text messages each teen sent. Jeremy’ texts + Aprils' texts = 2580 x + y = 2580 April sent twice as many texts as Jeremy. y = 2x We will solve the system graphically. (Intersection of both lines.) Jeremy sent 860 text messages and April sent 1,720 text messages.

Solve the following system graphically. y – 5.2 = –3.1x y + 3.1x = –7 We will graph both equations using the standard window. Observe that both lines in the system of equations have slope –3.1. Recall: parallel lines have equal slopes and different y-intercepts. Since these lines are parallel, there will be no point of intersection and the given system has no solution. In other words, both equations, independently, would have an infinite number of solutions, but there is no point that satisfies both equations simultaneously.

Classification of Linear Systems One SolutionNo SolutionInfinite Number of Solutions Unique point of intersection Parallel lines (no intersection) Lines coincide (infinite points of intersection) Consistent (has a solution) Inconsistent (has no solution) Consistent (all points are solutions) Independent lines Dependent lines

Which of the following applies to the graph of the linear system whose graph is shown below? a. The solution set consists of the points (1, 0) and (0, 4). b. This is a consistent and independent system. c. The system is inconsistent and dependent. d. The system has infinitely many solutions. The lines coincide, which means that all the points on one line will be exactly the same points for the second one. Therefore, we say that this system has infinitely many solutions. Answer is: d

Applications involving Break-Even Related Functions Revenue: Total amount of money received by a business for its products or services. If a business produces and sells x items, R(x) = price  quantity Cost: Fixed costs (e.g., rent, equipment, etc.) plus Variable costs (e.g., labor, raw materials, utilities, etc.) C(x) = fixed cost + variable cost Profit: Amount of money made after all costs and expenses are paid. P(x) = Revenue – Cost or R(x) – C(x) Break-even point: Expenses or costs and the revenue are equal; that is, there is neither a profit nor a loss (the profit is 0). Cost = Revenue or C(x) = R(x)

The production company of Avatar (2009) reported a total cost of $300 million production budget. The film had an approximate cost of $2.50 per projected ticket. The national average ticket price for all movies during January of 2010 was $7.50. Sources: ww.nytimes.com; a. Find the revenue function, R(x), and the cost function, C(x), as functions of the number of tickets, x, sold. R(x) = price  quantity: R(x) = 7.5x C(x) = fixed cost + variable cost: C(x) = 300,000, x b. Solve the system graphically to find how many tickets the movie had to sell to break even. We are looking for the point where R(x) = C(x). The movie must have sold 60,000,000 tickets to break even.

Using your textbook, practice the problems assigned by your instructor to review the concepts from Section 2.1.