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1 Learning Objectives for Section 4.1 After this lesson, you should be able to solve systems of linear equations in two variables by graphing solve these.

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Presentation on theme: "1 Learning Objectives for Section 4.1 After this lesson, you should be able to solve systems of linear equations in two variables by graphing solve these."— Presentation transcript:

1 1 Learning Objectives for Section 4.1 After this lesson, you should be able to solve systems of linear equations in two variables by graphing solve these systems using substitution solve these systems using elimination by addition solve applications of linear systems. Review: Systems of Linear Equations in Two Variables

2 2

3 3 System of Linear Equations System of Linear Equations: refers to more than one equation being graphed on the same set of coordinate axes. Solution(s) of a System: the point(s) (ordered pair) at which the lines ________________. The solution of a system is an ordered pair(s) that will satisfy all of the equations in the system.

4 4 Solving: Graphically 1.On a single set of coordinate axes, graph each equation. Label the equation of each line. 2.Find the coordinates of the point where the graphs intersect. These coordinates give the solution of the system. Label this point. 3.If the graphs have no point in common, the system has no solution. 4.Check the solution in both of the original equations.

5 5 Solve by Graphing Example 1: Solve the following system by graphing: x y Check your solution!

6 6 Checking… Solution:

7 7 Consistent Systems Consistent system- a system of equations that has a solution. (the system has at least one point of intersection) one solution existsinfinitely many solutions exist

8 8 Inconsistent Systems Inconsistent system- a system of equations that has no solutions. (the lines are parallel) no solution exists

9 9 Independent and Dependent Equations Independent Equations- the equations graph different lines Dependent Equations- the equations graph the same line one solution existsno solution exists infinitely many solutions exist

10 10 Solving a System on the Calculator Step 1: Graph each equation. (Use your calculator.) 4x – y = 9 x –3y = 16 Step 2: Find the coordinates of the point of intersection. 2 nd CALC5: intersect First curve? (make sure cursor is on the line for y1.) ENTER Second curve? (make sure cursor on the line for y2.) ENTER Guess? (cursor to where you feel the intersection is) ENTER *Make sure the equations are in slope-intercept form!

11 11 Calculator Example Continued… Solution: What type of system is this? What type of equations are these?

12 12 Example Example: Solve the system by graphing on your calculator.

13 13 Example with Dependent Equations Example: Solve the system by graphing on your calculator.

14 14 Solving: Method of Substitution 1)Solve one of the equations for either x or y. 2)Substitute that result into the other equation to obtain an equation in a single variable (either x or y). 3)Solve the equation for that variable. 4)Substitute this value into any convenient equation to obtain the value of the remaining variable. 5)Check solution in BOTH ORIGINAL equations. 6)Write your solution as an ordered pair. If there is no solution, state that the system is inconsistent.

15 15 Example Example: Solve the system using substitution and check.

16 16 Another Example Example: Solve the system using substitution and check.

17 17 Solving: Method of Elimination by Addition 1)Write both equations in the general form Ax + By = C. 2)Multiply the terms of one or both of the equations by nonzero constants to make the coefficients of one variable differ only in sign. 3)Add the equations and solve for the variable. 4)Substitute the value into one of the ORIGINAL equations to find the value of the other variable. 5)Check the solution in BOTH ORIGINAL equations. 6)Write your solution as an ordered pair. If there is no solution, state that the system is inconsistent.

18 18 Addition Method Example Example: Solve the system using addition method and check.

19 19 Another Addition Method Example Example: Solve the system using addition method and check.

20 20 Another Example Example: Solve the system using addition method.

21 21 Examples Example: Solve the system using ANY method.

22 22 Examples Example: Solve the system using ANY method.

23 23 Special Cases Summary When solving a system of two linear equations in two variables algebraically: If an identity is obtained, such as 0 = 0, then the system has an infinite # of solutions.  The equations are dependent  The system is consistent.

24 24 Special Cases Summary When solving a system of two linear equations in two variables algebraically: If a contradiction is obtained, such as 0 = 7, then the system has no solution.  The system is inconsistent.  The equations are independent.

25 25 Application A man walks at a rate of 3 miles per hour and jogs at a rate of 5 miles per hour. He walks and jogs a total distance of 3.5 miles in 0.9 hours. For how many minutes does the man jog? Solution: Let x represent the _____________________________________ Let y represent the _____________________________________

26 26 Application (continued)

27 27 Supply and Demand The quantity of a product that people are willing to buy during some period of time is related to its price. Generally, the higher the price, the less the demand; the lower the price, the greater the demand.

28 28 Supply and Demand (continued) Similarly, the quantity of a product that a supplier is willing to sell during some period of time is also related to the price. Generally, a supplier will be willing to supply more of a product at higher prices and less of a product at lower prices.

29 29 Supply and Demand (continued) The simplest supply and demand model is a linear model where the graphs of a demand equation and a supply equation are straight lines.

30 30 Supply and Demand (continued) In supply and demand problems we are often interested in finding the price at which supply will equal demand. This is called the equilibrium price, and the quantity sold at that price is called the equilibrium quantity.

31 31 Supply and Demand (continued) If we graph the the supply equation and the demand equation on the same axis, the point where the two lines intersect is called the EQUILIBRIUM POINT. Its horizontal coordinate is the value of the equilibrium quantity (q). Its vertical coordinate is the value of the equilibrium price (p). (q, p)

32 32 Supply and Demand Example Example: Suppose that the supply equation for long-life light bulbs is given by (supply)p = 1.04 q , and that the demand equation for the bulbs is (demand) p = -0.81q where q is in thousands of cases and p represents the price per bulb in dollars. Find the equilibrium price and quantity.

33 33 Supply and Demand Example Method I: Solve the system algebraically We want to find the price at which the supply is equal to the demand. We can do this by __________________________________ ________________________________________________. Supply: p = 1.04 q Demand: p = -0.81q + 7.5

34 34 Supply and Demand Example (continued)

35 35 Supply and Demand Example Method II: Solve the system graphically Graph the two equations in the same coordinate system using a graphing calculator and find the ______________________ _________________________________________________. p = 1.04 q p = -0.81q y1= 1.04 x y2= -0.81x Notice x represents the quantity and y represents the price.

36 36 Supply and Demand (Example continued) If we graph the two equations on a graphing calculator and find the intersection point, we see the graph below. Thus the equilibrium point is (7.854, 1.14). The equilibrium quantity is ____________ cases and the equilibrium price is $____________ per bulb. Supply Curve Demand Curve

37 37 Another Example of a Linear System A restaurant serves two types of fish dinners- small for $5.99 each and large for $8.99. One day, there were 134 total orders of fish, and the total receipts for these 134 orders was $ How many small dinners and how many large dinners were ordered?

38 38

39 39 Break-Even Analysis P 187 #56 A small plant manufactures riding lawn mowers. The plant has fixed costs (leases, insurance, etc.) of $65,000 per day and variable costs (labor, materials, etc.) of $1100 per mower produced. The mowers are sold for $1,600 each. A) Find the cost and revenue functions. Let x = ____________________________________________ Cost function: Revenue function:

40 40 Break-Even Analysis (continued) B) How many units must be manufactured and sold each day for the company to break even? C) Write the profit function, P(x).

41 41 Break-Even Analysis (continued) D) How many units must be manufactured and sold each day for the company to make a profit?


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