1. H.Melikian/12101 System of Linear Equations in two variables (4.1) Dr.Hayk Melikyan Departmen of Mathematics and CS melikyan@nccu.edu 1. Solve by graphing 2. Solve using substitution 3. Solve by elimination by addition 4. Applications

2. H.Melikian/12102 Opening example v A restaurant serves two types of fish dinners- small for $5.99 each and a large order for $8.99. One day, there were 134 total orders of fish and the total receipts for these 134 orders was $1024.66. How many small orders and how many large fish plates were ordered?

3. H.Melikian/12103 Systems of Two Equations in Two variables v Given the linear system v ax + by = c v dx + ey = f v A solution is an ordered pair that will satisfy each equation (make a true equation when substituted into that equation). The solution set is the set of all ordered pairs that satisfy both equations. In this section, we wish to find the solution set of a system of linear equations.

4. H.Melikian/12104 Solve by graphing v One method to find the solution of a system of linear equations is to graph each equation on a coordinate plane and to determine the point of intersection (if it exists). The drawback of this method is that it is not very accurate in most cases, but does give a general location of the point of intersection. Lets take a look at an example: v Solve the system by graphing: 3x + 5y = -9 x+ 4y = -10

5. H.Melikian/12105 Solve by graphing v 3x + 5y = -9 v x+ 4y = -10 v Intercept method: v If x = 0, y= -9/5 v If y = 0, x = - 3 v Plot points and draw line v Second line: v (0, -5/2), ( -10,0) v From the graph we see that the point of intersection is (2,-3). v Check: 3(2)+5(-3)= -9 and 2+4(-3)= -10 both check. (2,-3)

6. H.Melikian/12106 Another example: v Now, you try one: v Solve the system by graphing: v 2x+3 =y v X+2y = -4 v The solution is (-2,-1)

7. H.Melikian/12107 Method of Substitution v Although the method of graphing is intuitive, it is not very accurate in most cases. There is another method that is 100% accurate- it is called the method of substitution. This method is an algebraic one. This method works well when the coefficients of x or y are either 1 or -1. For example, let’s solve the previous system v 2x + 3 = y v x + 2y = -4 using the method of substitution. This steps for this method are as follows: 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.

8. H.Melikian/12108 Method of substitution 2x+3 = y x +2y=-4 x + 2(2x+3)=-4 x +4x+6 = -4 5x+6 = -4 5x= -10 x = -2 v If x = -2 then from the first equation, we have 2(-2)+3 = y or v -1 = y. Our solution is v (-2, -1)

9. H.Melikian/12109 Another example: v Solve the system using substitution: 3x-2y=-7 y = 2x -3 v Solution:

10. H.Melikian/121010 Terminology: v 1. A consistent linear system is one that has one or more solutions. –A) If a consistent system has exactly one solution then it is said to be independent. An independent system will occur when two lines have different slopes. –B) if a consistent system has more than one solution, then it is said to be dependent. A dependent system will occur when two lines have the same slope and the same y intercept. In other words, the two equations are identical. The graphs of the lines will coincide with one another and there will be an infinite number of points of intersection. v 2. An inconsistent linear system is one that has no solutions. This will occur when two lines have the same slope but different y intercepts. In this case, the lines will be parallel and will never intersect.

11. H.Melikian/121011 An example: v Determine if the system is consistent, independent, dependent or inconsistent: v 1) 2x – 5y = 6 v -4x + 10y = -1 v Solve each equation for y to obtain the slope intercept form of the equation: v v Since each equation has the same slope but different y intercepts, they will not intersect. This is an inconsistent system

12. H.Melikian/121012 Elimination by Addition v The method of substitution is not preferable if none of the coefficients of x and y are 1 or -1. For example, substitution is not the preferred method for the system below: 2x – 7y=3 -5x + 3y = 7 –A better method is elimination by addition: First of all, we need to know what operations can be used to produce equivalent systems. They are as follows: 1. Two equations can be interchanged. 2. An equation is multiplied by a non-zero constant. 3. An equation is multiplied by a non-zero constant and then added to another equation.

13. H.Melikian/121013 Elimination by Addition v For our system, we will seek to eliminate the x variable. The coefficients of the x variables are 2 and -5. The least common multiple of 2 and 5 is 10. Our goal is to obtain coefficients of x that are additive inverses of each other. v We can accomplish this by multiplying the first equation by 5 Multiply second equation by 2. v Next, we can add the two equations to eliminate the x-variable. v Solve for y v Substitute y value into original equation and solve for x v Write solution as an ordered pair v S: 2x – 7y = 32x – 7y = 3

14. H.Melikian/121014 Solve using elimination by addition v Solve 2x-5y = 6 -4x+10y=-1 v 1. Eliminate x by multiplying equation 1 by 2. v 2. Add two equations v 3. Upon adding the equations, both variables are eliminated producing the false equation v 0 = 11 v 4. Conclusion: If a false equation arises, the system is inconsistent and there is no solution. v Solution:

15. H.Melikian/121015 Applications v 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. How long does the man jog? v Solution: Let x represent the amount of time spent walking and y represent the amount of time spent jogging. Since the total time spent walking and jogging is 0.9 hours, we have the equation v x +y = 0.9. We are given the total distance traveled as 3.5 miles. Since Distance = Rate x time, we have [distance walking] + [distance jogging] = [total distance]. Distance walking = 3x and distance jogging = 5y. Then [distance walking] plus [distance jogging] 3x+5y= 3.5.

16. H.Melikian/121016 Application continued v We can solve the system using substitution. v 1. Solve the first equation for y v 2. Substitute this expression into the second equation. v 3. Solve 2 nd equation for x v 4. Use this x value to find the y value v 5. Answer the question. v 6. Time spent jogging is 0.4 hours. v Solution:

17. H.Melikian/121017 Now, solve the opening example: A restaurant serves two types of fish dinners- small for $5.99 each and a large order for $8.99. One day, there were 134 total orders of fish and the total receipts for these 134 orders was $1024.66. How many small orders and how many large fish plates were ordered? (60 small orders and 84 large orders)

18. H.Melikian/121018 SYSTEMS OF TWO EQUATIONS IN TWO VARIABLES Given the LINEAR SYSTEM ax + by = h cx + dy = k where a, b, c, d, h, and k are real constants, a pair of numbers x = x 0 and y = y 0 [also written as an ordered pair ( x 0, y 0 )] is a SOLUTION to this system if each equation is satisfied by the pair. The set of all such ordered pairs is called the SOLUTION SE T for the system. To SOLVE a system isto find its solution set.

19. H.Melikian/121019 SYSTEMS OF LINEAR EQUATIONS: BASIC TERMS A system of linear equations is CONSISTENT if it has one or more solutions and INCONSISTENT if no solutions exist. Furthermore, a consistent system is said to be INDEPENDENT if it has exactly one solution (often referred to as the UNIQUE SOLUTION ) and DEPENDENT if it has more than one solution

20. H.Melikian/121020 The system of two linear equations in two variables ax + by = h cx + dy = k can be solved by: (a)graphing; (b)substitution; (c)elimination by addition.

21. H.Melikian/121021 POSSIBLE SOLUTIONS TO A LINEAR SYSTEM The linear system v ax + by = h cx + dy = k must have: v (a)exactly one solution (consistent and independent); or v (b)no solution (inconsistent); or v (c)infinitely many solutions (consistent and dependent).

22. H.Melikian/121022 This method is algebraic method and leads to exact solution if it exists. In this method 1. Choose one of two equations in system and solve one variable in terms of other. 2. Then substitute the result in second equation and solve the resulting equation. 3. Substitute result of second step back in to first to complete solution.

23. H.Melikian/121023 Example ( substitution method) y = 2x - 3 (1) x + 2y = 14 (2) By substituting y from (1) into (2), we get: x + 2(2x - 3) = 14 x + 4x - 6 = 14 5x = 20 x = 4 Now, substituting x = 4 into (1), we have: y = 2(4) - 3 y = 5 Solution: x = 4 y = 5 ( 4, 5)

24. H.Melikian/121024 Example ( addition method) 2m - n = 10 (1) m - 2n = -4 (2) Multiply (1) by -2 and add to (2) to obtain: -3m = -24 m = 8 Substituting m = 8 into (2), we get: 8 - 2n = -4 -2n = -12 n = 6 Solution: m = 8 or (8,6) n = 6

25. H.Melikian/121025 Equivalent Systems Two systems of linear equations are EQUIVALENT if they have exactly the same solution set. A system of linear equations is transformed into an equivalent system if: (a)two equations are interchanged; (b)an equation is multiplied by a nonzero constant; (c)a constant multiple of one equation is added to another equation.