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Systems of Linear Equations: Substitution and Elimination.

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Presentation on theme: "Systems of Linear Equations: Substitution and Elimination."— Presentation transcript:

1 Systems of Linear Equations: Substitution and Elimination

2 A system of equations is a collection of two or more equations, each containing one or more variables. A solution of a system of equations consists of values for the variables that reduce each equation of the system to a true statement. When a system of equations has at least one solution, it is said to be consistent; otherwise it is called inconsistent. To solve a system of equations means to find all solutions of the system.

3 An equation in n variables is said to be linear if it is equivalent to an equation of the form whereare n distinct variables, are constants, and at least one of the a’s is not zero.

4 If each equation in a system of equations is linear, then we have a system of linear equations.

5 Two Linear Equations Containing Two Variables

6 If the graph of the lines in a system of two linear equations in two variables intersect, then the system of equations has one solution, given by the point of intersection. The system is consistent and the equations are independent. Solution y x

7 If the graph of the lines in a system of two linear equations in two variables are parallel, then the system of equations has no solution, because the lines never intersect. The system is inconsistent. x y

8 If the graph of the lines in a system of two linear equations in two variables are coincident, then the system of equations has infinitely many solutions, represented by the totality of points on the line. The system is consistent and dependent. x y

9 Two Algebraic Methods for Solving a System 1. Method of substitution 2. Method of elimination

10 STEP 1: Solve for x in (2) Use Method of Substitution to solve: (1) (2) add x and subtract 2 on both sides

11 STEP 2: Substitute for x in (1) STEP 3: Solve for y

12 STEP 4: Substitute y = -11/5 into result in Step1. Solution:

13 STEP 5: Verify solution in

14 Use graphing utility to solve the previous system of equations Solve for y in both equations. This is equivalent to writing both equations in slope-intercept form. Equation (1) in slope intercept form is Y 1 = -2x + 7. Equation (2) in slope intercept form is Y 2 = (- x - 2)/3 Graph both using graphing utility and use INTERSECT, to obtain the solution (4.6, -2.2) = (23/5, -11/5). (1) (2)

15 Rules for Obtaining an Equivalent System of Equations 1. Interchange any two equations of the system. 2. Multiply (or divide) each side of an equation by the same nonzero constant. 3. Replace any equation in the system by the sum (or difference) and a nonzero multiple of any other equation in the system.

16 Multiply (2) by 2 Replace (2) by the sum of (1) and (2) Equation (2) has no solution. System is inconsistent. Use Method of Elimination to solve: (1) (2)

17 Multiply (2) by 2 Replace (2) by the sum of (1) and (2) The original system is equivalent to a system containing one equation. The equations are dependent. Use Method of Elimination to solve: (1) (2)

18 This means any values x and y, for which 2x -y =4 represent a solution of the system. Thus there is infinitely many solutions and they can be written as or equivalently

19 Exactly one solution (consistent system with independent equations). No solution (inconsistent system). Infinitely many solutions (consistent system with dependent equations). A system of three linear equations containing three variables has either

20 Solve: (1) (2) (3) Step 1: Multiply (1) by -2 get -2x - y - 2z =-8 and add to (2). (1) (2) (3)

21 Step 2: Add (1) to (3). Step 3: Add (2) to (3). (1) (2) (3) (1) (2) (3)

22 Step 4: Now that z=3 substitute that into (2) to solve for y. Step 5: Substitute z=3 and y=-1 into (1) and solve for x. Solution (x, y, z) = (2, -1, 3).

23 Bob and Mary went to the movies. Bob purchased 3 medium bags of popcorns and 2 large drinks and paid $11.00. Mary purchased 2 medium bags of popcorns and 4 large drinks and ended up paying $12.00. What was the price of a medium bag of popcorn and what was the price of a large drink? Denote p the price of a medium bag of popcorn and d the price of a large drink. Then we have 3p + 2d = 11 2p + 4d = 12

24 3p + 2d = 11 2p + 4d = 12 -6p - 4d = -22 2p + 4d = 12 -4p = -10 p = 2.50 Substituting back and solving for d we get d = 1.75. (multiply the first by -2) (add the first to the second) Medium bag of popcorn costs $2.50 and a large drink costs $1.75.


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