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Systems of Equations and Inequalities 6 Systems of Equations and Inequalities 6.1 Functions and Equations in Two Variables 6.2 Systems of Equations and Inequalities in Two Variables 6.3 Systems of Linear Equations in Three Variables 6.4 Solutions to Linear Systems Using Matrices 6.5 Properties and Applications of Matrices 6.6 Inverses of Matrices 6.7 Determinants Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Functions and Equations in Two Variables 6.1 Evaluate functions of two variables Understand basic concepts about systems of equations in two variables Apply the method of substitution Apply graphical and numerical methods to systems of equations Solve problems involving joint variation Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Introduction Many quantities in everyday life depend on more than one variable. Examples Area of a rectangle requires both width and length. Heat index is the function of temperature and humidity. Wind chill is determined by calculating the temperature and wind speed. Grade point average is computed using grades and credit hours. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Functions in Two Variables The arithmetic operations of addition, subtraction, multiplication, and division are computed by functions of two inputs. The addition function of f can be represented symbolically by f(x,y) = x + y, where z = f(x,y). The independent variables are x and y. The dependent variable is z. The z output depends on the inputs x and y. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example For each function, evaluate the expression and interpret the result. f(5, –2) where f(x,y) = xy A(6,9), where calculates the area of a triangle with a base of 6 inches and a height of 9 inches. Solution f(5, –2) = (5)(–2) = –10. A(6,9) = If a triangle has a base of 6 inches and a height of 9 inches, the area of the triangle is 27 square inches. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example The equation V = lwh is the volume of a rectangular box. Solve V = lwh for l. Find l when w = 6.5 ft, h = 9 ft, and V = 187.2 ft3 . Solution a) b) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Systems of Equations A linear equation in two variables can be written in the form ax + by = k, where a, b, and k are constants, and a and b are not equal to 0. A pair of equations is called a system of linear equations because they involve solving more than one linear equation at once. A solution to a system of equations consists of an x-value and a y-value that satisfy both equations simultaneously. The set of all solutions is called the solution set. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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Example Solve the system symbolically. Solution Step 1: Solve one of the equations for one of the variables. Step 2: Substitute for y in the second equation. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued Step 3: Substitute x = 1 into the equation from Step 1. We find that Check: The ordered pair is (1, 2) since the solutions check in both equations. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solve the system. Solution Solve the second equation for y. Substitute 4x + 2 for y in the first equation, solving for x. The equation 4 = 4 is an identity that is always true and indicates that there are infinitely many solutions. The two equations are equivalent. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example The volume of a cylindrical container with a radius r and height h is computed by The lateral surface area S of the container, excluding the circular top and bottom, is computed by Write a system of equations whose solutions is the dimensions for the cylinder with a volume of 50 cubic centimeters and a lateral surface area of 130 square centimeters. Solve the system of equations graphically and symbolically. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example continued Solution The equation V(r,h) = 50 and S(r,h) = 130 must be satisfied. This results in the following system of nonlinear equations. Graphic Solution Let r correspond to x and h to y. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued Symbolic Solution Because Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Joint Variation A quantity may depend on more than one variable. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example The area of a triangle varies jointly as the lengths of the base and the height. A triangle with base 20 inches and height of 8 inches has area 80 square inches. Find the area of a triangle with base 9 centimeters and height 12 centimeters. Solution Let A represent the area, b the base, and h the height of the triangle. Then A = kbh for some number k. Since A = 80 when b is 20 and h is 8, Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued The familiar formula for the area of a triangle is found. So, when b = 9 centimeters and h = 12 centimeters, Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Systems of Equations and Inequalities in Two Variables 6.2 Recognize different types of linear systems Apply the elimination method Solve systems of linear and nonlinear inequalities Learn basic concepts of linear programming Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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Example Use elimination to solve each system of equations, if possible. Identify the system as consistent or inconsistent. If the system is consistent, state whether the equations are dependent or independent. Support your results graphically. a) 3x  y = 7 b) 5x  y = 8 c) x  y = 5 5x + y = 9 5x + y = 8 x  y =  2 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example continued Solution a) Eliminate y by adding the equations. Find y by substituting x = 2 in either equation. The solution is (2, 1). The system is consistent and the equations are independent. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued b) If we add the equations we obtain the following result. The equation 0 = 0 is an identity that is always true. The two equations are equivalent. There are infinitely many solutions. {(x, y)| 5x  y = 8} Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued c) If we subtract the second equation from the first, we obtain the following result. The equation 0 = 7 is a contradiction that is never true. Therefore there are no solutions, and the system is inconsistent. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solve the system by using elimination. Solution Multiply the first equation by 3 and the second equation by 4. Addition eliminates the y-variable. Substituting x = 3 in 2x + 3y = 12 results in 2(3) + 3y = 12 or y = 2 The solution is (3, 2). Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solve the system of equations. Solution Multiply the first equation by 3 and add the second equation, the variable y2 is eliminated. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued Substituting in the first equation to find y. A graph of the equations is shown to the right. The two points of intersection are shown. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Systems of Linear and Nonlinear Inequalities The graph of a linear inequality is a half-plane, which may include the boundary. The boundary line is included when the inequality includes a less than or equal to or greater than or equal to symbol. To determine which part of the plane to shade, select a test point. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Graph the solution set to the inequality x + 4y > 4. Graph the line x + 4y = 4 using a dashed line. Use a test point to determine which half of the plane to shade. Test Point x + 4y > 4 True or False? (4, 2) 4 + 4(2) > 4 True (0, 0) 0 + 4(0) > 4 False Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Graph the solution set to the inequality x2 + y ≤ 4. Solution Graph the parabola determined by y = 4  x2 using a solid line. Use a test point to determine which half of the plane to shade. Test Point x2 + y ≤ 4 True or False? (0, 0) 02 + 0 ≤ 4 True (3, 1) 32 + 1 ≤ 4 False Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solve the system of inequalities by shading the solution set. Use the graph to identify one solution. x + y ≤ 3 2x + y  4 Solution Solve each inequality for y. y ≤ x + 3 (shade below line) y  2x + 4 (shade above line) The point (4, 2) is a solution. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Linear Programming Linear programming is a procedure used to optimize quantities such as cost and profit. A linear programming problem consists of a linear objective function and a system of linear inequalities called constraints. The solution set for the system of linear inequalities is called the set of feasible solutions. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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Example Suppose a small company manufactures two products—VCR’s and DVD players. Each VCR results in a $15 profit and each DVD player provides a profit of $50. Due to demand, the company must produce at least 10 and not more than 50 VCR’s per day. The number of VCR’s cannot exceed the number of DVD players, and the number of DVD players cannot exceed 60. How many of each type should the company manufacture to obtain the maximum profit? Solution Let x = VCR and y = DVD The total daily profit P is given by P = 15x + 50y Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example continued The company produces from 10 to 50 VCR’s per day, so the inequalities x  10 and x ≤ 50 must be satisfied. VCR’s cannot exceed DVD players and the number of DVD players cannot exceed 60 indicate that x ≤ y and y ≤ 60. The number of VCR’s and DVD players cannot be negative so x  0 and y  0. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example continued Graph the constraints. The shaded region is the set of feasible solutions. The maximum value of P is 3750 at vertex (50, 60). The maximum profit occurs when 50 VCR’s and 60 DVD players are manufactured. Vertex P = 15x + 50y (10, 10) 15(10) + 50(10) =650 (10, 60) 15(10) + 50(60) = 3150 (50, 60) 15(50) + 50(60) = 3750 (50, 50) 15(50) + 50(50) = 3250 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Systems of Linear Equations in Three Variables 6.3 Learn basic concepts about systems in three variables Solve systems using elimination and substitution Solve systems with no solution Solve systems with infinitely many solutions Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Basic Concepts The solution to a system of linear equations in three variables is an ordered triple, (x, y, z). A system of linear equations can have zero, one, or infinitely many solutions. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solving with Elimination and Substitution Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solve the following system. Solution Step 1: Eliminate the variable z from equation one and two and then from equation two and three. Equation 1 Equation 2 Equation 2 times 6 Equation 3 Add Add Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued Step 2: Take the two new equations and eliminate either variable. Find x using y = 2. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued Step 3: Substitute x = 1 and y = 2 in any of the given equations to find z. The solution is (1, 2, 2). Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Eight hundred twenty zoo admissions were sold last Wednesday, which generated $4285 in revenue. The prices of the tickets were $4 for students, $7 for adults and $5 for seniors. There were 110 more student tickets sold than adults. Find the number of each type of ticket sold. Solution Let x = student tickets, y = adults and z = seniors Write a system of three equations: Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued Step 1: Eliminate z from equation 1 and 2. Multiply equation one by five and subtract equation 2. Step 2: Use the third equation and the new equation to eliminate x. Subtract the equations. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued Step 2: Substitute y = 295 into the third equation to find x. Step 3: Substitute x = 405, y = 295 into equation one to find z. There were 405 students tickets, 295 adult tickets and 120 senior tickets sold. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solve the system. Solution Step 1 Multiply equation one by 2 and add to equation two. Subtract equation two and three. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued The two equations are inconsistent because the sum of 10x + 9y cannot be both 3 and 0. Step 3 is not necessary the system of equations has no solution. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solutions to Linear Systems Using Matrices 6.4 Represent systems of linear equations with matrices Learn row-echelon form Perform Gaussian elimination Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Matrices A matrix is a rectangular array of elements. The dimension of a matrix is m  n, if it has m rows and n columns. A square matrix has the same number of rows and columns. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Express the linear system with an augmented matrix. State the dimensions of the matrix. Solution The system has two equations with two variables. The augmented matrix has dimensions 2  3. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Write the linear system represented by the augmented matrix. Solution x represents the first column y represents the second column z represents the third column The vertical line corresponds to the location of the equals sign. The last column represents the constant terms. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Gaussian Elimination Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution Use Gaussian elimination and backward substitution to solve the linear system of equations. Solution Write the augmented matrix. We need a zero in the highlighted area. We can add rows 1 and 2 denoted R1 + R2. The row that is changing is written first. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued We need a 1 in the highlighted area. We need a zero in the highlighted area. We need a 1 in the highlighted area. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued Because there is a one in the highlighted box, the matrix is now in row-echelon form. z = 2, apply backward substitution to find x and y The solution of the system is (2, 1, 2). Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Gaussian elimination may be used to transform an augmented matrix into reduced row-echelon form. It takes a bit more time but eliminates the need to use backward substitution. Transform the matrix from the previous example into row-reduced echelon form. Need a 0 in the highlighted area. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued Need zeros in the highlighted areas. The solution is (2, 1, 2). Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Use Gaussian elimination to solve the system of linear equations, if possible. Solution The last row is a false statement. 0  4. There are no solutions to the system of equations. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Properties and Applications of Matrices 6.5 Learn matrix notation Find sums, differences, and scalar multiples of matrices Find matrix products Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Matrix Notation A general element of a matrix is denoted aij. The refers to an element in the ith row, jth column. For example, a3,1 would be an element in matrix A located in the third row, first column. The matrices are equal if corresponding elements are equal. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Let aij denote a general element in A, where Identify a23, a12, and a33 Solution Element a23 is located in row 2, column 3 = 3 a12 = 2 (row 1, column 2) a33= 4 (row 3, column 3) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example If and , find a) A + B b) B  A Solution a) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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Example If possible, perform the indicated operations using a) A + 2B b) 3A  2B Solution a) b) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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Example If possible, compute each product using a) AC b) BA Solution a) The dimension of A is 2  3 and the dimension of C is 2  2. Therefore AC is undefined. The number of columns in A does not match the number of rows in C. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued b) BA B A BA 3  2 2  3 = 3  3 3  2 2  3 = 3  3 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example If find AB. Solution Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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Inverses of Matrices 6.6 Understand matrix inverses Find inverses symbolically Represent linear systems with matrix equations Solve linear systems with matrix inverses Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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Example Determine if B is the inverse of A, where Solution For B to be the inverse of A, it must satisfy the equations AB = I2 and BA = I2. B is the inverse of A. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Find A-1 if Solution Begin by forming a 2  4 augmented matrix. Perform row operations to obtain the identity matrix on the left side. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Represent the system of linear equations in the form AX = B. Solution The equivalent matrix equation is Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Write the linear system as a matrix equation in the form AX= B. Find A-1 and solve for X. Solution This inverse was found in a previous example. The solution to the system is (2, 3) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Write the linear system as a matrix equation of the form AX = B. Find A-1 and solve for X. Solution We can find the inverse by hand or with a graphing calculator. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Determinants 6.7 Define and calculate determinants Use determinants to find areas of regions Apply Cramer’s rule Learn about limitations on the method of cofactors and Cramer’s rule Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Insert both green boxes page 825 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Determine if A-1 exists by computing the determinant of the matrix A. a) b) Solution a) b) A-1 does exist A-1 does not exist Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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Example Find the minor M11 and cofactor A11 for matrix A. Solution To obtain M11 begin by crossing out the first row and column of A. The minor is equal to det B = 6(5)  (3)(7) = 9 Since A11 = (1)1+1M11, A11 can be computed as follows: A11 = (1)2(9) = 9 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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Example Find det A, if Solution To find the determinant of A, we can select any row or column. If we begin expanding about the first column of A, then det A = a11A11 + a21A21 + a31A31. A11 = 9 from the previous example A21 = 12 A31 = 24 det A = a11A11 + a21A21 + a31A31 = (8)(9) + (4)(12) + (2)(24) = 72 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Find the determinant of A and B using technology. Solution The determinant of A was calculated by hand in a previous example. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Area of Regions Determinants may be used to find the area of a triangle. If a triangle has vertices (a1, b1), (a2, b2), and (a3, b3), then its area is equal to the absolute value of D, where If the vertices are entered into the columns of D in a counterclockwise direction, then D will be positive. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Find the area of the parallelogram. Solution View the parallelogram as two triangles. The area is equal to the sum of the two triangles. The area is 1 + 7 = 8. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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Example Use Cramer’s rule to solve the linear system. Solution In this system a1 = 1, b1 = 4, c1 = 3, a2 = 2, b2 = 9 and c2 = 5 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued E = 7, F = 1 and D = 1 The solution is Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley