BLITZER LEARNING GUIDE SAMPLE Section 5.1 Systems of Linear Equations in Two Variables Procrastination makes you sick! Researchers compared college students who were procrastinators and nonprocrastinators. Early in the semester, procrastinators reported fewer symptoms of illness, but late in the semester, they reported more symptoms than their nonprocrastinating peers. In this section of the textbook, you will identify when both groups have the same number of symptoms as the point of intersection of two lines. Objective #1: Decide whether an ordered pair is a solution of a linear system. Solved Problem #1 1. Determine if the ordered pair (7, 6) is a solution of Pencil Problem #1 1. пѓ¬ 2 x пЂ 3 y пЂЅ пЂ4 the system: пѓ пѓ® 2x пЂ« y пЂЅ 4 Determine if the ordered pair (2, 3) is a solution of пѓ¬ x пЂ« 3 y пЂЅ 11 the system: пѓ пѓ® x пЂ 5 y пЂЅ пЂ13 To determine if (7, 6) is a solution to the system, replace x with 7 and y with 6 in both equations. 2 x пЂ 3 y пЂЅ пЂ4 2(7) пЂ 3(6) пЂЅ пЂ4 14 пЂ 18 пЂЅ пЂ4 пЂ4 пЂЅ пЂ4, true 2x пЂ« y пЂЅ 4 2(7) пЂ« 6 пЂЅ 4 14 пЂ« 6 пЂЅ 4 20 пЂЅ 4, false The ordered pair does not satisfy both equations, so it is not a solution to the system. Objective #2: Solve linear systems by substitution. Solved Problem #2 2. пѓ¬3 x пЂ« 2 y пЂЅ 4 Solve by the substitution method: пѓ пѓ® 2x пЂ« y пЂЅ 1 Pencil Problem #2 2. пѓ¬x пЂ« y пЂЅ 4 Solve by the substitution method: пѓ y пЂЅ 3x пѓ® Solve 2 x пЂ« y пЂЅ 1 for y. 2x пЂ« y пЂЅ 1 y пЂЅ пЂ2 x пЂ« 1 Copyright В© 2014 Pearson Education Inc. 1 College Algebra 6e 3x пЂ« 2 y Substitute: пЂЅ4 y пЂ¶пЂ·пЂё 3x пЂ« 2(пЂ2 x пЂ« 1) пЂЅ 4 3x пЂ 4 x пЂ« 2 пЂЅ 4 пЂx пЂ« 2 пЂЅ 4 пЂx пЂЅ 2 x пЂЅ пЂ2 Find y. y пЂЅ пЂ2 x пЂ« 1 y пЂЅ пЂ2(пЂ2) пЂ« 1 yпЂЅ5 The solution is пЂЁ пЂ2,5пЂ© . The solution set is пЃ»пЂЁ пЂ2,5пЂ©пЃЅ . Objective #3: Solve linear systems by addition. Solved Problem #3 3. пѓ¬4 x пЂ« 5 y пЂЅ 3 Solve the system: пѓ пѓ®2x пЂ 3y пЂЅ 7 Pencil Problem #3 3. пѓ¬3x пЂ 4 y пЂЅ 11 Solve the system: пѓ пѓ® 2 x пЂ« 3 y пЂЅ пЂ4 Multiply each term of the second equation by вЂ“2 and add the equations to eliminate x. 4x пЂ« 5y пЂЅ 3 пЂ4 x пЂ« 6 y пЂЅ пЂ14 11 y пЂЅ пЂ11 y пЂЅ пЂ1 Back-substitute into either of the original equations to solve for x. 2x пЂ 3y пЂЅ 7 2 x пЂ 3( пЂ1) пЂЅ 7 2x пЂ« 3 пЂЅ 7 2x пЂЅ 4 xпЂЅ2 The solution set is 2 пЃ»пЂЁ 2, пЂ1пЂ©пЃЅ . Copyright В© 2014 Pearson Education Inc. Section 5.1 Objective #4: Identify systems that do not have exactly one ordered-pair solution. Solved Problem #4 пѓ¬ 5x пЂ 2 y пЂЅ 4 4a. Solve the system: пѓ пѓ®пЂ10 x пЂ« 4 y пЂЅ 7 Pencil Problem #4 x пЂЅ 9пЂ 2y пѓ¬ 4a. Solve the system: пѓ пѓ® x пЂ« 2 y пЂЅ 13 Multiply the first equation by 2, and then add the equations. 10 x пЂ 4 y пЂЅ 8 пЂ10 x пЂ« 4 y пЂЅ 7 0 пЂЅ 15 Since there are no pairs пЂЁ x, y пЂ© for which 0 will equal 15, the system is inconsistent and has no solution. The solution set is пѓ† or пЃ» пЃЅ . x пЂЅ 4y пЂ8 пѓ¬ 4b. Solve the system: пѓ 5 20 x y пЂ пЂЅ пЂ40 пѓ® y пЂЅ 3x пЂ 5 пѓ¬ 4b. Solve the system: пѓ 21 пЂ 35 пЂЅ 7y x пѓ® Substitute 4 y пЂ 8 for x in the second equation. 5 x пЂ 20 y пЂЅ пЂ40 x пЃЅ 5(4 y пЂ 8) пЂ 20 y пЂЅ пЂ40 20 y пЂ 40 пЂ 20 y пЂЅ пЂ40 пЂ40 пЂЅ пЂ40 Since пЂ40 пЂЅ пЂ40 for all values of x and y, the system is dependent. пЃ»пЂЁ x, y пЂ© x пЂЅ 4 y пЂ 8пЃЅ or пЃ»пЂЁ x, y пЂ© 5x пЂ 20 y пЂЅ пЂ40пЃЅ. The solution set is Objective #5: Solve problems using systems of linear equations. Solved Problem #5 5. A company that manufactures running shoes has a fixed cost of $300,000. Additionally, it costs $30 to produce each pair of shoes. The shoes are sold at $80 per pair. 5a. Write the cost function, C, of producing x pairs of running shoes. Pencil Problem #5 5. A company that manufactures small canoes has a fixed cost of $18,000. Additionally, it costs $20 to produce each canoe. The selling price is $80 per canoe. 5a. Write the cost function, C, of producing x canoes. per pair fixed costs пЂ¶ пЂґпЂ·пЂґ пЂё $30пЃЅ C ( x ) пЂЅ 300,000 пЂ« 30 x Copyright В© 2014 Pearson Education Inc. 3 College Algebra 6e 5b. Write the revenue function, R, from the sale of x pairs of running shoes. 5b. Write the revenue function, R, from the sale of x canoes. $80 per pair R( x ) пЂЅ пЃЅ 80 x 5c. Determine the break-even point. Describe what this means. 5c. Determine the break-even point. Describe what this means. пѓ¬ y пЂЅ 300,000 пЂ« 30 x The system is пѓ пѓ® y пЂЅ 80 x The break-even point is where R( x ) пЂЅ C ( x ). R( x ) пЂЅ C ( x ) 80 x пЂЅ 300,000 пЂ« 30 x 50 x пЂЅ 300,000 x пЂЅ 6000 Back-substitute to find y: y пЂЅ 80 x y пЂЅ 80(6000) y пЂЅ 480, 000 The break-even point is (6000, 480,000). This means the company will break even when it produces and sells 6000 pairs of shoes. At this level, both revenue and costs are $480,000. Answers for Pencil Problems (Textbook Exercise references in parentheses): 1. The ordered pair is a solution to the system. (5.1 #1) 2. пЃ»пЂЁ1,3пЂ©пЃЅ 3. пЃ»пЂЁ1, пЂ2пЂ©пЃЅ 4a. пѓ† or 4b. (5.1 #5) (5.1 #27) пЃ» пЃЅ пЃ»пЂЁ x, y пЂ© (5.1 #31) пЃЅ y пЂЅ 3x пЂ 5 or 5a. C пЂЁ x пЂ© пЂЅ 18,000 пЂ« 20 x пЃ»пЂЁ x, y пЂ© 21x пЂ 35 пЂЅ 7 y (5.1 #61a) пЃЅ (5.1 #33) 5b. R пЂЁ x пЂ© пЂЅ 80 x (5.1 #61b) 5c. Break-even point: (300, 24,000). Which means when 300 canoes are produced the company will break-even with cost and revenue at $24,000. (5.1 #61c) 4 Copyright В© 2014 Pearson Education Inc. Section 5.2 Systems of Linear Equations in Three Variables Hit the BRAKES! Did you know that a mathematical model can be used to describe the relationship between the number of feet a car travels once the brakes are applied and the number of seconds the car is in motion after the brakes are applied? In the Exercise Set of this section, using data collected by a research firm, you will be asked to write the mathematical model that describes this situation. Objective #1: Verify the solution of a system of linear equations in three variables. Solved Problem #1 1. Show that the ordered triple (пЂ1, пЂ4,5) is a solution of the system: пѓ¬ x пЂ 2 y пЂ« 3 z пЂЅ 22 пѓЇ пѓ 2x пЂ 3y пЂ z пЂЅ 5 пѓЇ 3x пЂ« y пЂ 5z пЂЅ пЂ32 пѓ® Pencil Problem #1 1. Determine if the ordered triple (2, пЂ1, 3) is a solution of the system: пѓ¬ x пЂ« yпЂ«zпЂЅ 4 пѓЇ пѓx пЂ 2 y пЂ z пЂЅ 1 пѓЇ 2 x пЂ y пЂ z пЂЅ пЂ1 пѓ® Test the ordered triple in each equation. x пЂ 2 y пЂ« 3 z пЂЅ 22 (пЂ1) пЂ 2 ( пЂ4) пЂ« 3 (5) пЂЅ 22 22 пЂЅ 22, true 2x пЂ 3y пЂ z пЂЅ 5 2(пЂ1) пЂ 3( пЂ4) пЂ (5) пЂЅ 5 5 пЂЅ 5, true 3x пЂ« y пЂ 5z пЂЅ пЂ32 3( пЂ1) пЂ« ( пЂ4) пЂ 5(5) пЂЅ пЂ32 пЂ32 пЂЅ пЂ32, true The ordered triple (вЂ“1, вЂ“4, 5) makes all three equations true, so it is a solution of the system. Copyright В© 2014 Pearson Education Inc. 1 College Algebra 6e Objective #2: Solve systems of linear equations in three variables. Solved Problem #2 2. пѓ¬ x пЂ« 4 y пЂ z пЂЅ 20 пѓЇ Solve the system: пѓ 3x пЂ« 2 y пЂ« z пЂЅ 8 пѓЇ2 x пЂ 3 y пЂ« 2 z пЂЅ пЂ16 пѓ® Pencil Problem #2 2. пѓ¬ 4 x пЂ y пЂ« 2 z пЂЅ 11 пѓЇ Solve the system: пѓ x пЂ« 2 y пЂ z пЂЅ пЂ1 пѓЇ 2 x пЂ« 2 y пЂ 3 z пЂЅ пЂ1 пѓ® Add the first two equations to eliminate z. x пЂ« 4 y пЂ z пЂЅ 20 3x пЂ« 2 y пЂ« z пЂЅ 8 4x пЂ« 6 y пЂЅ 28 Multiply the first equation by 2 and add it to the third equation to eliminate z again. 2 x пЂ« 8 y пЂ 2 z пЂЅ 40 2 x пЂ 3 y пЂ« 2 z пЂЅ пЂ16 4x пЂ« 5y пЂЅ 24 Solve the system of two equations in two variables. 4 x пЂ« 6 y пЂЅ 28 4 x пЂ« 5 y пЂЅ 24 Multiply the second equation by вЂ“1 and add the equations. 4 x пЂ« 6 y пЂЅ 28 пЂ4 x пЂ 5 y пЂЅ пЂ24 yпЂЅ4 Back-substitute 4 for y to find x. 4 x пЂ« 6 y пЂЅ 28 4 x пЂ« 6(4) пЂЅ 28 4 x пЂ« 24 пЂЅ 28 4x пЂЅ 4 x пЂЅ1 Back-substitute into an original equation. 3x пЂ« 2 y пЂ« z пЂЅ 8 3(1) пЂ« 2(4) пЂ« z пЂЅ 8 11 пЂ« z пЂЅ 8 z пЂЅ пЂ3 The solution is пЂЁ1, 4, пЂ3пЂ© and the solution set is 2 пЃ»пЂЁ1, 4, пЂ3пЂ©пЃЅ. Copyright В© 2014 Pearson Education Inc. Section 5.2 Objective #3: Solve problems using systems in three variables. Solved Problem #3 3. Find the quadratic function y пЂЅ ax 2 пЂ« bx пЂ« c whose graph passes through the points (1, 4), (2,1), and (3, 4). Pencil Problem #3 3. Find the quadratic function y пЂЅ ax 2 пЂ« bx пЂ« c whose graph passes through the points (пЂ1, 6), (1, 4), and (2, 9). Use each ordered pair to write an equation. y пЂЅ ax 2 пЂ« bx пЂ« c (1, 4) : 4 пЂЅ a пЂЁ1пЂ© пЂ« b пЂЁ1пЂ© пЂ« c 2 4 пЂЅ aпЂ«bпЂ«c (2,1) : y пЂЅ ax 2 пЂ« bx пЂ« c 1 пЂЅ a пЂЁ2пЂ© пЂ« b пЂЁ2пЂ© пЂ« c 2 1 пЂЅ 4 a пЂ« 2b пЂ« c (3, 4) : y пЂЅ ax 2 пЂ« bx пЂ« c 4 пЂЅ a пЂЁ 3пЂ© пЂ« b пЂЁ 3пЂ© пЂ« c 2 4 пЂЅ 9a пЂ« 3b пЂ« c The system of three equations in three variables is: пѓ¬ a пЂ«bпЂ«c пЂЅ 4 пѓЇ пѓ 4 a пЂ« 2b пЂ« c пЂЅ 1 пѓЇ 9a пЂ« 3b пЂ« c пЂЅ 4 пѓ® пѓ¬ a пЂ«bпЂ«c пЂЅ 4 пѓЇ Solve the system: пѓ4a пЂ« 2b пЂ« c пЂЅ 1 пѓЇ 9a пЂ« 3b пЂ« c пЂЅ 4 пѓ® Multiply the first equation by вЂ“1 and add it to the second equation: пЂ a пЂ b пЂ c пЂЅ пЂ4 4 a пЂ« 2b пЂ« c пЂЅ 1 3a пЂ« b пЂЅ пЂ3 (continued on next page) Copyright В© 2014 Pearson Education Inc. 3 College Algebra 6e Multiply the first equation by вЂ“1 and add it to the third equation: пЂ a пЂ b пЂ c пЂЅ пЂ4 9a пЂ« 3b пЂ« c пЂЅ 4 8a пЂ« 2b пЂЅ 0 Solve this system of two equations in two variables. 3a пЂ« b пЂЅ пЂ3 8a пЂ« 2b пЂЅ 0 Multiply the first equation by вЂ“2 and add to the second equation: пЂ6a пЂ 2b пЂЅ 6 8a пЂ« 2b пЂЅ 0 2a пЂЅ 6 aпЂЅ3 Back-substitute to find b: 3a пЂ« b пЂЅ пЂ3 3(3) пЂ« b пЂЅ пЂ3 9 пЂ« b пЂЅ пЂ3 b пЂЅ пЂ12 Back-substitute into an original equation to find c: aпЂ«bпЂ«c пЂЅ 4 (3) пЂ« (пЂ12) пЂ« c пЂЅ 4 пЂ9 пЂ« c пЂЅ 4 c пЂЅ 13 The quadratic function is y пЂЅ 3x 2 пЂ 12 x пЂ« 13 . Answers for Pencil Problems (Textbook Exercise references in parentheses): 1. not a solution (5.2 #1) 2. {(2, пЂ1,1)} (5.2 #7) 3. y пЂЅ 2 x 2 пЂ x пЂ« 3 (5.2 #19) 4 Copyright В© 2014 Pearson Education Inc. Section 5.3 Partial Fractions WhereвЂ™s the вЂњUNDOвЂќ Button? We have learned how to write a sum or difference of rational expressions as a single rational expression and saw how this skill is necessary when solving rational inequalities. However, in calculus, it is sometimes necessary to write a single rational expression as a sum or difference of simpler rational expressions, undoing the process of adding or subtracting. In this section, you will learn how to break up a rational expression into sums or differences of rational expressions with simpler denominators. Objective #1: Decompose P , where Q has only distinct linear factors. Q Solved Problem #1 1. Find the partial fraction decomposition of 5x пЂ 1 . ( x пЂ 3)( x пЂ« 4) Write a constant over each distinct linear factor in the denominator. 5x пЂ 1 A B пЂЅ пЂ« ( x пЂ 3)( x пЂ« 4) x пЂ 3 x пЂ« 4 Multiply by the LCD, ( x пЂ 3)( x пЂ« 4), to eliminate fractions. Then simplify and rearrange terms. 5x пЂ 1 A B ( x пЂ 3)( x пЂ« 4) пЂЅ ( x пЂ 3)( x пЂ« 4) пЂ« ( x пЂ 3)( x пЂ« 4) ( x пЂ 3)( x пЂ« 4) xпЂ3 xпЂ«4 5 x пЂ 1 пЂЅ A( x пЂ« 4) пЂ« B( x пЂ 3) 5 x пЂ 1 пЂЅ Ax пЂ« 4 A пЂ« Bx пЂ 3B 5 x пЂ 1 пЂЅ ( A пЂ« B ) x пЂ« (4 A пЂ 3B ) Equating the coefficients of x and equating the constant terms, we obtain a system of equations. пѓ¬ AпЂ« B пЂЅ 5 пѓ пѓ®4 A пЂ 3B пЂЅ пЂ1 Multiplying the first equation by 3 and adding it to the second equation, we obtain 7A = 14, so A = 2. Substituting 2 for A in either equation, we obtain B = 3. The partial fraction decomposition is 5x пЂ 1 2 3 пЂЅ пЂ« . ( x пЂ 3)( x пЂ« 4) x пЂ 3 x пЂ« 4 Copyright В© 2014 Pearson Education Inc. 1 College Algebra 6e Pencil Problem #1 1. Find the partial fraction decomposition of 3x пЂ« 50 . ( x пЂ 9)( x пЂ« 2) Objective #2: Decompose P , where Q has repeated linear factors. Q Solved Problem #2 2. Find the partial fraction decomposition of xпЂ«2 . x ( x пЂ 1)2 Include one fraction for each power of x в€’ 1. xпЂ«2 A B C пЂЅ пЂ« пЂ« x ( x пЂ 1)2 x x пЂ 1 ( x пЂ 1)2 Multiply by the LCD, x ( x пЂ 1)2 , to eliminate fractions. Then simplify and rearrange terms. x ( x пЂ 1) 2 xпЂ«2 A B C пЂЅ x( x пЂ 1)2 пЂ« x( x пЂ 1)2 пЂ« x ( x пЂ 1)2 2 x x пЂ1 x ( x пЂ 1) ( x пЂ 1)2 x пЂ« 2 пЂЅ A( x пЂ 1)2 пЂ« Bx( x пЂ 1) пЂ« Cx x пЂ« 2 пЂЅ A( x 2 пЂ 2 x пЂ« 1) пЂ« Bx ( x пЂ 1) пЂ« Cx x пЂ« 2 пЂЅ Ax 2 пЂ 2 Ax пЂ« A пЂ« Bx 2 пЂ Bx пЂ« Cx 2 0 x 2 пЂ« x пЂ« 2 пЂЅ ( A пЂ« B ) x 2 пЂ« ( пЂ2 A пЂ B пЂ« C ) x пЂ« A Copyright В© 2014 Pearson Education Inc. Section 5.3 Equating the coefficients of like terms, we obtain a system of equations. AпЂ« B пЂЅ 0 пѓ¬ пѓЇ пѓ пЂ2 A пЂ B пЂ« C пЂЅ 1 пѓЇ AпЂЅ 2 пѓ® We see immediately that A = 2. Substituting 2 for A in the first equation, we obtain B = в€’2. Substituting these values into the second equation, we obtain C = 3. The partial fraction decomposition is xпЂ«2 2 пЂ2 3 2 2 3 пЂЅ пЂ« пЂ« or пЂ пЂ« . 2 2 x x пЂ 1 x x пЂ 1 x ( x пЂ 1) ( x пЂ 1) ( x пЂ 1)2 Pencil Problem #2 2. Find the partial fraction decomposition of x2 . ( x пЂ 1)2 ( x пЂ« 1) Copyright В© 2014 Pearson Education Inc. 3 College Algebra 6e Objective #3: Decompose P , where Q has a nonrepeated prime quadratic factor. Q Solved Problem #3 8 x 2 пЂ« 12 x пЂ 20 . ( x пЂ« 3)( x 2 пЂ« x пЂ« 2) Use a constant over the linear factor and a linear expression over the prime quadratic factor. A Bx пЂ« C 8 x 2 пЂ« 12 x пЂ 20 пЂЅ пЂ« 2 2 ( x пЂ« 3)( x пЂ« x пЂ« 2) x пЂ« 3 x пЂ« x пЂ« 2 3. Find the partial fraction decomposition of Multiply by the LCD, ( x пЂ« 3)( x 2 пЂ« x пЂ« 2), to eliminate fractions. Then simplify and rearrange terms. ( x пЂ« 3)( x 2 пЂ« x пЂ« 2) 8 x 2 пЂ« 12 x пЂ 20 A Bx пЂ« C пЂЅ ( x пЂ« 3)( x 2 пЂ« x пЂ« 2) пЂ« ( x пЂ« 3)( x 2 пЂ« x пЂ« 2) 2 2 xпЂ«3 ( x пЂ« 3)( x пЂ« x пЂ« 2) x пЂ« xпЂ«2 8 x 2 пЂ« 12 x пЂ 20 пЂЅ A( x 2 пЂ« x пЂ« 2) пЂ« ( Bx пЂ« C )( x пЂ« 3) 8 x 2 пЂ« 12 x пЂ 20 пЂЅ Ax 2 пЂ« Ax пЂ« 2 A пЂ« Bx 2 пЂ« 3Bx пЂ« Cx пЂ« 3C 8 x 2 пЂ« 12 x пЂ 20 пЂЅ ( A пЂ« B ) x 2 пЂ« ( A пЂ« 3B пЂ« C ) x пЂ« (2 A пЂ« 3C ) Equating the coefficients of like terms, we obtain a system of equations. AпЂ« B пЂЅ 8 пѓ¬ пѓЇ пѓ A пЂ« 3B пЂ« C пЂЅ 12 пѓЇ 2 A пЂ« 3C пЂЅ пЂ20 пѓ® Multiply the second equation by в€’3 and add to the third equation to obtain в€’A в€’ 9B = в€’56. Add this result to the first equation in the system above to obtain в€’8B = в€’48, so B = 6. Substituting this value into the first equation, we obtain A = 2. Substituting the value of A into the third equation, we obtain C = в€’8. The partial fraction decomposition is 8 x 2 пЂ« 12 x пЂ 20 2 6x пЂ 8 . пЂЅ пЂ« 2 2 ( x пЂ« 3)( x пЂ« x пЂ« 2) x пЂ« 3 x пЂ« x пЂ« 2 Pencil Problem #3 3. Find the partial fraction decomposition of 4 5x 2 пЂ« 6 x пЂ« 3 . ( x пЂ« 1)( x 2 пЂ« 2 x пЂ« 2) Copyright В© 2014 Pearson Education Inc. Section 5.3 Objective #4: P , where Q has a prime, repeated quadratic factor Q Solved Problem #4 4. Find the partial fraction decomposition of 2 x3 пЂ« x пЂ« 3 . ( x 2 пЂ« 1)2 Include one fraction with a linear numerator for each power of x 2 пЂ« 1. 2 x 3 пЂ« x пЂ« 3 Ax пЂ« B Cx пЂ« D пЂЅ 2 пЂ« ( x 2 пЂ« 1)2 x пЂ« 1 ( x 2 пЂ« 1)2 Multiply by the LCD, ( x 2 пЂ« 1)2 , to eliminate fractions. Then simplify and rearrange terms. ( x 2 пЂ« 1)2 Ax пЂ« B Cx пЂ« D 2 x3 пЂ« x пЂ« 3 пЂЅ ( x 2 пЂ« 1)2 2 пЂ« ( x 2 пЂ« 1)2 2 2 2 x пЂ«1 ( x пЂ« 1) ( x пЂ« 1)2 2 x 3 пЂ« x пЂ« 3 пЂЅ ( Ax пЂ« B )( x 2 пЂ« 1) пЂ« Cx пЂ« D 2 x 3 пЂ« x пЂ« 3 пЂЅ Ax 3 пЂ« Ax пЂ« Bx 2 пЂ« B пЂ« Cx пЂ« D 2 x 3 пЂ« x пЂ« 3 пЂЅ Ax 3 пЂ« Bx 2 пЂ« ( A пЂ« C ) x пЂ« ( B пЂ« D ) Equating the coefficients of like terms, we obtain a system of equations. AпЂЅ2 пѓ¬ пѓЇ B пЂЅ0 пѓЇ пѓ пѓЇ AпЂ«C пЂЅ1 пѓЇпѓ® B пЂ« D пЂЅ 3 We immediately see that A = 2 and B = 0. By performing appropriate substitutions, we obtain C = в€’1 and D = 3. The partial fraction decomposition is 2 x3 пЂ« x пЂ« 3 2x пЂx пЂ« 3 . пЂЅ 2 пЂ« 2 2 2 ( x пЂ« 1) x пЂ« 1 ( x пЂ« 1) 2 Copyright В© 2014 Pearson Education Inc. 5 College Algebra 6e Pencil Problem #4 4. Find the partial fraction decomposition of x3 пЂ« x2 пЂ« 2 . ( x 2 пЂ« 2)2 Answers for Pencil Problems (Textbook Exercise references in parentheses): 1. 2. 3. 4. 6 7 4 (5.3 #11) пЂ xпЂ9 xпЂ«2 1 3 1 пЂ« пЂ« 4( x пЂ« 1) 4( x пЂ 1) 2( x пЂ 1) 2 (5.3 #27) 2 3x пЂ 1 пЂ« (5.3 #31) x пЂ« 1 x2 пЂ« 2 x пЂ« 2 x пЂ«1 2x (5.3 #37) пЂ 2 2 x пЂ« 2 ( x пЂ« 2) 2 Copyright В© 2014 Pearson Education Inc. Section 5.4 Systems of Nonlinear Equations in Two Variables Do you feel SAFE???? Scientists debate the probability that a вЂњdoomsday rockвЂќ will collide with Earth. It has been estimated that an asteroid crashes into Earth about once every 250,000 years, and that such a collision would have disastrous results. Understanding the path of Earth and the path of a comet is essential to detecting threatening space debris. Orbits about the sun are not described by linear equations. The ability to solve systems that do not contain linear equations provides NASA scientists watching for troublesome asteroids with a way to locate possible collision points with EarthвЂ™s orbit. Objective #1: Recognize systems of nonlinear equations in two variables. Solved Problem #1 1a. True or false: A solution of a nonlinear system in two variables is an ordered pair of real numbers that satisfies at least one equation in the system. Pencil Problem #1 1a. True or false: A system of nonlinear equations cannot contain a linear equation. False; a solution must satisfy all equations in the system. 1b. True or false: The solution of a system of nonlinear equations corresponds to the intersection points of the graphs in the system. 1b. True or false: The graphs of the equations in a nonlinear system could be a parabola and a circle. True; each solution will correspond to an intersection point of the graphs. Copyright В© 2014 Pearson Education Inc. 1 College Algebra 6e Objective #2: Solve nonlinear systems by substitution. Solved Problem #2 2. Pencil Problem #2 Solve by the substitution method: x пЂ« 2y пЂЅ 0 пѓ¬пѓЇ пѓ 2 2 пѓЇпѓ®( x пЂ 1) пЂ« ( y пЂ 1) пЂЅ 5 2. Solve by the substitution method: пѓ¬пѓЇ x пЂ« y пЂЅ 2 пѓ 2 пѓЇпѓ® y пЂЅ x пЂ 4 x пЂ« 4 Solve the first equation for x: x пЂ« 2 y пЂЅ 0 x пЂЅ пЂ2 y Substitute the expression вЂ“2y for x in the second equation and solve for y. ( x пЂ 1)2 пЂ« ( y пЂ 1)2 пЂЅ 5 x пЃЅ (пЂ2 y пЂ 1)2 пЂ« ( y пЂ 1)2 пЂЅ 5 4 y2 пЂ« 4 y пЂ«1пЂ« y2 пЂ 2 y пЂ«1 пЂЅ 5 5y2 пЂ« 2 y пЂ 3 пЂЅ 0 (5 y пЂ 3)( y пЂ« 1) пЂЅ 0 5 y пЂ 3 пЂЅ 0 or y пЂ« 1 пЂЅ 0 yпЂЅ 3 5 y пЂЅ пЂ1 or If y пЂЅ 53 , x пЂЅ пЂ2 пЂЁ 53 пЂ© пЂЅ пЂ 65 . If y пЂЅ пЂ1, x пЂЅ пЂ2( пЂ1) пЂЅ 2. пЂЁ 2, пЂ 1пЂ© Check in both original equations. x пЂ«2y пЂЅ 0 2 пЂ« 2( пЂ1) пЂЅ 0 0 пЂЅ 0, true пЂЁ Check пЂ 65 , 53 ( x пЂ 1) 2 пЂ« ( y пЂ 1)2 пЂЅ 5 пЂЁ 53 пЂ© пЂЅ 0 пЂЁпЂ 65 пЂ 1пЂ© пЂ« пЂЁ 35 пЂ 1пЂ© 2 пЂ пЂ« пЂЅ0 6 5 (2 пЂ 1)2 пЂ« (пЂ1 пЂ 1)2 пЂЅ 5 1пЂ« 4 пЂЅ 5 5 пЂЅ 5, true пЂ© in both original equations. x пЂ« 2y пЂЅ 0 пЂ 65 пЂ« 2 ( x пЂ 1)2 пЂ« ( y пЂ 1)2 пЂЅ 5 6 5 121 25 0 пЂЅ 0, true 2 пЂЅ5 4 пЂ« 25 пЂЅ5 125 25 пЂЅ5 5 пЂЅ 5, true The solution set is пЃ»пЂЁпЂ 6 3 , 5 5 пЂ© , пЂЁ2, пЂ 1пЂ©пЃЅ. Objective #3: Solve nonlinear systems by addition. 2 Copyright В© 2014 Pearson Education Inc. Section 5.4 Solved Problem #3 3. Solve by the addition method: пѓ¬пѓЇ y пЂЅ x2 пЂ« 5 пѓ 2 2 пѓЇпѓ® x пЂ« y пЂЅ 25 Pencil Problem #3 3. Solve by the addition method: пѓ¬пѓЇ x 2 пЂ« y 2 пЂЅ 13 пѓ 2 2 пѓЇпѓ® x пЂ y пЂЅ 5 Arrange the first equation so that variable terms appear on the left, and constants appear on the right. Add the resulting equations to eliminate the x 2 -terms and solve for y. пЂ x2 пЂ« y пЂЅ 5 x 2 пЂ« y 2 пЂЅ 25 y 2 пЂ« y пЂЅ 30 Solve the resulting quadratic equation. y 2 пЂ« y пЂ 30 пЂЅ 0 ( y пЂ« 6)( y пЂ 5) пЂЅ 0 y пЂ«6 пЂЅ 0 or yпЂ5пЂЅ 0 y пЂЅ пЂ6 or yпЂЅ5 If y пЂЅ пЂ6, x 2 пЂ« (пЂ6)2 пЂЅ 25 x 2 пЂ« 36 пЂЅ 25 x 2 пЂЅ пЂ11 When y пЂЅ пЂ6 there is no real solution. If y пЂЅ 5, x 2 пЂ« (5) 2 пЂЅ 25 x 2 пЂ« 25 пЂЅ 25 x2 пЂЅ 0 xпЂЅ0 Check (0,5) in both original equations. y пЂЅ x2 пЂ« 5 x 2 пЂ« y 2 пЂЅ 25 5 пЂЅ (0) 2 пЂ« 5 5 пЂЅ 5, true 02 пЂ« 52 пЂЅ 25 25 пЂЅ 25, true The solution set is пЃ»(0,5)пЃЅ . Objective #4: Solve problems using systems of nonlinear equations. Copyright В© 2014 Pearson Education Inc. 3 College Algebra 6e Solved Problem #4 4. Pencil Problem #4 Find the length and width of a rectangle whose perimeter is 20 feet and whose area is 21 square feet. 4. The sum of two numbers is 10 and their product is 24. Find the numbers. пѓ¬2 x пЂ« 2 y пЂЅ 20 The system is пѓ xy пЂЅ 21. пѓ® Solve the second equation for x: xy пЂЅ 21 xпЂЅ Substitute the expression 21 y 21 for x in the first equation y and solve for y. 2 x пЂ« 2 y пЂЅ 20 пѓ¦ 21 пѓ¶ 2 пѓ§ пѓ· пЂ« 2 y пЂЅ 20 пѓЁ yпѓё 42 пЂ« 2 y пЂЅ 20 y 42 пЂ« 2 y 2 пЂЅ 20 y 2 y 2 пЂ 20 y пЂ« 42 пЂЅ 0 y 2 пЂ 10 y пЂ« 21 пЂЅ 0 ( y пЂ 7)( y пЂ 3) пЂЅ 0 y пЂ 7 пЂЅ 0 or yпЂ3пЂЅ 0 y пЂЅ 7 or yпЂЅ3 21 пЂЅ 3. 7 21 If y пЂЅ 3, x пЂЅ пЂЅ 7. 3 If y пЂЅ 7, x пЂЅ The dimensions are 7 feet by 3 feet. Answers for Pencil Problems (Textbook Exercise references in parentheses): 1a. false (5.4 #1) 3. 4 1b. true (5.4 #27) пЃ»пЂЁпЂ3, пЂ2пЂ© , пЂЁ пЂ3, 2пЂ© , пЂЁ3, пЂ2пЂ© , пЂЁ3, 2пЂ©пЃЅ 2. (5.4 #19) пЃ»пЂЁ1,1пЂ© , пЂЁ2,0пЂ©пЃЅ (5.4 #3) 4. 6 and 4 (5.4 #43) Copyright В© 2014 Pearson Education Inc. Section 5.5 Systems of Inequalities Does Your Weight Fit You? This chapter opened by noting that the modern emphasis on thinness as the ideal body shape has been suggested as a major cause of eating disorders. In this section, the textbook will demonstrate how systems of linear inequalities in two variables can enable you to establish a healthy weight range for your height and age. Objective #1: Graph a linear inequality in two variables. Solved Problem #1 1a. Graph: 4 x в€’ 2 y в‰Ґ 8 . Pencil Problem #1 1a. Graph: x в€’ 2 y > 10 . First, graph the equation 4 x в€’ 2 y = 8 with a solid line. Find the x вЂ“intercept: 4x в€’ 2 y = 8 4 x в€’ 2(0) = 8 4x = 8 x=2 Find the y вЂ“intercept: 4x в€’ 2 y = 8 4(0) в€’ 2 y = 8 в€’2 y = 8 y = в€’4 Next, use the origin as a test point. 4x в€’ 2 y в‰Ґ 8 4(0) в€’ 2(0) в‰Ґ 8 0 в‰Ґ 8, false Since the statement is false, shade the halfplane that does not contain the test point. Copyright В© 2013 Pearson Education Inc. 1 College Algebra 6e 1b. Graph: x в‰¤ 1 . 1b. Graph: y > 1 . Graph the line y = 1 with a dashed line. Since the inequality is of the form y > a, shade the halfplane above the line. Objective #2: Graph a nonlinear inequality in two variables.. Solved Problem #2 2. Graph: x 2 пЂ« y 2 п‚і 16. Pencil Problem #2 2. Graph: x 2 пЂ« y 2 пЂѕ 25. The graph of x 2 пЂ« y 2 пЂЅ 16 is a circle of radius 4 centered at the origin. Use a solid circle because equality is included in в‰Ґ. The point (0, 0) is not on the circle, so we use it as a test point. The result is 0 в‰Ґ 16, which is false. Since the point (0, 0) is inside the circle, the region outside the circle belongs to the solution set. Shade the region outside the circle. 2 Copyright В© 2013 Pearson Education Inc. Section 5.5 Objective #3: Use mathematical models involving systems of linear inequalities. Solved Problem #3 3. The healthy weight region for men and women ages 19 to 34 can be modeled by the following system of linear inequalities: Pencil Problem #3 3. The healthy weight region for men and women ages 35 and older can be modeled by the following system of linear inequalities: пѓ¬ 4.9 x в€’ y в‰Ґ 165 пѓ пѓ®3.7 x в€’ y в‰¤ 125 пѓ¬5.3x в€’ y в‰Ґ 180 пѓ пѓ® 4.1x в€’ y в‰¤ 14 Show that (66, 130) is a solution of the system of inequalities that describes healthy weight for this age group. Show that (66, 160) is a solution of the system of inequalities that describes healthy weight for this age group. Substitute the coordinates of (66, 130) into both inequalities of the system. пѓ¬ 4.9 x в€’ y в‰Ґ 165 пѓ пѓ®3.7 x в€’ y в‰¤ 125 4.9 x в€’ y в‰Ґ 165 4.9(66) в€’ 130 в‰Ґ 165 193.4 в‰Ґ 165, true 3.7 x в€’ y в‰¤ 125 3.7(66) в€’ 130 в‰¤ 125 114.2 в‰¤ 125, true (66, 130) is a solution of the system. Objective #4: Graph a system of linear inequalities. Solved Problem #4 4. Graph the solution set of the system: пѓ¬ x в€’ 3y < 6 пѓ пѓ® 2 x + 3 y в‰Ґ в€’6 Pencil Problem #4 4. Graph the solution set of the system: пѓ¬ y > 2x в€’ 3 пѓ пѓ® y < в€’x + 6 Graph the line x в€’ 3 y = 6 with a dashed line. Graph the line 2 x + 3 y = в€’6 with a solid line. For x в€’ 3 y < 6 use a test point such as (0, 0). x в€’ 3y < 6 0 в€’ 3(0) < 6 0 < 6, true Since the statement is true, shade the half-plane that contains the test point. Copyright В© 2014 Pearson Education Inc. 3 College Algebra 6e For 2 x + 3 y в‰Ґ в€’6 use a test point such as (0, 0). 2 x + 3 y в‰Ґ в€’6 2(0) + 3(0) в‰Ґ в€’6 0 в‰Ґ в€’6, true Since the statement is true, shade the half-plane that contains the test point. The solution set of the system is the intersection (the overlap) of the two halfplanes. Answers for Pencil Problems (Textbook Exercise references in parentheses): 1a. (5.5 #3) 2. (5.5 #15) 5.3 x в€’ y в‰Ґ 180 5.3(66) в€’ 160 в‰Ґ 180 3. 189.8 в‰Ґ 180, true 1b. (5.5 #9) 4.1x в€’ y в‰¤ 14 4.1(66) в€’ 160 в‰¤ 140 110.6 в‰¤ 140, true (5.5 #77) 4. 4 Copyright В© 2013 Pearson Education Inc. Section 5.6 Linear Programming MAXIMUM Output with MINIMUM Effort! Many situations in life involve quantities that must be maximized or minimized. Businesses are interested in maximizing profit and minimizing costs. In the Exercise Set for this section of the textbook, you will encounter a manufacturer looking to maximize profits from selling two models of mountain bikes. But there is a limited amount of time available to assemble and paint these bikes. We will learn how to balance these constraints and determine the proper number of each bike that should be produced. Objective #1: Write an objective function describing a quantity that must be maximized or minimized. Solved Problem #1 1. A company manufactures bookshelves and desks for computers. Let x represent the number of bookshelves manufactured daily and y the number of desks manufactured daily. The companyвЂ™s profits are $25 per bookshelf and $55 per desk. Write the objective function that models the companyвЂ™s total daily profit, z, from x bookshelves and y desks. Pencil Problem #1 1. A television manufacturer makes rear-projection and plasma televisions. The profit per unit is $125 for the rear-projection televisions and $200 for the plasma televisions. Let x = the number of rearprojection televisions manufactured in a month and y = the number of plasma televisions manufactured in a month. Write the objective function that models the total monthly profit. The total profit is 25 times the number of bookshelves, x, plus 55 times the number of desks, y. profit пЃЅ z = $25 for each bookshelf пЃЅ 25 x $55 for each desk + пЃЅ 55 y The objective function is z = 25 x + 55 y . Copyright В© 2014 Pearson Education Inc. 1 College Algebra 6e Objective #2: Use inequalities to describe limitations in a situation. Solved Problem #2 2. Recall that the company in Solved Problem #1 manufactures bookshelves and desks for computers. x represents the number of bookshelves manufactured daily and y the number of desks manufactured daily. 2a. Write an inequality that models the following constraint: To maintain high quality, the company should not manufacture more than a total of 80 bookshelves and desks per day. Pencil Problem #2 2. Recall that the manufacturer in Pencil Problem #1 makes rear-projection and plasma televisions. x represents the number of rear-projection televisions manufactured monthly and y the number of plasma televisions manufactured monthly. 2a. Write an inequality that models the following constraint: Equipment in the factory allows for making at most 450 rear-projection televisions in one month. x + y в‰¤ 80 2b. Write an inequality that models the following constraint: To meet customer demand, the company must manufacture between 30 and 80 bookshelves per day, inclusive. 2b. Write an inequality that models the following constraint: Equipment in the factory allows for making at most 200 plasma televisions in one month. 30 в‰¤ x в‰¤ 80 2c. Write an inequality that models the following constraint: The company must manufacture at least 10 and no more than 30 desks per day. 2c. Write an inequality that models the following constraint: The cost to the manufacturer per unit is $600 for the rear-projection televisions and $900 for the plasma televisions. Total monthly costs cannot exceed $360,000. 10 в‰¤ y в‰¤ 30 2d. Summarize what you have described about this company by writing the objective function for its profits (from Solved Problem #1) and the three constraints. 2d. Summarize what you have described about this company by writing the objective function for its profits (from Pencil Problem #1) and the three constraints. Objective function: z = 25 x + 55 y . пѓ¬ x + y в‰¤ 80 пѓЇ Constraints: пѓ30 в‰¤ x в‰¤ 80 пѓЇ10 в‰¤ y в‰¤ 30 пѓ® 2 Copyright В© 2014 Pearson Education Inc. Section 5.6 Objective #3: Use linear programming to solve problems. Solved Problem #3 3a. For the company in Solved Problems #1 and 2, how many bookshelves and how many desks should be manufactured per day to obtain maximum profit? What is the maximum daily profit? Pencil Problem #3 3a. For the company in Pencil Problems #1 and 2, how many rear-projection and plasma televisions should be manufactured per month to obtain maximum profit? What is the maximum monthly profit? Graph the constraints and find the corners, or vertices, of the region of intersection. Find the value of the objective function at each corner of the graphed region. Corner Objective Function ( x, y ) z = 25 x + 55 y z = 25(30) + 55(10) (30,10) = 750 + 550 = 1300 z = 25(30) + 55(30) (30,30) = 750 + 1650 = 2400 z = 25(50) + 55(30) (50,30) = 1250 + 1650 = 2900 (Maximum) z = 25(70) + 55(10) (70,10) = 1750 + 550 = 2300 The maximum value of z is 2900 and it occurs at the point (50, 30). In order to maximize profit, 50 bookshelves and 30 desks must be produced each day for a profit of $2900. Copyright В© 2014 Pearson Education Inc. 3 College Algebra 6e 3b. Find the maximum value of the objective function z = 3 x + 5 y subject to the constraints: пѓ¬ x в‰Ґ 0, y в‰Ґ 0 пѓЇ пѓx + y в‰Ґ 1 пѓЇx + y в‰¤ 6 пѓ® 3b. Find the maximum value of the objective function z = 4 x + y subject to the constraints: пѓ¬ x в‰Ґ 0, y в‰Ґ 0 пѓЇ пѓ 2 x + 3 y в‰¤ 12 пѓЇx + y в‰Ґ 3 пѓ® Graph the region that represents the intersection of the constraints: Find the value of the objective function at each corner of the graphed region. Corner Objective Function ( x, y ) z = 3x + 5 y (0,1) z = 3(0) + 5(1) = 5 (1, 0) (0, 6) (6, 0) z = 3(1) + 5(0) = 3 z = 3(0) + 5(6) = 30 (Maximum) z = 3(6) + 5(0) = 18 The maximum value is 30. Answers for Pencil Problems (Textbook Exercise references in parentheses): 1. z = 125 x + 200 y (5.6 #15a) 2a. x в‰¤ 450 (5.6 #15b) 2c. 600 x + 900 y в‰¤ 360, 000 (5.6 #15b) 2b. y в‰¤ 200 (5.6 #15b) пѓ¬ x в‰¤ 450 пѓЇ 2d. z = 125 x + 200 y ; пѓ y в‰¤ 200 (5.6 #15b) пѓЇ600 x + 900 y в‰¤ 360, 000 пѓ® 3a. 300 rear-projection and 200 plasma televisions; Maximum profit: $77,500 (5.6 #15e) 4 Copyright В© 2014 Pearson Education Inc. 3b. 24 (5.6 #7)