Math141SolsFinalS11

Math141SolsFinalS11 - SOLUTIONS TO THE FINAL (CHAPTERS 7,...

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SOLUTIONS TO THE FINAL (CHAPTERS 7, 8, AND 9) MATH 141 – SPRING 2011 – KUNIYUKI 250 POINTS TOTAL Show all work, simplify as appropriate, and use “good form and procedure” (as in class). Box in your final answers! No notes or books allowed. A scientific calculator is allowed. To maximize chances for partial credit, please be neat and indicate any elementary row operations (EROs) you use! Clarity is important. I might not grade “messes.” 1) Find the intersection point(s) of the graphs of 3 x 2 ± y = 0 and 5 x + y ± 2 = 0 in the usual xy -plane by solving a system, as we have done in class. Do not rely on graphing, “trial-and-error,” guessing, or point-plotting as a basis for your method. Show all work! Write all solutions as ordered pairs of the form x , y () . If there are none, write “NONE.” (14 points) We want to find all real solutions of the system: 3 x 2 ± y = 0 5 x + y ± 2 = 0 ² ³ ´ µ . Both the Substitution and the Addition / Elimination Methods can be used here. Let’s use the Substitution Method. We can, for example, solve the first equation for y in terms of x : 3 x 2 ± y = 0 3 x 2 = y y = 3 x 2 We can then plug (substitute) our expression for y into the second equation and solve for x : 5 x + y ± 2 = 0 and y = 3 x 2 ² 5 x + 3 x 2 ± 2 = 0 3 x 2 + 5 x ± 2 = 0 3 x ± 1 x + 2 = 0 3 x ± 1 = 0 or x = 1 3 x + 2 = 0 x = ± 2
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Plug (substitute) these x -values into the equation y = 3 x 2 to find the corresponding y -values. x = 1 3 ± y = 3 1 3 ² ³ ´ µ · 2 = 3 1 3 2 ² ³ ´ µ · = 1 3 ± 1 3 , 1 3 ² ³ ´ µ · is a tentative solution. x = ± 2 ² y = 3 ± 2 () 2 = 12 ²± 2,12 is a tentative solution. Check the proposed solutions in the given system. 3 1 3 ± ² ³ ´ µ 2 · 1 3 ± ² ³ ´ µ = 0 Checks out 5 1 3 ± ² ³ ´ µ + 1 3 ± ² ³ ´ µ · 2 = 0 Checks out ¸ ¹ º º » º º The solution set is: 1 3 , 1 3 ± ² ³ ´ µ , · 2, 12 ¸ ¹ º » º ¼ ½ º ¾ º . These solutions correspond to intersection points (in black below) for the graphs of the given equations. The graph of the first equation, 3 x 2 ± y = 0 , or y = 3 x 2 , is the parabola in blue. The graph of the second equation, 5 x + y ± 2 = 0, or y = ± 5 x + 2 , is the line in red.
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2) Write the PFD (Partial Fraction Decomposition) for 3 x ± 13 x 2 ± 4 x + 4 . You must find the unknowns in the PFD Form. Show all work, as we have done in class! (17 points) Fortunately, the given rational expression is proper, since the numerator has degree 1 and the denominator has degree 2. Factor the denominator over ± : 3 x ± 13 x 2 ± 4 x + 4 = 3 x ± 13 x ± 2 () 2 Set up the PFD Form: 3 x ± 13 x ± 2 2 = A x ± 2 + B x ± 2 2 Multiply both sides by the LCD, which is the denominator on the left, x ± 2 2 . 3 x ± 13 = Ax ± 2 + B This is the basic equation. Plug in (substitute) x = 2 and solve for B : 32 ± 13 = 0 + B ± 7 = B B = ± 7 Rewrite the basic equation: 3 x ± 13 = ± 2 ± 7 Use the “Match Coefficients” Method: 3 x ± 13 = Ax ± 2 A ± 7 3 x + ± 13 = A x + ± 2 A ± 7 By matching the coefficients of x : 3 = A A = 3
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Alternatively, you could have matched the constant terms: ± 13 = ± 2 A ± 7 ± 6 = ± 2 A A = 3 PFD: 3 x ± 13 x 2 ± 4 x + 4 = A x ± 2 + B x ± 2 () 2 = 3 x ± 2 + ± 7 x ± 2 2 = 3
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This note was uploaded on 09/08/2011 for the course MATH 141 taught by Professor Staff during the Fall '11 term at Mesa CC.

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Math141SolsFinalS11 - SOLUTIONS TO THE FINAL (CHAPTERS 7,...

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