math_201_(fall_2009)_(smith)_test_01(key)

math_201_(fall_2009)_(smith)_test_01(key) - Solutions to...

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Unformatted text preview: Solutions to Math 201 Exam 1 from spring 2007 1. The given equation is 2 3 y + 1 2 ( y- 3) = y + 1 4 . Since the denominators all divide into 12, it will help to first multiply both sides by 12 in order to eliminate all the fractions: original equation: 2 3 y + 1 2 ( y- 3) = y + 1 4 BS 12: 12 2 3 y + 12 1 2 ( y- 3) = 12 y + 1 4 8 y + 6( y- 3) = 3( y + 1) multiply stuff out: 8 y + 6 y- 18 = 3 y + 3 14 y- 18 = 3 y + 3 BS- 3 y : 11 y- 18 = 3 BS +18: 11 y = 21 BS / 11: y = 21 11 . 2. The given equation is A = P + P rt . In order to solve for r , treat all other variables ( A , P and t ) as if they are constants (numbers). So you can think of the equation as if it said this: 10 = 7 + 7 r 5 . In order to solve this last equation, you could follow these steps: multiply the 7 and 5 together in the 7 r 5 term to get 10 = 7 + 35 r .; subtract 7 from both sides to get 3 = 35 r ; divide both sides by 35 to get 3 35 = r . So to solve the original equation, follow the same steps with all the variables in place: original equation: A = P + P rt multiply P ans t together in the P rt term: A = P + ( P t ) r BS- P : A- P = ( P t ) r BS / ( P t ): A- P P t = r . 3. The given equation is x + 6- x = 4 . Since this is a radical equation, a good strategy is to isolate the radical, then square both sides, and proceed from there: original equation: x + 6- x = 4 BS+ x : x + 6 = x + 4 (BS) 2 : ( x + 6 ) 2 = ( x + 4) 2 x + 6 = x 2 + 8 x + 16 Its now quadratic, so get a zero on one side and then try to factor: BS- x- 6: 0 = x 2 + 7 x + 10 factor: 0 = ( x + 2)( x + 5) possible answers: x =- 2, x =- 5 Remember that when solving any radical equation, you should check your answers, because squaring both sides can result in extraneous solutions. So we shoud check each answer in the original equation: When x =- 2, the equation says p (- 2) + 6- (- 2) = 4 , which is equivalent to 4 + 2 = 4 , which is true. So x =- 2 is a solution. When x =- 5, the equation says p (- 5) + 6- (- 5) = 4 , which is equivalent to 1 + 5 = 4 , which is false. So x =- 5 is not a solution. So the only solution to the original equation is x =- 2 . 4. The given equation is x 4- 6 x 2 + 5 = 0 . Since x 4 is the same as ( x 2 ) 2 , we are supposed to recognize that if we replace every x 2 in the equation by a u , the equation becomes u 2- 6 u + 5 = 0 , which is a quadratic equation that we can solve for u . And of course, once we know the solution value(s) for u , we can figure out what the solution values for x have to be. So here goes. Factoring the left-hand side of the above equation gives us ( u- 5)( u- 1) = 0 , so u = 1 or u = 5. Since u = x 2 , we need x 2 = 1 and x 2 = 5, which gives us solutions (for x ) of x = 1 or x = 5 ....
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This note was uploaded on 10/20/2010 for the course MATH 201 taught by Professor Smith during the Fall '08 term at Washington State University .

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math_201_(fall_2009)_(smith)_test_01(key) - Solutions to...

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