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Unformatted text preview: MA 35100 HOMEWORK ASSIGNMENT #1 SOLUTIONS Problem 1. pg. 5; prob. 10 In Exercises 1 through 10, find all solutions of the linear systems using elimination as discussed in this section. Then check your solutions. x + 2 y + 3 z = 1 2 x + 4 y + 7 z = 2 3 x + 7 y + 11 z = 8 Solution: The method in the text was to eliminate x , then eliminate y , then solve for z , and work backwards to solve for y and x . First we eliminate x from the second and third equations. Subtract twice the first equation from the second, then subtract three times the first equation from the third: x + 2 y + 3 z = 1 z = 0 y + 2 z = 5 From the second equation, we find that z = 0. Substituting this into the third equation, we find that y = 5. Finally, substituting this into the first equation, we find that x + 2 5 + 3 0 = 1, so that x = 9. Hence x = 9, y = 0, and z = 0. We check that this is indeed the solution: 9 + 2 5 + 3 0 = 9 + 10 = 1 2 ( 9) + 4 5 + 7 0 = 18 + 20 = 2 3 ( 9) + 7 5 + 11 0 = 27 + 35 = 8 X Problem 2. pg. 5; prob. 11 In Exercises 11 through 13, find all solutions of the linear systems. Represent your solutions graphically, as intersections of lines in the x y plane. x 2 y = 2 3 x + 5 y = 17 Solution: Subtract three times the first equation from the second: x 2 y = 2 11 y = 11 From the second equation, we find that y = 1. Substituting this into the first equation, we find that x 2 1 = 2, so that x = 4. Hence the solution is x = 4 and y = 1. Figure ?? contains a graphical representation of the intersection of these two lines. Problem 3. pg. 7; prob. 43 1 Figure 1. Graph of the intersection of x 2 y = 2 and 3 x + 5 y = 172 2 4 6 8 10 x21 1 2 3 4 y x,y 4,1 x 2y 2 3x 5y 17 Find a system of linear equations with three unknowns x , y , z whose solutions are x = 6 + 5 t, y = 4 + 3 t, and z = 2 + t. Solution: We seek a system of equations which only involve x , y , and z , not t . It suffices to eliminate t from the expressions above. From the third equation we may express t in terms of z i.e. t = z 2. Upon substituting this value into the expressions for x and y , we find the desired system: x = 6 + 5( z 2) y = 4 + 3( z 2) = x 5 z = 4 y 3 z = 2 Problem 4. pg. 7; prob. 44 Boris and Marina are shopping for chocolate bars. Boris observes, If I add half my money to yours, it will be enough to buy two chocolate bars. Marina naively asks, If I add half my money to yours, how many can we buy? Boris replies, One chocolate bar. How much money did Boris have? Solution: This is somewhat of a trick question. Not enough information is given to find explicit numbers, but we will see that explicit numbers are not necessary....
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 Spring '10
 EGoins
 Linear Algebra, Algebra, Linear Systems

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