This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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....
View
Full Document
 Spring '10
 EGoins
 Linear Algebra, Algebra, Linear Systems, Elementary algebra, Boris, ﬁrst equation

Click to edit the document details