Unformatted text preview: Chapter 1 ISM: Linear Algebra 3—52
y—3z —2 41‘ This system does the job. 42. Let b = Boris' money, m = Marina's money, and c = cost of a chocolate bar. , with solution [5, m] = [[1,2c}. Boris has no money. 43. Let us start by reducing the system: s+2y+32 =39 s+2y+32 =39 m+3y+32 =34 —I —a y—z =—5 3.1: + 3p + z = 23 —3[I} —4y — 8.2 = —91
Note that the last two equations are exactly those we get when we substitute
:3 — 39 25; 3.2. either way, we end up with the system _4y _ 3.2 = _91 . 44. a. We set up two equations here, with our variables: 51 = servings of rice, 1'2 = servings of yogurt. 3m, +1232 :60
31031 +3332 :30!) Solving this system reveals that 1'1 = 8,32 = 3.
331 +123:2 = P‘ So our system is: b. Again, we set up our equations: l3ﬂs1 +2133; 2 C and reduce them to ﬁnd that 1'1: —% + 505, while :32 = % — %. 45. Let 2:1 = number of one—dollar hills, 2:: = the number of ﬁve—dollar hills, and at; = the . . _ $1 + $2 + $3 = 32
number of ten—dollar bills. Then our system looks hke. I] + 512 + 1111'; = lﬂl] , which reduces to give us solutions that ﬁt: 2:1 =15+%33, $2 = 1'1"— 333, where 33 can be
chosen freely. Now let’s keep in mind that 2:1, 52, and $3 must be positive integers and
see what conditions this imposes on the variable 3;. We see that since 1'1 and 1:2 must be
integers, 33 must be a multiple of 4. Furthermore, 13 must be positive, and $2 = 17—%1.'3 12 ...
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 Spring '09
 Derksen
 Algebra

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