The ABC Ink Company is a small family owned company that sells packages of ink cartridge refills for smartpens. The Xavier set contains one blue ink refill and one black ink refill. The Yvonne set includes two blue ink refills, three black ink refills, and one red ink refill. The Zena set includes four blue ink refills, five black ink refills, and one red ink refill. The company has sold most of its stock and has found that it has only 11 blue ink cartridge refills, 14 black ink cartridge refills, and 3 red ink cartridge refills. How many of each set should the company package to sell in order to use all of the remaining ink cartridges so that there will be none left in inventory.

a. Identify the variables of the problem using x, y and z.

b. Determine the system of equations to be solved.

Solve the system of equations using Gauss-Jordan Elimination.

Step-by-step explanation:

suppose the number of Xavier sets = x

suppose the number of Yvonne sets = y

suppose the number of Zena sets = z

                       Blue          Black           Red

Xavier(x)             1                 1                 0

Yvonne(y)           2                3                 1

Zena(z)                4                5                 1

refills                   11                14               3

value of x and y depend upon the value of z

suppose

Hence following ways company package to sell

5 xavier, 3 yvonnes , o zena or

3 xavier, 2 yvonnes, 1 zena or

1 xavier, 1 yvonnes, 2 zena

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

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