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Sagot :
Answer:
1.
R= Number of orange juice cases.
G = Number of greapefruit juice cases.
T = Number of tomato juice cases.
Equations:
9R+10G+12T=398
6R+4G+4T=164
R+2G+T=58
2.
R=6
G=20
T=12
Step-by-step explanation:
1.
The problem is talking about three types of products and it wants us to find how many of each the factory prepares. Since this is the data we need to know, then we set them to be our variables:
R= Number of orange juice cases.
G = Number of greapefruit juice cases.
T = Number of tomato juice cases.
Next, we can use the provided information to build our equations. First, we start with the Sterilizing machine equation: 9 minutes for orange juice, 10 minutes for grape juice and 12 minutes for tomato juice. So we use the sterilizing values for each of the product, to build our first equation:
9R+10G+12T=398
next, we build the filling machine equation, so like on the previous equation, we use the provided data for filling to build the second equation:
6R+4G+4T=164
and finally we build the labeling machine equation so we get:
R+2G+T=58
so we need to solve the following system of equations:
9R+10G+12T=398
6R+4G+4T=164
R+2G+T=58
2.
The very first thing we need to do in order to solve this problem by using the Gauss-Jordan elimination method is to build our matrix based on the system of equations we got on the previous part.
[tex]\left[\begin{array}{cccc}9&10&12&398\\6&4&4&164\\1&2&1&58\\\end{array}\right][/tex]
The idea is to end up with an identity matrix on the first three columns, that will directly give us the answer to the system of equations. So we can start by dividing the first row into 9: [tex]\frac{R_{1}}{9}[/tex] So we get:
[tex]\left[\begin{array}{cccc}1&\frac{10}{9}&\frac{4}{3}&\frac{398}{9}\\6&4&4&164\\1&2&1&58\\\end{array}\right][/tex]
Next, we can multiply the first row by -6 and add it to the second row to get the new second row. [tex]-6R_{1}+R_{2}[/tex]
so we get:
-6 -20/3 -8 -796/3
6 4 4 164
---------------------------------
0 -8/3 -4 -304/3
our matrix now looks like this:
[tex]\left[\begin{array}{cccc}1&\frac{10}{9}&\frac{4}{3}&\frac{398}{9}\\0&-\frac{8}{3}&-4&-\frac{304}{3}\\1&2&1&58\\\end{array}\right][/tex]
Next, we can subtract R3 from R1 so we get:
1 10/9 4/3 398/9
-1 -2 -1 -58
------------------------------
0 -8/9 1/3 -124/9
So the matrix looks like this now:
[tex]\left[\begin{array}{cccc}1&\frac{10}{9}&\frac{4}{3}&\frac{398}{9}\\0&-\frac{8}{3}&-4&-\frac{304}{3}\\0&-\frac{8}{9}&\frac{1}{3}&-\frac{124}{9}\\\end{array}\right][/tex]
Now, we can multiply the second row by -3/8 so we get:
[tex]\left[\begin{array}{cccc}1&\frac{10}{9}&\frac{4}{3}&\frac{398}{9}\\0&1&\frac{3}{2}&38\\0&-\frac{8}{9}&\frac{1}{3}&-\frac{124}{9}\\\end{array}\right][/tex]
Now, we can subtract: [tex]R_{1}-\frac{10}{9}R_{2}[/tex] so we get:
0 -10/9 -5/3 -380/9
1 10/9 4/3 398/9
------------------------------
1 0 -1/3 2
So the matrix will now look like this:
[tex]\left[\begin{array}{cccc}1&0&-\frac{1}{3}&2\\0&1&\frac{3}{2}&38\\0&-\frac{8}{9}&\frac{1}{3}&-\frac{124}{9}\\\end{array}\right][/tex]
Next, we do the following operation: [tex]R_{3}+\frac{8}{9}R_{2}[/tex]
0 8/9 4/3 304/9
0 -8/9 1/3 -124/9
------------------------------
0 0 5/3 20
So our matrix will now look like this:
[tex]\left[\begin{array}{cccc}1&0&-\frac{1}{3}&2\\0&1&\frac{3}{2}&38\\0&0&\frac{5}{3}&20\\\end{array}\right][/tex]
We next multiply R3 by 3/5 so we get:
[tex]\left[\begin{array}{cccc}1&0&-\frac{1}{3}&2\\0&1&\frac{3}{2}&38\\0&0&1&12\\\end{array}\right][/tex]
and now we do: [tex]\frac{R_{3}}{3}+R_{1}[/tex]
0 0 1/3 4
1 0 -1/3 2
------------------------------
1 0 0 6
So our matrix will now look like this:
[tex]\left[\begin{array}{cccc}1&0&0&6\\0&1&\frac{3}{2}&38\\0&0&1&12\\\end{array}\right][/tex]
and now we do the following: [tex]R_{2}-\frac{3}{2}R_{3}[/tex]
so we get:
0 0 -3/2 -18
0 1 3/2 38
------------------------------
0 1 0 20
for our final matrix to be:
[tex]\left[\begin{array}{cccc}1&0&0&6\\0&1&0&20\\0&0&1&12\\\end{array}\right][/tex]
so now we can retrieve the corresponding answers:
R=6
G=20
T=12
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