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galben [10]
3 years ago
7

Mark, Jessica & Nate each downloaded music from the same website. Mark downloaded 10 songs in total consisting of pop, rock,

& hip hop. Jessica downloaded 5 times as many pop songs, twice as many rock and 3 times as many hip hop songs as Mark. She downloaded 28 songs total. If Nate downloaded 20 songs total with 3 times as many pop songs, 3 times as many rock songs, and the same number of hip hop songs as Mark, how many songs of each type did Mark download?
Mathematics
2 answers:
KengaRu [80]3 years ago
6 0
P + r + h = 10 <== mark
5p + 2r + 3h = 28 <== jessica
3p + 3r + h = 20 <== nate

p + r + h = 10
3p + 3r + h = 20...multiply by -1
--------------------
p + r + h = 10
-3p - 3r - h = -20 (result of multiplying by -1)
-------------------add
-2p - 2r = -10

p + r + h = 10...multiply by -3
5p + 2r + 3h = 28
---------------------
-3p - 3r - 3h = -30 (result of multiplying by -3)
5p + 2r + 3h = 28
--------------------add
2p - r = -2

-2p - 2r = -10
2p - r = -2
---------------add
-3r = -12
r = -12/-3
r = 4

2p - r = -2
2p - 4 = -2
2p = -2 + 4
2p = 2
p = 1

p + r + h = 10
1 + 4 + h = 10
5 + h = 10
h = 10 - 5
h = 5

Jessica downloaded : 5p + 2r + 3h = 28
5(1) + 2(4) + 3(5) = 28
5 + 8 + 15 = 28
28 = 28 (correct)......she got 5 pop, 8 rock, and 15 hip hop

Nate downloaded : 3p + 3r + h = 20
3(1) + 3(4) + 5 = 20
3 + 12 + 5 = 20
20 = 20 (correct)...he got 3 pop, 12 rock, and 5 hip hop

Mark downloaded : p + r + h = 10
1 + 4 + 5 = 10
10 = 10 (correct)....he got 1 pop, 4 rock, and 5 hip hop <===

sveta [45]3 years ago
4 0

Mark downloaded 1 pop songs, 4 rock songs, and 5 hip hop songs

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\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

Step-by-step explanation:

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Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.

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Eq. 2 : x1 + x2 + x3 = 6

E1. 3 : 4x1 + 6x2 + 5x3 = 7

Coefficients for x1 ; x2 ; x3

From eq. 1 : 1 ; 2 ; 5

From eq. 2 : 1 ; 1 ; 1

From eq. 3 : 4 ; 6 ; 5

So matrix A is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]

And the vector of vriables (X) is:

\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]

Now we can find the resulting vector (B) using the 'resulting values' from each equation:

\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

In conclusion, AX=B is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

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