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Crank
3 years ago
7

Solve the following system of equations. Write each of your answers as a fraction reduced to lowest terms. In other words, write

the numbers in the exact form that the row operation tool gives them to you when you use the tool in fraction mode. No decimal answers are permitted.
x + 5y + 8z = 1

3x + 2y + 2z = 5

-2x - 7y + 2z = 5
x = ____
y = ____
z = _____
Mathematics
1 answer:
morpeh [17]3 years ago
8 0

Answer:

x = 57/28

y = -95/84

z = 97/168

Step-by-step explanation:

Use the application in the next link: https://www.zweigmedia.com/RealWorld/tutorialsf1/scriptpivotold.html

Start with the expanded array:

\left[\begin{array}{cccc}1&5&8&1\\3&2&2&5\\-2&-7&2&5\\\end{array}\right]

then using the tool provided, make row operations until you find the solution:

r2 = r2-3r1

\left[\begin{array}{cccc}1&5&8&1\\0&-13&-22&2\\-2&-7&2&5\\\end{array}\right]

r3 = r3+2r1

\left[\begin{array}{cccc}1&5&8&1\\0&-13&-22&2\\0&3&18&7\\\end{array}\right]

r2 = r2*(-1/13)

\left[\begin{array}{cccc}1&5&8&1\\0&1&22/13&-2/13\\0&3&18&7\\\end{array}\right]

r1 = r1- r2*5

\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&22/13&-2/13\\0&3&18&7\\\end{array}\right]

r3 = r3+ r2*-3

\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&22/13&-2/13\\0&0&168/13&97/13\\\end{array}\right]

r3 = r3*13/168

\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&22/13&-2/13\\0&0&1&97/168\\\end{array}\right]

r2 = r2- r3*22/13

\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&0&-95/84\\0&0&1&97/168\\\end{array}\right]

r2 = r2+ r3*6/13

\left[\begin{array}{cccc}1&0&0&57/28\\0&1&0&-95/84\\0&0&1&97/168\\\end{array}\right]

Here you have a reduced array an therefore the answers to each variable are on each row:

\left[\begin{array}{c}x\\y\\z\end{array}\right]

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Step-by-step explanation:

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3 years ago
Suppose you draw a card from a well shuffled deck of playing cards what is probability of drawing a ace
Tasya [4]
So there are a totally of 52 cards right?(not counting jokers because jokers are not in official cards)
And there are 4 aces right?
So there are 52 outcomes and 4 of them are aces.
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So the total number is outcomes is 52 and becomes the denominator and 4 the number of aces become the numerator.
So 4/52 right?
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3 years ago
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2 years ago
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Anna [14]

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