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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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