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hram777 [196]
3 years ago
7

Write a compound inequality that represents each situation. Graph your solution all real numbers that are greater than –8 but le

ss than 8.
Mathematics
2 answers:
VikaD [51]3 years ago
7 0
All real numbers greater then -8 but less then 8
-8 < x < 8 <==
nevsk [136]3 years ago
7 0
D. because i know what im talking about and the answer below me confirmed what i thought to be true so dont judge


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Solve each of the following systems by Gauss-Jordan elimination. (b) X1-2x2+ x3- 4x4=1 X1+3x2 + 7x3 + 2x4=2 -12x2-11x3- 16x4 5 (
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Answer:

a) The set of solutions is \{(0,-3x_3,x_3): x_3\; \text{es un real}\} y b) the set of solutions is \{(-6,\frac{-41}{17}-\frac{30}{17}x_4 , \frac{37}{17}+\frac{8}{17} x_4 ,x_4): x_4\;\text{es un real}\}.

Step-by-step explanation:

a) Let's first find the echelon form of the matrix \left[\begin{array}{ccc}5&2&6\\-2&1&3\end{array}\right].

  • We add \frac{2}{5} from row 1 to row 2 and we obtain the matrix \left[\begin{array}{ccc}5&2&6\\0&\frac{9}{5} &\frac{27}{5}\end{array}\right]
  • From the previous matrix, we multiply row 1 by \frac{1}{5} and the row 2 by \frac{5}{9} and we obtain the matrix \left[\begin{array}{ccc}1&\frac{2}{5} &\frac{6}{5} \\0&1&3\end{array}\right]. This matrix is the echelon form of the initial matrix.

The system has a free variable (x3).

  • x2+3x3=0, then x2=-3x3
  • 0=x1+\frac{2}{5}x2+\frac{6}{5}x3=

       x1+\frac{2}{5}(-3x3)+\frac{6}{5}x3=

      x1-\frac{6}{5}x3+\frac{6}{5}x3

     then x1=0.

The system has infinite solutions of the form (x1,x2,x3)=(0,-3x3,x3), where x3 is a real number.

b) Let's first find the echelon form of the aumented matrix \left[\begin{array}{ccccc}1&-2&1&-4&1\\1&3&7&2&2\\0&-12&-11&-16&5\end{array}\right].

  • To row 2 we subtract row 1 and we obtain the matrix \left[\begin{array}{ccccc}1&-2&1&-4&1\\0&5&6&6&1\\0&-12&-11&-16&5\end{array}\right]
  • From the previous matrix, we add to row 3, \frac{12}{5} of row 2 and we obtain the matrix \left[\begin{array}{ccccc}1&-2&1&-4&1\\0&5&6&6&1\\0&0&\frac{17}{5}&\frac{-8}{5}&\frac{37}{5}   \end{array}\right].
  • From the previous matrix, we multiply row 2 by \frac{1}{5} and the row 3 by \frac{5}{17} and we obtain the matrix \left[\begin{array}{ccccc}1&-2&1&-4&1\\0&1&\frac{6}{5} &\frac{6}{5}&\frac{1}{5}\\0&0&1&\frac{-8}{17}&\frac{37}{17} \end{array}\right]. This matrix is the echelon form of the initial matrix.

The system has a free variable (x4).

  • x3-\frac{8}{17}x4=\frac{37}{17}, then x3=\frac{37}{17}+ \frac{8}{17}x4.x2+[tex]\frac{6}{5}x3+\frac{6}{5}x4=\frac{1}{5}, x2+\frac{6}{5}(\frac{37}{17}+\frac{8}{17}x4)+[tex]\frac{6}{5}x4=\frac{1}{5}, then

      x2=\frac{-41}{17}-\frac{30}{17}x4.

  • x1-2x2+x3-4x4=1, x1+\frac{82}{17}+\frac{60}{17}x4+\frac{37}{17}+\frac{8}{17}x4-4x4=1, then x1=1-\frac{119}{17}=-6

The system has infinite solutions of the form (x1,x2,x3,x4)=(-6,\frac{-41}{17}-\frac{30}{17}x4,\frac{37}{17}+ \frac{8}{17}x4,x4), where x4 is a real number.

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