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what is 3/18 + 4/19

My teacher says im doing this problem wrong and I don't see how so could you please help me figure it out....thanks o yeah my name is trey

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What is the sum of 1/9, 2/3, and 5/18? A. 8/30 B. 12/9 C. 4 ...


What is the sum of 1/9, 2/3, and 5/18? ... What is the sum of 1/9, 2/3, and 5/18? A. 8/30 B. 12/9 C. 4/15 D. 19/18 1. Ask for details ; Follow; Report; by baccqirl985 ...
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Solving an Equation - Avvanta


Solving an Equation Here is a typical equation: 3X ... 3 times 6 = 18: X/4 = 5: ... = 4 because 4 - 1 = 3: X = 8, because
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What is the sum of 1/9, 2/3, and 5/18? A. 4/15 B. 12/9 C. 8 ...


What is the sum of 1/9, 2/3, and 5/18? A. 4/15 B. 12/9 C. 8/30 D. 19/18 ... Weegy: Teamwork is a joint action by two or more people or a group, [ in which ...
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18. A New Slant on Suffering (1 Peter 3:13-4:6) | Bible.org


The first is verses 13-16; the second verses 3:18-4:6. ... Second, it involves preparation (see Exodus 19:15), ability, and resolve (1 Peter 4:5).
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Ephesians 3:14-19 NIV - A Prayer for the Ephesians - For this ...


Ephesians 3:14-19 New International Version (NIV) ... Ephesians 3:18: Job 11:8, 9; Ps 103:11; Ephesians 3:19: Php 4:7; Ephesians 3:19: Col 2:10; Ephesians 3:19: Eph 1:23;
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Fractions - Math League


The fractions 1/2, 2/4, 3/6, ... we have 9 1/2 = 19/2 and 5 3/4 = 23/4. ... 1 ÷ 3 3/5 = 1/1 ÷ 18/5 = 1/1 × 5/18 = ...
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1 Peter 3:18-22 A Difficult Passage Explained And ... - Bible.org


Lesson 18: A Difficult Passage Explained and Applied (1 Peter 3:18-22) ... (3:18); B. Christ bore ... Connecting the verb “went” in 3:19 with the same verb in 3 ...
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Arithmetic Homework Help : Multiplication of Fractions ...


2/19 x 9 = Answer: 18/19. 9 = 9/1. So, 2/19 x 9 = (2 x 9)/(19 x 1) = 18/19. 2. ... Try the Quiz : Arithmetic Homework Help : Multiplication of Fractions.
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www.mathpuzzle.com


... 26 2,5,19,3,6; 8,12,1,10; 4,14,16; 15,17; 25 2,6,18,4,3; 1,11,9,14; 5,10,17; 13,16; 19 2,6,18,4,3; 11,9,1,13; 5,12,15; 16,19; 29 2,7,6,14,3; 1,21,4,8; 11,5 ...
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NFPA 101-2012 Life Safety Code TIA Log No.: 1075 Reference ...


Life Safety Code ® TIA Log No.: 1075 ... Revise 18.3.2.5.3 (11) – (12), 19.3.2.5.3 (11) – (13), related 18.3.4 and 19.3.4 alarm system provisions, and associated ...
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Suggested Questions And Answer :


what are the possible combination of making 30 with 1, 3, 5, 7, 9

I started off by figuring how many ways to use nines: 30   3 nines, 27 total   2 nines, 18 total   1 nine, 9 total   0 nines, 0 total I then figured how many ways to use sevens: 30   3 nines, 27 total     0 sevens, 27 total   2 nines, 18 total     1 seven, 25 total     0 sevens, 18 total   1 nine, 9 total     3 sevens, 30 total     2 sevens, 23 total     1 seven, 16 total     0 sevens, 9 total   0 nines, 0 total     4 sevens, 28 total     3 sevens, 21 total     2 sevens, 14 total     1 seven, 7 total     0 sevens, 0 total I then figured how many ways to use fives, then threes, ending up with this: 30   3 nines, 27 total     0 sevens, 27 total       0 fives, 27 total         1 three, 30 total         0 threes, 27 total   2 nines, 18 total     1 seven, 25 total       1 five, 30 total         0 threes, 30 total       0 fives, 25 total         1 three, 28 total         0 threes, 25 total     0 sevens, 18 total       2 fives, 28 total         0 threes, 28 total       1 five, 23 total         2 threes, 29 total         1 three, 26 total         0 threes, 23 total       0 fives, 18 total         4 threes, 30 total         3 threes, 27 total         2 threes, 24 total         1 three, 21 total         0 three, 18 total   1 nine, 9 total ​    3 sevens, 30 total       0 fives, 30 total         0 threes, 30 total     2 sevens, 23 total       1 five, 28 total         0 threes, 28 total       0 fives, 23 total         2 threes, 29 total         1 three, 26 total         0 threes, 23 total     1 seven, 16 total       2 fives, 26 total         1 three, 29 total         0 threes, 26 total       1 five, 21 total         3 threes, 30 total         2 threes, 27 total         1 three, 24 total         0 threes, 21 total       0 fives, 16 total         4 threes, 28 total         3 threes, 25 total         2 threes, 22 total         1 three, 19 total     0 threes, 16 total     0 sevens, 9 total       4 fives, 29 total     0 threes, 29 total       3 fives, 24 total     2 threes, 30 total     1 three, 27 total     0 threes, 24 total       2 fives, 19 total     3 threes, 28 total     2 threes, 25 total     1 three, 22 total     0 threes, 19 total       1 five, 14 total     5 threes, 29 total     4 threes, 26 total         3 threes, 23 total     2 threes, 20 total     1 three, 17 total     0 threes, 14 total       0 fives, 9 total     7 threes, 30 total     6 threes, 27 total     5 threes, 24 total     4 threes, 21 total     3 threes, 18 total     2 threes, 15 total     1 three, 12 total     0 threes, 9 total   0 nines, 0 total     4 sevens, 28 total       0 fives, 28 total     0 threes, 28 total     3 sevens, 21 total       1 five, 26 total     1 three, 29 total     0 threes, 26 total       0 fives, 21 total     3 threes, 30 total     2 threes, 27 total     1 three, 24 total     0 threes, 21 total     2 sevens, 14 total       3 fives, 29 total     0 threes, 29 total       2 fives, 24 total     2 threes, 30 total     1 three, 27 total     0 threes, 24 total       1 five, 19 total     3 threes, 28 total     2 threes, 25 total     1 three, 22 total     0 threes, 19 total       0 fives, 14 total     5 threes, 29 total     4 threes, 26 total     3 threes, 23 total     2 threes, 20 total     1 three, 17 total     0 threes, 14 total     1 seven, 7 total       4 fives, 27 total     1 three, 30 total     0 threes, 27 total       3 fives, 22 total     2 threes, 28 total     1 three, 25 total     0 threes, 22 total       2 fives, 17 total     4 threes, 29 total     3 threes, 26 total     2 threes, 23 total     1 three, 20 total     0 threes, 17 total       1 five, 12 total     6 threes, 30 total     5 threes, 27 total     4 threes, 24 total     3 threes, 21 total     2 threes, 18 total     1 three, 15 total     0 threes, 12 total       0 fives, 7 total     7 threes, 28 total     6 threes, 25 total     5 threes, 22 total     4 threes, 19 total     3 threes, 16 total     2 threes, 13 total     1 three, 10 total     0 threes, 7 total     0 sevens, 0 total       6 fives, 30 total     0 threes, 30 total       5 fives, 25 total     1 three, 28 total     0 threes, 25 total       4 fives, 20 total     3 threes, 29 total     2 threes, 26 total     1 three, 23 total     0 threes, 20 total       3 fives, 15 total     5 threes, 30 total     4 threes, 27 total     3 threes, 24 total     2 threes, 21 total     1 three, 18 total     0 threes, 15 total       2 fives, 10 total     6 threes, 28 total     5 threes, 25 total     4 threes, 22 total     3 threes, 19 total     2 threes, 16 total     1 three, 13 total     0 threes, 10 total       1 five, 5 total     8 threes, 29 total     7 threes, 26 total     6 threes, 23 total     5 threes, 20 total     4 threes, 17 total     3 threes, 14 total     2 threes, 11 total     1 three, 8 total     0 threes, 5 total       0 fives, 0 total     10 threes, 30 total     9 threes, 27 total     8 threes, 24 total     7 threes, 21 total     6 threes, 18 total     5 threes, 15 total     4 threes, 12 total     3 threes, 9 total     2 threes, 6 total     1 three, 3 total     0 threes, 0 total (more to follow)
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what is the answer to 2x-y-2z=1, 4x+y-z=5, x+y+4z=-5 using matrices

what is the answer to 2x-y-2z=1, 4x+y-z=5, x+y+4z=-5 using matrices how to solve this math problem using matrices?   2  -1  -2  │  1   4   1  -1  │  5   1   1   4  │ -5 Multiply line 2 by 2.   2  -1  -2  │  1   8   2  -2  │ 10   1   1   4  │ -5 Subtract line 2 from line 1, replace line 1.  -6  -3   0  │ -9   8   2  -2  │ 10   1   1   4  │ -5 Multiply line 2 by 2.  -6  -3   0   │ -9  16   4  -4  │ 20    1   1   4  │ -5 Add line 3 to line 2, replace line 2.  -6   -3   0  │ -9  17   5   0  │ 15    1   1   4  │ -5 Multiply line 3 by 6.  -6  -3   0  │  -9  17   5   0  │  15   6   6  24  │ -30 Add line 1 to line 3, replace line 3.  -6  -3   0  │  -9  17   5   0  │  15   0   3  24  │ -39 Multiply line 1 by 5; multiply line 2 by 3. -30 -15   0  │ -45  51  15   0  │  45    0   3  24  │ -39 Add line 2 to line 1, replace line 1.  21    0   0  │   0  51  15   0  │  45    0   3  24  │ -39 Divide line 1 by 21.   1     0   0  │   0  51  15   0  │  45    0   3  24  │ -39 Multiply line 1 by 51.  51    0   0  │   0  51  15   0  │  45    0   3  24  │ -39 Subtract line 1 from line 2, replace line 2.  51   0   0  │   0   0  15   0  │  45   0   3  24  │ -39 Divide line 2 by 5.  51   0   0  │   0    0   3   0  │   9   0   3  24  │ -39 Subtract line 2 from line 3, replace line 3.  51   0   0  │   0    0   3   0  │   9   0   0  24  │ -48 Three-in-one: divide line 1 by 51, divide line 2 by 3, divide line 3 by 24.   1   0   0  │   0   0   1   0  │   3   0   0   1  │  -2 This shows that x = 0, y = 3, z = -2  
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how many ways can you have coins that total 18p using 1p , 2p, 5p, and 10p coins

in how many ways can you have coins that total exactly 18pence using 1p, 2p, 5p and 10p coins but you may wish to use as many of each sort as you wish. Start with the biggest coin, then the next biggest, etc. This is so that you always end up adding on just single coins of 1p. 10    we can only have 1*10 because 2*10 is greater than 18. 10 + 5   we can only add on 1*5 because adding on 2*5 will make the sum greater than 18 10 + 5 + 2   again we can only add on 1*2 10 + 5 + 2 + 1   our first arrangement  (1*10, 1*5, 1*2, 1*1) Now we modify the 2p intp 2*1p, the 5p into 2*2p + 1p and the 10 p into 2*5p (as well as 2p's and 1p). 10 + 5 + (1+1) + 1    our 2nd arrangement   (1*10, 1*5, 0*2, 3*1) Now modify the 5, in the 1st arrangement. 10 + (2+2+1) + 2 + 1    3rd arrangement    (1*10, 0*5, 3*2, 2*1) 10 + (2+2+1) + (1+1) + 1    etc.   (1*10, 0*5, 2*2, 4*1) 10 + (2+(1+1)+1) + (1+1) + 1    etc.  (1*10, 1*5, 1*2, 1*1) 10 + ((1+1)+(1+1)+1) + (1+1) + 1     (1*10, 0*5, 0*2, 8*1) Now modify the 10, in the 1st arrangement. (5+5) + 5 + 2 + 1   our 7th arrangement     (0*10, 3*5, 1*2, 1*1) (5+5) + 5 + (1+1) + 1               (0*10, 3*5, 0*2, 3*1) (5+5) + (2+2+1) + (1+1) + 1      (0*10, 2*5, 2*2, 4*1) (5+5) + (2+(1+1)+1) + (1+1) + 1    (0*10, 2*5, 1*2, 6*1) (5+5) + ((1+1)+(1+1)+1) + (1+1) + 1    (0*10, 2*5, 0*2, 8*1)  -- 11th arrangment (5+(2+2+1)) + ((1+1)+(1+1)+1) + (1+1) + 1    (0*10, 1*5, 2*2, 9*1) (5+(2+(1+1)+1)) + ((1+1)+(1+1)+1) + (1+1) + 1    (0*10, 1*5, 1*2, 11*1) (5+((1+1)+(1+1)+1)) + ((1+1)+(1+1)+1) + (1+1) + 1    (0*10, 1*5, 0*2, 13*1)   --- 14th arrangememt ((2+2+1)+((1+1)+(1+1)+1)) + ((1+1)+(1+1)+1) + (1+1) + 1    (0*10, 0*5, 2*2, 14*1) ((2+(1+1)+1)+((1+1)+(1+1)+1)) + ((1+1)+(1+1)+1) + (1+1) + 1    (0*10, 0*5, 1*2, 16*1) (((1+1)+(1+1)+1)+((1+1)+(1+1)+1)) + ((1+1)+(1+1)+1) + (1+1) + 1    (0*10, 0*5, 0*2, 18*1) --17th arrangement So, in total there are 17 arrangements of 1p, 2p, 5p and 10p coins to give 18p
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how do you solve this using a gausian reduction using the three equations? 4x+z=3,2x-y=2,and 3y+2z=0

how do you solve this using a gausian reduction using the three equations? 4x+z=3,2x-y=2,and 3y+2z=0 I think gaussian reduction is similar to back substitution, but with all the equation shaving different variables... just not sure how to do it. You create a matrix using the constants in the equations. Below, you will see approximately what the matrix would look like. Unfortunately, it is impossible to get it to display properly on this page. ┌                       ┐ │ 4   0   1  |   3  │ │ 2  -1  0   |  2   │ │ 0   3   2  |   0   │ └                        ┘ The idea is to perform the same math procedures on these matrix rows that you would perform on the full equations. We want the first row to be  1 0 0 | a, meaning that whatever value appears as the a entry is the value of x. The second row has to be  0 1 0 | b,  and the third row has to be  0 0 1 | c. Multiply row 2 by 2. ┌                      ┐ │ 4   0   1  |   3  │ │ 4  -2   0  |  4   │ │ 0   3   2  |   0   │ └                       ┘ Now subtract row 2 from row 1, and replace row 2 with the result. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  2   1   |  -1   │ │ 0   3   2  |   0   │ └                       ┘ Multiply row 2 by 2. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  4    2  |  -2   │ │ 0   3   2  |   0   │ └                       ┘ Subtract row 3 from row 2, replacing row 2. ┌                        ┐ │ 4   0   1  |   3   │ │ 0  1    0  |  -2   │ │ 0   3   2  |   0   │ └                        ┘ Multiply row 2 by 3 and subtract row 3, replacing row 3. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  1    0  |  -2   │ │ 0   0  -2  |   -6  │ └                       ┘ Divide row 3 by -2. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  1    0  |  -2   │ │ 0   0   1  |   3   │ └                       ┘ Subtract row 3 from row 1, replacing row 1. ┌                      ┐ │ 4   0   0  |   0  │ │ 0  1    0  |  -2  │ │ 0   0  1  |   3   │ └                      ┘ Divide row 1 by 4. ┌                      ┐ │ 1   0   0  |   0  │ │ 0  1    0  |  -2  │ │ 0   0   1  |   3  │ └                      ┘ Row 1 shows that x = 0 Row 2 shows that y = -2 Row 3 shows that z = 3 Plug those values into the original equations to check the answer. 4x + z = 3 4(0) + 3 = 3 0 + 3 = 3 3 = 3 2x – y = 2 2(0) – (-2) = 2 0 + 2 = 2 2 = 2 3y + 2z = 0 3(-2) + 2(3) = 0 -6 + 6 = 0 0 = 0
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is there a solution to (a) x1 + 2x2 + 3x3 = 1 2x1 + 3x2 + 4x3 = 3 x1 + 2x2 + x3 = 3

Matrix format (Gauss method): ( 1 2 3 | 1 ) ( 2 3 4 | 3 ) ( 1 2 1 | 3 ) R3-R1: ( 1 2 3 | 1 ) ( 2 3 4 | 3 ) ( 0 0 -2 | 2 ) R3/-2: ( 1 2 3 | 1 ) ( 2 3 4 | 3 ) ( 0 0 1 | -1 ) R2-R1: ( 1 2 3 | 1 ) ( 1 1 1 | 2 ) ( 0 0 1 | -1 ) R1⇔R2: ( 1 1 1 | 2 ) ( 1 2 3 | 1 ) ( 0 0 1 | -1 ) R2-R1: ( 1 1 1 | 2 ) ( 0 1 2 | -1 ) ( 0 0 1 | -1 ) R2-R3: ( 1 1 1 | 2 ) ( 0 1 1 | 0 ) ( 0 0 1 | -1 ) R1-R2: ( 1 0 0 | 2 ) ( 0 1 1 | 0 ) ( 0 0 1 | -1 ) R2-R3: ( 1 0 0 | 2 ) ( 0 1 0 | 1 ) ( 0 0 1 | -1 ) x1=2; x2=1; x3=-1. So there is a unique solution.
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Solve by Gauss-Jordan elimination (a) x1 + 2x2 + 3x3 = 1 2x1 + 3x2 + 4x3 = 3 x1 + 2x2 + x3 = 3

Matrix format: ( 1 2 3 | 1 ) ( 2 3 4 | 3 ) ( 1 2 1 | 3 ) R3-R1: ( 1 2 3 | 1 ) ( 2 3 4 | 3 ) ( 0 0 -2 | 2 ) R3/-2: ( 1 2 3 | 1 ) ( 2 3 4 | 3 ) ( 0 0 1 | -1 ) R2-R1: ( 1 2 3 | 1 ) ( 1 1 1 | 2 ) ( 0 0 1 | -1 ) R1⇔R2: ( 1 1 1 | 2 ) ( 1 2 3 | 1 ) ( 0 0 1 | -1 ) R2-R1: ( 1 1 1 | 2 ) ( 0 1 2 | -1 ) ( 0 0 1 | -1 ) R2-R3: ( 1 1 1 | 2 ) ( 0 1 1 | 0 ) ( 0 0 1 | -1 ) R1-R2: ( 1 0 0 | 2 ) ( 0 1 1 | 0 ) ( 0 0 1 | -1 ) R2-R3: ( 1 0 0 | 2 ) ( 0 1 0 | 1 ) ( 0 0 1 | -1 ) x1=2; x2=1; x3=-1.
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show that Z15 is isomorphic to Z5 external Z3

Z15 can be represented as a pair of quantities a and b so that a has a range of 0 to 4 (Z5) and b has a range of 0 to 2 (Z3). So we have a scheme of representation: (0,0)=0 (1,1)=11 (2,2)=7 (0,3)=3 (1,4)=14 (2,0)=10 (0,1)=6 (1,2)=2 (2,3)=13 (0,4)=9 (1,0)=5 (2,1)=1 (0,2)=12 (1,3)=8 (2,4)=4. In general (X,Y) maps to (5X+6Y) modulo 15, where X is confined to modulo 3 and Y to modulo 5.  EXAMPLES:  ADDITION: (2,3)+(1,2)=(13+2) mod 15 = 15 mod 15 = 0 (2,3)+(1,2)=((2+1) mod 3, (3+2) mod 5)=(0,0)=0 (1,4)+(0,2)=(14+12) mod 15 = 26 mod 15 = 11 (1,4)+(0,2)=((1+0) mod 3, (4+2) mod 5)=(1,1)=11 SUBTRACTION: (2,3)-(1,2)=(13-2) mod 15 = 11 (2,3)-(1,2)=((2-1) mod 3, (3-2) mod 5)=(1,1)=11 (1,4)-(0,2)=(14-12) mod 15 = 2 (1,4)-(0,2)=((1-0) mod 3, (4-2) mod 5)=(1,2)=2 MULTIPLICATION (2,3)*(1,2)=(13*2) mod 15 = 26 mod 15 = 11 (2,3)*(1,2)=(2,3)+(2,3)=(1,1)=11 (1,4)*(0,2)=(14*12) mod 15 = 168 mod 15 = 3 (1,4)*(0,2)=((1,4)+(1,4)+(1,4))+((1,4)+(1,4)+(1,4))+((1,4)+(1,4)+(1,4))+((1,4)+(1,4)+(1,4)) =(0,2)+(0,2)+(0,2)+(0,2)=(0,3)=3 Or is this cheating?! The point is that the mapping shows for each of all the combinations of Z3 and Z5 there is one and only one Z15 element, and no elements of Z15 have been omitted, so demonstrating isomorphism.
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solve these equations using matrices without a calculator. x=y=z=6, 2x=y-4z=-15, 5x-3y+z=-10

x-y-z=6 2x-y-4z=-15 5x-3y+z=-10 Your equations, as written, had typos in them. I've assumed that some of the equal-signs should have been minus-signs and altered then appropriately. In matrix form the equations would be AX = b Where A is the matrix 1 -1 -1 2 -1 -4 5 -3 1 X is your unknown column vector [x y z] and b is a scalar column vector [ 6 -15 -10]. The solution is given by X = A^(-1)b, where A^(-1) is the inverse matrix of A. We now do Gauss-Jordan elimination on A to get its inverse. We write this out as, 1 -1 -1     |  1   0   0  --- [Row 1] 2 -1 -4     |  0   1   0  --- [Row 2] 5 -3 1      |  0   0   1  --- [Row 3] R2 - 2*R1, R3 - 5*R1,   1 -1 -1     |  1   0   0  --- [Row 1] 0  1 -2     |  -2  1   0  --- [Row 2] 0  2  6     |  -5  0   1  --- [Row 3] R1 + R2, R3 - 2*R2   1  0 -3     |  -1   1   0  --- [Row 1] 0  1 -2     |  -2   1   0  --- [Row 2] 0  0 10    |  -1  -2   1  --- [Row 3] R3/10,   1  0 -3     |  -1     1     0    --- [Row 1] 0  1 -2     |  -2     1     0    --- [Row 2] 0  0  1     | -0.1 -0.2  0.1  --- [Row 3] R1 + 3*R3, R2 + 2*R3,   1  0  0     |  -1.3   0.4   0.3  --- [Row 1] 0  1  0     |  -2.2   0.6   0.2  --- [Row 2] 0  0  1     |  -0.1  -0.2   0.1  --- [Row 3] Now that we have an identity marix on the lhs, then the rhs is the inverse matrix, so A^(-1) =    |  -1.3   0.4   0.3  |                  |  -2.2   0.6   0.2  |                  |  -0.1  -0.2   0.1  | And X = A^(-1)b        X  =    |  -1.3   0.4   0.3 | |   6  | = |-1.3*6 +0.4*(-15) + 0.3*(-10)  |                  |  -2.2   0.6   0.2  | | -15 |    |-2.2*6 +0.6*(-15) + 0.2*(-10) |                  |  -0.1  -0.2   0.1  | | -10 |    |-0.1*6 -0.2*(-15) + 0.1*(-10)  |        X  =   | -7.8 - 6 - 3   |                  | -13.2 - 9 - 2 |                  | -0.6 + 3 - 1  |        X  =   | -16.8  |                  | -24.2 |                  |    1.4 | The solution is: x = -16.8, y = -24.2, z = 1.4
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x+y+z=9,2x-3y+4z=13,2x-3y+4z=13,3x+4y+5z=40. solve

There are 4 equations but only 3 unknowns. This means one equation is unnecessary or there is inconsistency. Call the equations in order A, B, C and D. B and C are the same so we can remove one. Let's remove C. 3A+B: 3x+3y+3z+2x-3y+4z=27+13; 5x+7z=40. So 5x=40-7z. We have x in terms of z. D-3A: y+2z=40-27=13. So y=13-2z. We have y in terms of z. We can substitute for x and y in one equation (choose A) to leave z as the only variable: (40-7z)/5+13-2z+z=9. Multiply through by 5: 40-7z+65-10z+5z=45; -12z+60=0, so 12z=60 and z=5. We can now find x and y: 5x=40-7z=40-35=5, making x=1. y=13-2z=13-10=3. So the solution is x=1, y=3 and z=5. Substitute these values in the original equations to check them out.   Gauss-Jordan method: Write the equations in matrix format: [ 1 1 1 | 9 ] [ 2 -3 4 | 13 ] [ 3 4 5 | 40 ] R2→R2-2R1 [ 1 1 1 | 9 ] [ 0 -5 2 | -5] [ 3 4 5 | 40 ] R3→R3-3R1: [ 1 1 1 | 9 ] [ 0 -5 2 | -5 ] [ 0 1 2 | 13 ] R3→5R3+R2: [ 1 1 1 | 9 ] [ 0 -5 2 | -5 ] [ 0 0 12 | 60 ] R2→-R2+R3 then R2→(1/5)R2 and R3→(1/12)R3: [ 1 1 1 | 9 ] [ 0 1 2 | 13 ] [ 0 0 1 | 5 ] R2→R2-R3: [ 1 1 1 | 9 ] [ 0 1 1 | 8 ] [ 0 0 1 | 5 ] R1→R1-R2: [ 1 0 0 | 1 ] [ 0 1 1 | 8 ] [ 0 0 1 | 5 ] R2→R2-R3: [ 1 0 0 | 1 ] [ 0 1 0 | 3 ] [ 0 0 1 | 5 ] From this identity matrix x=1, y=3 and z=5.  
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how to solve gauss jordan reduction using three equations?

Augmented matrix format: ( 2 -3 1 | 2 ) ( 1 1 1 | 8 ) ( 3 -1 -1 | 0 ) R1⇔R2: ( 1 1 1 | 8 ) ( 2 -3 1 | 2 ) ( 3 -1 -1 | 0 ) R2-2R1 then R2*-1: ( 1 1 1 | 8 ) ( 0 5 1 | 14 ) ( 3 -1 -1 | 0 ) R3-3R1 then R3*(-1/4): ( 1 1 1 | 8 ) ( 0 5 1 | 14 ) ( 0 1 1 | 6 ) R2-R3 then R2/4: ( 1 1 1 | 8 ) ( 0 1 0 | 2 ) ( 0 1 1 | 6 ) R1-R3: ( 1 0 0 | 2 ) ( 0 1 0 | 2 ) ( 0 1 1 | 6 ) R3-R2: ( 1 0 0 | 2 ) ( 0 1 0 | 2 ) ( 0 0 1 | 4 ) (x,y,z)=(2,2,4)
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