Guide :

# 34×216÷9-1=170

How can this problem equal 170

## Research, Knowledge and Information :

### 170 - 32'' - #34 - 216 cm/85 (10 - 9-12½) - bushwear.co.uk

... #34 170 32'' 216 cm/85" (10" - 9-12½) Clear All. Filter. Category. Stalking, Hunting & Shooting ... #34 (1) #35 (1) #37 (1) #38 (1) #39 (1) #4 (2) #45 (1) ... (1 ...

### how do you figure this problem out 34.216÷9-1=170 (287832 ...

how do you figure this problem out 34.216÷9-1=170. ... 34.216÷9-1=170 is an arithmetic statement that is either ... For instance if the problem were X/9-1=170, ...

### ip - Pastebin.com

162.114.170.1. 64.12.98.165. 62.156.144.202. ... 216.105.34.73. 74.113.142.171. 122.70.158.0. ... 216.87.162.170. 12.150.31.107. 92.62.51.2.

### EGGER HIGHLANDS 0 3 2 3 9 25 40 25 9 34 35 137 172 33 239 279 ...

... 1 1 19 27 46 13 59 170 ... 216 189 45 234 520 403 923 227 1,384 1,600 358 palm city 3 6 6 1 4 16 30 9 4 13 10 21 31 29 73 103 9 paradise hills 0 1 0 2 1 39 43 34 ...

### 1/2/97 -- IP Addresses - Scripting News

153.34.216.17; 204.157.237.69; 204.184.18.104; 199.174.230.90; ... 146.202.9.242; 204.176.224.9; 207.34.98.244; 206.170.212.88; 205.149.161.52; 152.163.237.89; 199 ...

### 6 37 18 9.4 23 212.4 9 6 3 19 2.16 15 20 2.328 9 19 1.18 0 1 ...

... 6 37 18 9 .4 23 212.4 9 6 3 19 2.16 15 20 2.328 9 19 1.18 0 1 19 6 .5 8 75 20 6 .138 20 1.36 27 170 .5 9 13 20 .75 8 2 172.20 1 18 6 ... 14 9 5 216 .4 34 7 ...

### 194.170.47.137 178.124.156.12 110.20.27.170 62.20.93.216 190 ...

216.112.34.210. 192.192.18.243. ... 75.104.1.170. 112.127.232.214. 62.92.54.34. 216.136.119.82. ... 216.3.242.170. 112.163.205.227. 38.103.150.152.

### Web Sites Sharing IP Addresses - IPs Hosting 225 to 249 Domains

Web Sites Sharing IP Addresses: ... 216.122.202.34: 248: tendrils.com: 211.167.74.170: 248: ... 216.116.124.170: 246: ww07.hostica.com:

### 1.9.216.34 IP Address Location | SG IP network tools

1.9.216.34 IP address Information. The IP address 1.9.216.34 was found in Kuala Lumpur, Kuala Lumpur, Malaysia. It is allocated to TM Net, Tmnet, Telekom Malaysia Bhd..

### 8 19 5 0 4 10 9 34 10 4 0 25 8 0 4 73 4 39 1 6 4 170 5 26 4 1 ...

8 19 5 0 4 10 9 34 10 4 0 25 8 0 4 73 4 39 1 6 4 170 5 26 4 1 6 0 1 1 ... 10 16 21716 17 7 2 28 6 9 6 10 25 9 4 2 26 0 4 216 4 5 4 0 5 7 9 5 1 1.7 3 3 .35 231 1 ...

## 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)

### 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

### 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

### 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

### 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.

### 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.

### 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.

### 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

### 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.

### 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)