Find solution using Two-Phase method
MIN Z = x1 + x2
subject to
2x1 + x2 >= 4
x1 + 7x2 >= 7
and x1,x2 >= 0

Solution:
Problem is

Min `Z` `=` `` `` `x_1` ` + ` `` `x_2`

subject to

`` `2` `x_1` ` + ` `` `x_2` ≥ `4` `` `` `x_1` ` + ` `7` `x_2` ≥ `7`

and `x_1,x_2 >= 0; `

The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate

1. As the constraint-1 is of type '`>=`' we should subtract surplus variable `S_1` and add artificial variable `A_1`

2. As the constraint-2 is of type '`>=`' we should subtract surplus variable `S_2` and add artificial variable `A_2`

After introducing surplus,artificial variables

Min `Z` `=` `` `` `A_1` ` + ` `` `A_2`

subject to

`` `2` `x_1` ` + ` `` `x_2` ` - ` `` `S_1` ` + ` `` `A_1` = `4` `` `` `x_1` ` + ` `7` `x_2` ` - ` `` `S_2` ` + ` `` `A_2` = `7`

and `x_1,x_2,S_1,S_2,A_1,A_2 >= 0`

Iteration-1		`C_j`	`0`	`0`	`0`	`0`	`1`	`1`
`B`	`C_B`	`X_B`	`x_1`	`x_2` Entering variable	`S_1`	`S_2`	`A_1`	`A_2`	MinRatio `(X_B)/(x_2)`
`A_1`	`1`	`4`	`2`	`1`	`-1`	`0`	`1`	`0`	`(4)/(1)=4`
`A_2` Leaving variable	`1`	`7`	`1`	`(7)`  (pivot element)	`0`	`-1`	`0`	`1`	`(7)/(7)=1``->`
`Z=0` `0=` `Z_j=sum C_B X_B`		`Z_j` `Z_j=sum C_B x_j`	`3` `3=1xx2+1xx1` `Z_j=sum C_B x_1`	`8` `8=1xx1+1xx7` `Z_j=sum C_B x_2`	`-1` `-1=1xx(-1)+1xx0` `Z_j=sum C_B S_1`	`-1` `-1=1xx0+1xx(-1)` `Z_j=sum C_B S_2`	`1` `1=1xx1+1xx0` `Z_j=sum C_B A_1`	`1` `1=1xx0+1xx1` `Z_j=sum C_B A_2`
		`C_j-Z_j`	`-3` `-3=0-3`	`-8` `-8=0-8`   `uarr`	`1` `1=0-(-1)`	`1` `1=0-(-1)`	`0` `0=1-1`	`0` `0=1-1`

Negative minimum `C_j-Z_j` is `-8` and its column index is `2`. So, the entering variable is `x_2`.

Minimum ratio is `1` and its row index is `2`. So, the leaving basis variable is `A_2`.

`:.` The pivot element is `7`.

Entering `=x_2`, Departing `=A_2`, Key Element `=7`

`R_2`(new)`= R_2`(old)`-: 7`

`R_1`(new)`= R_1`(old)`- R_2`(new)

Iteration-2		`C_j`	`0`	`0`	`0`	`0`	`1`
`B`	`C_B`	`X_B`	`x_1` Entering variable	`x_2`	`S_1`	`S_2`	`A_1`	MinRatio `(X_B)/(x_1)`
`A_1` Leaving variable	`1`	`3` `3=4-1` `R_1`(new)`= R_1`(old)`- R_2`(new)	`(13/7)` `13/7=2-1/7` (pivot element) `R_1`(new)`= R_1`(old)`- R_2`(new)	`0` `0=1-1` `R_1`(new)`= R_1`(old)`- R_2`(new)	`-1` `-1=(-1)-0` `R_1`(new)`= R_1`(old)`- R_2`(new)	`1/7` `1/7=0-(-1/7)` `R_1`(new)`= R_1`(old)`- R_2`(new)	`1` `1=1-0` `R_1`(new)`= R_1`(old)`- R_2`(new)	`(3)/(13/7)=1.62``->`
`x_2`	`0`	`1` `1=7-:7` `R_2`(new)`= R_2`(old)`-: 7`	`1/7` `1/7=1-:7` `R_2`(new)`= R_2`(old)`-: 7`	`1` `1=7-:7` `R_2`(new)`= R_2`(old)`-: 7`	`0` `0=0-:7` `R_2`(new)`= R_2`(old)`-: 7`	`-1/7` `-1/7=(-1)-:7` `R_2`(new)`= R_2`(old)`-: 7`	`0` `0=0-:7` `R_2`(new)`= R_2`(old)`-: 7`	`(1)/(1/7)=7`
`Z=0` `0=0xx1` `Z_j=sum C_B X_B`		`Z_j` `Z_j=sum C_B x_j`	`13/7` `13/7=1xx13/7+0xx1/7` `Z_j=sum C_B x_1`	`0` `0=1xx0+0xx1` `Z_j=sum C_B x_2`	`-1` `-1=1xx(-1)+0xx0` `Z_j=sum C_B S_1`	`1/7` `1/7=1xx1/7+0xx(-1/7)` `Z_j=sum C_B S_2`	`1` `1=1xx1+0xx0` `Z_j=sum C_B A_1`
		`C_j-Z_j`	`-13/7` `-13/7=0-13/7`   `uarr`	`0` `0=0-0`	`1` `1=0-(-1)`	`-1/7` `-1/7=0-1/7`	`0` `0=1-1`

Negative minimum `C_j-Z_j` is `-13/7` and its column index is `1`. So, the entering variable is `x_1`.

Minimum ratio is `1.62` and its row index is `1`. So, the leaving basis variable is `A_1`.

`:.` The pivot element is `13/7`.

Entering `=x_1`, Departing `=A_1`, Key Element `=13/7`

`R_1`(new)`= R_1`(old)`xx7/13`

`R_2`(new)`= R_2`(old)`- 1/7 R_1`(new)

Iteration-3		`C_j`	`0`	`0`	`0`	`0`
`B`	`C_B`	`X_B`	`x_1`	`x_2`	`S_1`	`S_2`	MinRatio
`x_1`	`0`	`21/13` `21/13=3xx7/13` `R_1`(new)`= R_1`(old)`xx7/13`	`1` `1=13/7xx7/13` `R_1`(new)`= R_1`(old)`xx7/13`	`0` `0=0xx7/13` `R_1`(new)`= R_1`(old)`xx7/13`	`-7/13` `-7/13=(-1)xx7/13` `R_1`(new)`= R_1`(old)`xx7/13`	`1/13` `1/13=1/7xx7/13` `R_1`(new)`= R_1`(old)`xx7/13`
`x_2`	`0`	`10/13` `10/13=1-1/7xx21/13` `R_2`(new)`= R_2`(old)`- 1/7 R_1`(new)	`0` `0=1/7-1/7xx1` `R_2`(new)`= R_2`(old)`- 1/7 R_1`(new)	`1` `1=1-1/7xx0` `R_2`(new)`= R_2`(old)`- 1/7 R_1`(new)	`1/13` `1/13=0-1/7xx(-7/13)` `R_2`(new)`= R_2`(old)`- 1/7 R_1`(new)	`-2/13` `-2/13=(-1/7)-1/7xx1/13` `R_2`(new)`= R_2`(old)`- 1/7 R_1`(new)
`Z=0` `0=0xx21/13+0xx10/13` `Z_j=sum C_B X_B`		`Z_j` `Z_j=sum C_B x_j`	`0` `0=0xx1+0xx0` `Z_j=sum C_B x_1`	`0` `0=0xx0+0xx1` `Z_j=sum C_B x_2`	`0` `0=0xx(-7/13)+0xx1/13` `Z_j=sum C_B S_1`	`0` `0=0xx1/13+0xx(-2/13)` `Z_j=sum C_B S_2`
		`C_j-Z_j`	`0` `0=0-0`	`0` `0=0-0`	`0` `0=0-0`	`0` `0=0-0`

Since all `C_j-Z_j >= 0`

Hence, optimal solution is arrived with value of variables as :
`x_1=21/13,x_2=10/13`

Min `Z = 0`


we eliminate the artificial variables and change the objective function for the original,

Iteration-1		`C_j`	`1`	`1`	`0`	`0`
`B`	`C_B`	`X_B`	`x_1`	`x_2`	`S_1`	`S_2`	MinRatio
`x_1`	`1`	`21/13`	`1`	`0`	`-7/13`	`1/13`
`x_2`	`1`	`10/13`	`0`	`1`	`1/13`	`-2/13`
`Z=31/13` `31/13=1xx21/13+1xx10/13` `Z_j=sum C_B X_B`		`Z_j` `Z_j=sum C_B x_j`	`1` `1=1xx1+1xx0` `Z_j=sum C_B x_1`	`1` `1=1xx0+1xx1` `Z_j=sum C_B x_2`	`-6/13` `-6/13=1xx(-7/13)+1xx1/13` `Z_j=sum C_B S_1`	`-1/13` `-1/13=1xx1/13+1xx(-2/13)` `Z_j=sum C_B S_2`
		`C_j-Z_j`	`0` `0=1-1`	`0` `0=1-1`	`6/13` `6/13=0-(-6/13)`	`1/13` `1/13=0-(-1/13)`

Since all `C_j-Z_j >= 0`

Hence, optimal solution is arrived with value of variables as :
`x_1=21/13,x_2=10/13`

Min `Z = 31/13`