# Solve fails

I wonder why the Variable 'x_0' not found

```
gr=Polyhedron(ieqs=[(10,-1,0),(-12,0,1)])
p1=gr.plot()
x_0, x_1 = var('x_0 x_1')
lin=solve([70==6*x_0+18*x_1],x_1)
p2=plot(lin, (x,8,14))
p1+p2
```

Solve fails

I wonder why the Variable 'x_0' not found

```
gr=Polyhedron(ieqs=[(10,-1,0),(-12,0,1)])
p1=gr.plot()
x_0, x_1 = var('x_0 x_1')
lin=solve([70==6*x_0+18*x_1],x_1)
p2=plot(lin, (x,8,14))
p1+p2
```

1

This is very unclear. For `p2`

, if you want an implicit plot where `x_0`

is a parameter that belongs to the interval $[8,14]$, you can do:

```
sage: p2 = implicit_plot(70==6*x_0+18*x_1,(8,14),(-3,3))
```

You can also have a look to the `parametric_plot`

function.

But without more details about your actual goal, there is not much to say.

0

Hello, @Cyrille! Try the following:

```
gr = Polyhedron(ieqs=[(10,-1,0), (-12,0,1)])
p1 = gr.plot()
x_0, x_1 = var('x_0 x_1')
lin = solve([70==6*x_0+18*x_1], x_1, solution_dict=True)
p2 = plot(lin[0][x_1], (x_0,8,14))
p1 + p2
```

Notice I have added `solution_dict=True`

, which makes `lin`

a list of dictionaries of solutions:

```
[{x_1: -1/3*x_0 + 35/9}]
```

In order to plot it, first extract the dictionary from the list (that is done with `lin[0]`

); we get

```
{x_1: -1/3*x_0 + 35/9}
```

Finally, we extract the equation you want to plot (that is done with `lin[0][x_1]`

); we get

```
-1/3*x_0 + 35/9
```

That is what I used in the `plot`

command.

**Note:** It is not necessary to use `(x_0, 8, 14)`

in the `plot`

command; you can use `(x, 8, 14)`

. I did that to be consistent with notation.

There are a couple of things that bother me:

- The polyhedron is too small in the plot, because the line is too far. Is this a problem for you?
- This won't work if
`x_1`

has more than one solution.

Problem 2 is solved using tmontiel's suggestion.

There may be an alternative way to achieve the same thing. May I ask: what are you trying to achieve?

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Asked: ** 2019-10-18 06:50:09 +0100 **

Seen: **222 times**

Last updated: **Oct 18 '19**

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because there is an infinity of solutions x_0,x_1 in R

I was expecting thant $x_0$ will be taken as a parameter.