Effacer les filtres
Effacer les filtres

The expression to the left of the equals sign is not a valid target for an assignment

2 vues (au cours des 30 derniers jours)
both x and y has been introduced by syms. This seems to me perfectly correct. What is wrong here?
>> solve(5*x + 4*y = 3, x-6*y = 2)
solve(5*x + 4*y = 3, x-6*y = 2)
Error: The expression to the left of the equals sign is not a valid target for an assignment.
Then I tried this
>> solve('5*x + 4*y = 3, x-6*y = 2')
Warning: Support of character vectors will be removed in a future release. Character vectors can be used only for variable names and
numbers. Instead, to create symbolic expressions first create symbolic variables using 'syms'. To evaluate character vectors and
strings representing symbolic expressions, use 'str2sym'.
> In sym>convertExpression (line 1581)
In sym>convertChar (line 1486)
In sym>tomupad (line 1236)
In sym (line 215)
In solve>getEqns (line 406)
In solve (line 226)
Warning: Do not specify equations and variables as character vectors. Instead, create symbolic variables with syms.
> In solve>getEqns (line 446)
In solve (line 226)
ans =
struct with fields:
x: [1×1 sym]
y: [1×1 sym]
It says something about x and y. Would you please explain the situation in a simple way?
Thank you.

Réponse acceptée

Walter Roberson
Walter Roberson le 8 Nov 2020
Modifié(e) : Walter Roberson le 8 Nov 2020
syms x y
sol = solve(5*x + 4*y == 3, x-6*y == 2);
pretty(sol.x), pretty(sol.y)
13 -- 17 7 - -- 34
  2 commentaires
metin yilmaz
metin yilmaz le 8 Nov 2020
Would you please explain what was wrong with my attempt? what does
ans =
struct with fields:
x: [1×1 sym]
y: [1×1 sym]
mean?
Thank you.
Walter Roberson
Walter Roberson le 8 Nov 2020
You used
solve(5*x + 4*y = 3, x-6*y = 2)
but = is the assignment operator, not the comparison operator.

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Plus de réponses (1)

metin yilmaz
metin yilmaz le 8 Nov 2020
Modifié(e) : metin yilmaz le 8 Nov 2020
The reference is quite difficult for me to understand. I am someone with no background of data structures. Would you please explain it to me over this simple example?
>> solve(5*x + 4*y == 3, x-6*y == 2)
ans =
struct with fields:
x: [1×1 sym]
y: [1×1 sym]
2. But when I did s.x it gave me this notification. non-existent field.
solve(5*x + 4*y == 3, x-6*y == 2) ;s.x
Reference to non-existent field 'x'.
b=solve(5*x + 4*y == 3, x-6*y == 2)
b =
struct with fields:
x: [1×1 sym]
y: [1×1 sym]
>> b.x
ans =
13/17
>> b.y
ans =
-7/343. b=solve(5*x + 4*y == 3, x-6*y == 2)
b =
struct with fields:
x: [1×1 sym]
y: [1×1 sym]
>> b.x
ans =
13/17
>> b.y
ans =
-7/34

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