How many equations would you need if you had 9 variables to solve?

Answer 1

To get unique solutions, *at least* 9. Better said, you need 9 equations that aren't linear combinations of each other.

E.g., x + 2y = z, 2x + 3y = z, and 3x + 5y = 2z do not uniquely determine x, y, z since the third equation (the sum of the other two) gives no new information.

On the other hand, you need the equations to be consistent: x + y = 1, x + y = 2 has no solutions at all. To use linear algebra language, you need 9 consistent and linearly independent equations.

Answer 2

9

Answer 3

To "Solve" the problem and get a "unique solution", you need 9 equations.

Answer 4

Nine at a minimum. At its very simplest, equations like a=2, b=3. At more complex, like a + 2b = 7, you still need 9.

Answer 5

It depends on the problem.
May be 1 equation. or may be more.
Accordingly if a=1, b=a, c=a+b, d=c+b-a, How much you need to solve these four.
Just think.

Answer 6

For real solutions: 9.

Answer 7

you would only need one, you can have one equation with 9 different variables that you dont know like:

3=5x-y-3k+5gh-z/f+rwv