when is it impossible to find the value of x?

Answer 1

You have to be careful at answering this question. I shall demonstrate with two equations

Example 1:

4x + 5 = 4x + 5

It is true when you attempt to get x by itself x cancels on both sides leaving you with 5 = 5. However, since 5 = 5 x actually has a solution, all solutions. No matter what you substitute in for x the equation will always be true!

Example 2: When x truely has no solution!

4x + 3 = 4x + 5

Once again when you attempt to get x by itself by subtracting 4x on both sides you get 3 = 5. However you know 3 will never equal 5. This is a false statement making it in impossible to find a value which works for x.

Therefore the only time you can not find a value for x is when x cancels on both sides of the equation AND you get a false statement such as 3 = 5.

Answer 2

Not a trivial question at all, since there exists complex roots for any algebraic equation of the form f(x) = 0. But for equations like

x^2 + 1 = 0

there exists no real value for x.

The other answers given are valid, in that many equations have degenerate solutions for x, meaning that there isn't a finite set of values for x that satisfy the equation. For an easy example, take the equation:

0*x = 0

Any value of x will make this equation true. But that does not mean it's impossible to FIND any value of x in which this would be true! It's just that there's too many of them.

Likewise, infinity is accepted as a "value of x". So, for example, for the equation,

1/x = 0

the solution is x = infinity. This is not unacceptable as a "value of x" that satisfies this equation. What makes this question so vexatious is that even if there was given an equation of the form f(x) = 0 where f(x) is NOT an algebraic function (as for example an indefinite integral equation), it's far from trivial to say that solutions to it won't form another field, which then becomes another kind of mathematical "value". But this is fun to think about, the idea of trying to come up with ANY equation of the form f(x) = 0 where there simply exists no kind of x at all in any manner that would satisfy it.

Answer 3

When there are more variables in an equation than the total number of terms -1
this would make it so that when solving for X or one of the other variables,.... you would always have at least 2 other variables in the answer... making it impossible (or at least highly unlikely that one can isolate the variable w/o a system of equations)

*if X is canceled out of an equation.... one can then assume that the value of X is indenendent of the answer.... so all values of X solve the equation *except for when a denominator value would be zero as in 5/(x) = (4+1)/X all values of X solve this equation except for X=0

note: 4x + 3 = 3x + 5
you end up with x =2.....
had it been written as 4x+3 = 4x+5 then giving you 3=5
there is no solution, the equation was false... so if an equation is improperly stated... then you may never find a solution.

Answer 4

The value of x could also be impossible to find if there was an asymptote or a hole at that specific value of x.

Answer 5

When x cancels out of the equation you are trying to solve. This might show up as a divide by zero.

Answer 6

When you are not given enough information.
Or when there is not an equation to solve, like now.
Or when the x variable cancels out of the equation entirely.

Answer 7

When x is an undefined answer such as dividing a number by 0.

Answer 8

when in the course of trying to solve the problem you end up with a divide by zero situation.

Answer 9

When you are not given a function