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If there was an inner product between two vectors, say:

$\langle x , a \rangle$

with $x$ being a variable and $a$ a constant. Would the following equality be valid?

$\langle x , a \rangle b = \langle xb,a\rangle = x\langle b,a \rangle$

Assuming linear properties, I would think this is correct, but I'm not sure with $x$ being variable.

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    Mathematically, there is very little difference between variables and constants. That difference is only relevant when you're done calculating and it's time to interpret your result. – 2017-01-18
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    Well, of course $\;x\;$ **must** be a vector, which you call "variable" just because you want to. From a formal point of view, we don't care what **you** call your symbols, yet they must be things which belong to the correct set or linear space or whatever. An inner product on a linear space $\;V\;$ over a field $\;F\;$ is a function $\;V\times V\to F\;$ , so both its arguments must be in $\;V\;$ ... – 2017-01-18
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    What does $xb$ mean? – 2017-01-18

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Your inner product is defined like this $$ \langle ., . \rangle : V \times V \to F $$ for some vector space $V$ over some field $F$, two vectors from $V$ get a scalar from $F$ assigned.

Regarding your first equation: $$ \langle x , a \rangle b = \langle xb,a\rangle $$

A general inner product over complex numbers is $$ \langle x, y \rangle = y^+ M x $$ where $+$ is adjugation (not sure if this a correct English term), thus transposition and complex conjugation, and $M$ is some Hermitian positive-definite matrix, typically the identity matrix.

So $$ \langle x , a \rangle b = (a^+ M x) b = a^+ M (xb) = \langle xb,a\rangle $$ So your equality is true, where $x, a$ should be vectors from $V$, $b$ a scalar from $F$.

Regarding your second equation:

The second equation $$ \langle xb,a\rangle = x\langle b,a \rangle $$ is only true if $V = F$, otherwise $\langle b,a \rangle$ makes no sense.