Unless we are told something to the contrary, unknowns in equations such as this are always assumed to be integers.
If d is the greatest common divisor (gcd) of a and n, the linear congruence
$$\displaystyle ax\equiv b \bmod n$$
has a solution only if b is also divisible by d and in that case there will be an infinite number of solutions given by
$$\displaystyle x=x_{0}+\frac{nt}{d},$$
where $$x_{0}$$ is any solution whatever, and t is an integer, (positive or negative), or zero.
For the equation in question,
$$\displaystyle 7x = 1\bmod 26,\qquad d = 1,$$
so the general solution will be
$$\displaystyle x = x_{0}+\frac{26t}{1} = x_{0}+26t.$$
Finding a suitable $$x_{0}$$ is a trial and error thing, run through the multiples of 7 and 26 looking for the first pair for which the multiple of 7 is 1 greater than the multiple of 26 That happens to be 15 and 4, so the general solution is
$$\displaystyle x = 15 + 26t, \quad t\:\in \:Z.$$
The result given earlier comes from the theory of linear Diophantine equations.
If
$$\displaystyle d = \gcd(a,b),$$
the equation
$$\displaystyle ax+by = c$$
has integer solutions only if c is also divisible by d, and in that case there will be an infinite number of solutions given by
$$\displaystyle x = x_{0}+\frac{bt}{d},\quad y=y_{0}-\frac{at}{d} \quad (t\: \in Z),$$
and where $$\displaystyle \; x_{0},y_{0}$$ is any particular solution.
(For large values of a and b, $$\displaystyle \; x_{0},y_{0}$$ and d are usually found using Euclid's algorithm.)
( The equation
$$\displaystyle ax\equiv b \bmod n$$
is equivalent to
$$\displaystyle ax-b=kn \Rightarrow ax-kn=b,$$
which, with a change of letters and a change of sign, is the Diophantine equation above.)
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