Linear Congruences in One Variable.

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I am currently learning Linear Congruences in One Variable.

This sections starts with the example $$2x≡3\bmod 4$$ It just states that the congruence is not solvable.

I was wondering if any could show me how to solve this to get such answer.

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In the most direct formulation (definition of modulo), the question is asking

do there exist integers $x,y$ with $2x+4y=3$?

No matter what $x$ and $y$ are, the left-hand side of this equation will be even, and 3 is not even. Thus there is no $x$ that solves the original congruence.

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You may proceed as follows by assuming the congruence had a solution and leading this to a contradiction:

$$2x \equiv 3 \mod 4 $$ $$\stackrel{\cdot 2}{\Rightarrow} 4x \equiv 6 \mod 4 $$ $$\stackrel{4 \equiv 0 \mod 4}{\Leftrightarrow} 0 \equiv 2 \mod 4 \mbox{ Contradiction!}$$