Q.Solve the inequation $3x-5<7$ for $x\in\mathbb{R}$ and represent the solution on the number line.

Isolate $x$; since we only add and divide by the positive $3$, the sign is unchanged.

$3x-5<7\Rightarrow 3x<12\Rightarrow x<4$. On the number line: an open dot at $4$, shaded to the left.

Solve step by step, applying the rules of Section 1.

Add $5$ to both sides (sign unchanged):

Divide by $3$ — a positive number, so the sign stays $<$:

The solution set is $\{x\in\mathbb{R}:x<4\}=(-\infty,4)$, a round bracket at $4$ because $4$ is excluded.

Number line: place an open (hollow) dot at $4$ (strict $<$, so $4$ is excluded) and shade the ray to the left — through $3,2,1,0$ and beyond — capturing every real number strictly less than $4$, i.e. $(-\infty,4)$.

Check (independent verification): test $x=3$ (should satisfy): $3(3)-5=4<7$ ✓. Test $x=4$ (boundary, should fail strictly): $3(4)-5=7$, not $<7$ ✓ excluded. Test $x=5$ (should fail): $3(5)-5=10<7$ is false ✓.

Using a closed dot or a square bracket at $4$ — the sign is strict $<$, so $4$ is excluded (open dot, round bracket).