Reasoning: Standard form of linear equation in one variable is $ax + b = 0$.

Reasoning: In a linear equation, the highest power (exponent) of variable is $1$.

Reasoning: Only one variable $(x)$ with power $1\ \ \ \ \Longrightarrow \ \ \ \ $linear in one variable.

Reasoning: $12x + 17 = 65\ \ \ \ \Longrightarrow \ \ \ \ 12x = 48\ \ \ \ \Longrightarrow \ \ \ \ x = 4$

Reasoning: We want to find a number $(k)$ to subtract from the right side of:
$7x = 30$
So the new equation becomes:$7x = 30 - k$
Given that $x = 4$ is a solution, substitute:
$7(4) = 30 - k \Longrightarrow 28 = 30 - k$
$k = 30 - 28 \Longrightarrow k = 2$

Reasoning: $- \ 2x = \frac{2}{5} \Longrightarrow$ divide both sides by $- \ 2$ to isolate $x$.

Reasoning: $5|x| = 25\ \ \ \ \Longrightarrow \ \ \ |x| = 5\ \ \ \ \Longrightarrow \ \ \ x = \pm \ 5$

Reasoning: $|x| + 7 = 3\ \ \ \Longrightarrow \ \ \ \ |x| = - \ 4$ (impossible), so no solution.

Reasoning: $\sqrt{5x} = -10$ is impossible because square root cannot be negative.

Reasoning: We solve: $\sqrt{3x+1} = 5$
Square both sides: $3x + 1 = 25\ \ \ \Longrightarrow \ \ 3x = 24\ \ \ \Longrightarrow \ \ \ x = 8$

Reasoning: Given: $\frac{4}{x} - \frac{2}{x} = 5$
Combine like terms: $\frac{4\ - \ 2}{x} = 5\ \ \ \Longrightarrow \ \ \ \ \frac{2}{x} = 5$
$2 = 5x\ \ \ \Longrightarrow \ \ \ \ x = \frac{2}{5}$

Reasoning: Strict inequalities use only $\ > \ \ $or $\ <$, without equality signs.

Reasoning: If '$k$' is negative, multiplying both sides reverses the sign, resulting in $(kx > ky)$. So, the correct value is$\ k < 0$.

Reasoning: A compound relation combines two or more conditions into a single statement. The symbol $\leq \ $ stands for less than or equal to. It represents the compound statement $(x < 0)$ or $(x = 0)$.

Reasoning: $3 - \frac{1}{2}x \geq 0$ (Move $3$ to other side)
$\Longrightarrow - \frac{1}{2}x \geq - \ 3$ ; $($Multiply both sides by $- \ 2)$ ; [Inequality reverses]
$\Longrightarrow x \leq 6$

Reasoning: $2(x + 6) \leq 3(x + 4)\ \ \ \Longrightarrow \ \ \ \ 2x + 12\ \leq \ 3x + 12$
$12 \leq x + 12\ \ \ \ \ \Longrightarrow \ \ \ \ 0 \leq x\ \ \ \ \Longrightarrow \ \ \ \ x\ \geq \ 0$