1) The same number can be added to or subtracted from both sides of an inequality.

$$ \boxed {x < y \iff x \pm a < y \pm a }$$

For example: $$3 < 7 \iff 3+2 < 7+2 $$

$$-5 < -2 \iff -5-3 < -2-3 $$

2) Both sides of an inequality can be multiplied or divided by the same positive number.

$$ \boxed {x < y \iff x \cdot a < y \cdot a } \quad (a > 0 ) $$

For example: $$-2 < 5 \iff -2 \cdot 3 < 5 \cdot 3 $$

$$-4 < -18 \iff -4:4 < 18:4 $$

3) If both sides of an inequality are multiplied or divided by the same negative number, the direction of the inequality reverses.

$$ \boxed {x < y \iff x \cdot a > y \cdot a } \quad (a < 0 ) $$

For example: $$3 < 11 \iff 3 \cdot (-1) > 11 \cdot (-1) $$

$$-35 < 49 \iff -35 :(-7) > 49: (-7) $$

4) Inequalities with the same direction can be added side-by-side, but they cannot be subtracted.

$$ \boxed {x < y \quad \text{and} \quad a < b \Rightarrow x + a < y + b } $$

For example: Given $$-2 < x < 3$$ and $$-5 < y < -1$$, let’s find the intervals containing the values of \(x + y\) and \(x – y\).

5) Provided that \(x, y, a,\) and \(b\) are positive, inequalities with the same direction can be multiplied side-by-side, but they cannot be divided.

$$ \boxed {x

For example: \( 2 < x < 7 \quad \text{and} \quad 3 < y < 5 \Rightarrow 2 \cdot 3 < x \cdot y < 7 \cdot 5 \).

6) Inequalities possess the transitive property.

$$ \boxed {x < a \quad \text{and} \quad a

For example: \( -2 < 5 \quad \text{and} \quad 5< 7 \quad \Rightarrow \quad -2 < 7 \).

Since \(c > 0\), multiplying both sides of the inequality \(a < b \) by $c$ preserves its direction, yielding \( ac < bc \). Therefore, the claim \( ac > bc \) is incorrect.

\(\textbf{Answer: E} \)

Question 60:

$$ ab < \left| ab\right| $$ $$ bc < \left| bc\right| $$ Based on these conditions, which of the following options is always correct?   \[ \text{A)} \frac{a}{c} + \frac{c}{a} >0 \quad \text{B) } \frac{a}{b} + \frac{b}{a}>0\]

\[ \quad \text{C) } \frac{1}{a} + \frac{1}{b} + \frac{1}{c} <0 \quad \text{D) } abc>0\]

\[ \quad \text{E) } abc < 0 \]

Solution:

\[
ab < |ab| \Rightarrow ab < 0 \quad \text{is obtained.}
\]

\[
bc < |bc| \Rightarrow bc < 0 \quad \text{is obtained.} \] Analyzing the signs of $a, b,$ and $c$ based on these results gives: Table: \begin{array}{|c|c|c|c|} \hline & a & b & c \\ \hline 1 & + & – & + \\ \hline 2 & – & + & – \\ \hline \end{array} \[ \text{The table indicates that } a \text{ and } c \text{ share identical signs in both cases.} \] \[ \text{Consequently, because } \frac{a}{c} \text{ and } \frac{c}{a} \text{ are always positive, it follows that:} \] \[ \frac{a}{c} + \frac{c}{a} > 0 \quad \text{holds true at all times.}
\]

\(\textbf{Answer: A} \)

Question 61:

Given \(a \cdot 2^{-4} = b \) and \( 64 \le a \le 192 \), what is the maximum integer value that $b$ can take?

\[ \text{A)} 8 \quad \text{B) } 9 \quad \text{C) } 10 \quad \text{D) } 11 \quad \text{E)} 12 \]

Solution:

\( a \cdot 2^{-4} =b \Rightarrow a \cdot \frac{1}{16} = b \Rightarrow a=16b \). Substituting this relation into the given interval inequality yields:

$$ 64 \le a < 192 \Rightarrow 64 \le 16b < 192 $$

$$ \Rightarrow 4 \le b < 12 $$

Hence, the largest possible integer value for $b$ is 11.

\(\textbf{Answer: D} \)

Question 62:

Given \(-\frac{3}{2} \le x < 6 \quad \text{and} \quad -\frac{7}{3} < y \le \frac{10}{3} \), find the minimum integer value of the sum $2x + 3y$.

\[ \text{A)} -10 \quad \text{B) } -9 \quad \text{C) } -8 \quad \text{D) } -7 \quad \text{E)} -6 \]

Solution:

\[\begin{array}{l l }
\quad \quad -\frac{3}{2} \le x < 6 \quad \Rightarrow & -3 \le 2x < 12 \\
+ \quad -\frac{7}{3} < y \le \frac{10}{3} \quad \Rightarrow & -7 < 3y \le 10 \\ \hline \\
& -10 < 2x + 3y < 22
\end{array}\]

Thus, the smallest integer value for the expression $2x + 3y$ is -9.

\(\textbf{Answer: B} \)

Question 63: