Questions: 5x - 3 ≤ 7 - 3x 2x - 1/2 ≥ 7x + 7/6 (6 - x)/4 < (3 - 4)/2 1/9 < x^2 < 1/4 -1 < (1 + x)/(1 - x) ≤ 1 (3x - 1)/x ≤ 2x (x + 3)^2 < 2 x^2 - x - 2 ≥ 0 -(x + 5)/2 ≤ (12 + 3x)/4 -1 < (x^2 - 2)/(x + 1) < 2

Solution Steps

To solve the inequality \( 5x - 3 \leq 7 - 3x \), we rearrange the terms:

\[ 5x + 3x \leq 7 + 3 \implies 8x \leq 10 \implies x \leq \frac{10}{8} = \frac{5}{4} \]

Thus, the solution is:

\[ (-\infty < x) \land (x \leq \frac{5}{4}) \]

Rearranging the terms gives us:

\[ 2x - 7x \geq \frac{1}{2} + \frac{7}{6} \]

Finding a common denominator for the right side:

\[ \frac{1}{2} = \frac{3}{6} \implies \frac{3}{6} + \frac{7}{6} = \frac{10}{6} = \frac{5}{3} \]

Thus, we have:

\[ -5x \geq \frac{5}{3} \implies x \leq -\frac{1}{3} \]

The solution is:

\[ (-\infty < x) \land (x \leq -\frac{1}{3}) \]

First, simplify the right side:

\[ \frac{3 - 4}{2} = \frac{-1}{2} \]

Now, we have:

\[ \frac{6 - x}{4} < -\frac{1}{2} \]

Multiplying both sides by 4 (noting that this does not change the inequality since 4 is positive):

\[ 6 - x < -2 \implies -x < -8 \implies x > 8 \]

Thus, the solution is:

\[ (8 < x) \land (x < \infty) \]

Final Answer