Find the range of values of x which satisfy the inequalities $\frac{17-8x}{3}<\frac{2(3x-1)}{5}\le 4.$ Hence, state the smallest prime number that satisfies the inequalities.

1. Split the compound inequality into two separate inequalities.

$\frac{17-8x}{3}<\frac{2(3x-1)}{5}\quad \text{and}\quad \frac{2(3x-1)}{5}\le 4$

2. Solve the left inequality.

$\frac{17-8x}{3}<\frac{2(3x-1)}{5}$
Multiply by $15$:
$5(17-8x)<6(3x-1)$
$85-40x<18x-6$
$91<58x$
$x>\frac{91}{58}=\frac{7}{4}$

3. Solve the right inequality.

$\frac{2(3x-1)}{5}\le 4$
Multiply by $5$:
$2(3x-1)\le 20$
$6x-2\le 20$
$6x\le 22$
$x\le \frac{11}{3}$

4. Combine the two conditions to give the range of values of $x$.

From the two inequalities,
$x>\frac{7}{4}\quad \text{and}\quad x\le \frac{11}{3}.$
Therefore,
$\frac{7}{4}<x\le \frac{11}{3}.$

1. Use the inequality from part (a) to choose the smallest prime number satisfying it.

From part (a),
$x>\frac{7}{4}.$
The prime numbers are $2,3,5,\ldots$.

The smallest prime number greater than $\frac{7}{4}$ is
$2.$