Solve the inequalities $2x + 13 < 4(x + 2) \le x + 41$. Answer ___ Hence list all the prime integer values of x which satisfy the inequalities $2x + 13 < 4(x + 2) \le x + 41$. Answer ___

1. Expand the middle expression and split the compound inequality into two inequalities.

$2x+13<4(x+2)\leq x+41$

$4(x+2)=4x+8$

So solve both:

$2x+13<4x+8$ and $4x+8\leq x+41$

2. Solve the left inequality.

$2x+13<4x+8$

$13<2x+8$

$5<2x$

$x>\dfrac{5}{2}$

3. Solve the right inequality and combine both results.

$4x+8\leq x+41$

$3x+8\leq41$

$3x\leq33$

$x\leq11$

Combining with $x>\dfrac{5}{2}$ gives

$\dfrac{5}{2}<x\leq11$.

1. Use the inequality range from part (a) and identify the prime integer values in it.

From part (a),
$\frac{5}{2}<x\leq 11.$
The integers in this range are $3,4,5,6,7,8,9,10,11$.

The prime integers are
$3,\ 5,\ 7,\ 11.$