Solve the compound inequality 16 < 3y - 8 <= 40.

Step-by-step solution

1. 1

   Read the statement as two conditions joined by "and". The chain $16<3y-8\le40$ says $16<3y-8$ and $3y-8\le40$ hold at once. Rather than splitting them, you can operate on all three parts simultaneously — that is exactly what a chained inequality allows.

2. 2

   Undo the subtraction first by adding 8 to every part. Adding the same number to all three parts shifts the whole chain and cannot flip any sign:

   $16+8<3y-8+8\le 40+8$

   $24<3y\le 48$

3. 3

   Divide every part by 3 to isolate $y$. The divisor $3$ is positive, so the directions of both inequality signs are preserved — this is the step where a negative coefficient would have forced a reversal:

   $\frac{24}{3}<\frac{3y}{3}\le\frac{48}{3}$

   $8<y\le 16$

4. 4

   Keep track of which endpoint is included. The left relation stayed strict ($<$) and the right stayed non-strict ($\le$), so $y=16$ is a solution but $y=8$ is not.

5. 5

   Verify with one interior value and both endpoints. At $y=10$: $3(10)-8=22$ and $16<22\le40$ holds. At $y=16$: $3(16)-8=40$ and $40\le40$ holds. At $y=8$: $3(8)-8=16$ and $16<16$ is false — exactly as the open endpoint predicts.

6. 6

   Write the solution set in interval notation.

   $y\in(8,\,16]$