Solve the compound inequality 10 < 60(1 + x)/7 < 11.

Step-by-step solution

1. 1

   Work on all three parts at once. A chained inequality with the variable only in the middle can be solved without splitting it, provided every operation is applied to the left, middle and right simultaneously.

2. 2

   Multiply through by 7. Since $7 > 0$, both inequality signs are preserved:

   $70 < 60(1+x) < 77$

3. 3

   Divide through by 60. Again positive, so the directions hold:

   $\frac{70}{60} < 1 + x < \frac{77}{60} \quad\Longrightarrow\quad \frac{7}{6} < 1 + x < \frac{77}{60}$

4. 4

   Subtract 1 from every part. Using the common denominator $60$ on each side:

   $\frac{7}{6} - 1 = \frac{70 - 60}{60} = \frac{1}{6}, \qquad \frac{77}{60} - 1 = \frac{17}{60}$

   $\frac{1}{6} < x < \frac{17}{60}$

5. 5

   Check the endpoints and an interior point. At $x = \tfrac16 \approx 0.166667$: $\tfrac{60(7/6)}{7} = 10$ exactly, so the strict inequality correctly excludes it. At $x = \tfrac{17}{60} \approx 0.283333$: the middle equals $11$ exactly, also excluded. At $x = 0.22$: $\tfrac{60(1.22)}{7} = 10.457$, comfortably inside $(10, 11)$.