Solve the compound inequality 11 < 60(1 + x)/7 < 12.

Step-by-step solution

1. 1

   Read the structure before touching anything. This is a three-part inequality: the same quantity $\frac{60(1+x)}{7}$ is trapped between $11$ and $12$. It is really two inequalities at once, so every operation must be applied to all three parts, not just two.

2. 2

   Clear the denominator by multiplying all three parts by 7. Multiplying an inequality by a positive number preserves both directions, and $7>0$, so nothing flips:

   $77<60(1+x)<84$

3. 3

   Divide all three parts by 60. Again $60>0$, so the directions hold:

   $\frac{77}{60}<1+x<\frac{84}{60}=\frac{7}{5}$

   Reducing $\tfrac{84}{60}$ to $\tfrac{7}{5}$ now keeps the final arithmetic clean.

4. 4

   Subtract 1 from all three parts to isolate x. Adding or subtracting never changes an inequality's direction:

   $\frac{77}{60}-\frac{60}{60}<x<\frac{7}{5}-\frac{5}{5}$

   $\frac{17}{60}<x<\frac{2}{5}$

5. 5

   Check both endpoints by substitution. At $x=\tfrac{17}{60}$: $1+x=\tfrac{77}{60}$, so $\frac{60}{7}\cdot\frac{77}{60}=\frac{77}{7}=11$ — the lower bound exactly. At $x=\tfrac{2}{5}$: $1+x=\tfrac{7}{5}$, so $\frac{60}{7}\cdot\frac{7}{5}=12$ — the upper bound exactly. Because both are strict, the endpoints themselves are excluded, giving the open interval $\left(\tfrac{17}{60},\tfrac{2}{5}\right)$.

6. 6

   Sanity-check with a decimal test point. $\tfrac{17}{60}\approx0.2833$ and $\tfrac{2}{5}=0.4$. Taking $x=0.35$ inside the interval gives $\frac{60(1.35)}{7}=\frac{81}{7}\approx11.57$, comfortably between $11$ and $12$ ✓.