Solve the compound inequality 8 < 60(1 + x)/7 < 9.

Step-by-step solution

1. 1

   Predict where the answer sits. At $x=0$ the middle expression is $\tfrac{60}{7}=8.571$, which already lies inside the target window $(8,9)$. So $x=0$ is a solution, and the interval must straddle zero — one bound negative, one positive.

2. 2

   Multiply all three parts by 7. The multiplier is positive, so both directions hold:

   $56<60(1+x)<63$

3. 3

   Divide all three parts by 60, reducing as you go.

   $\frac{56}{60}<1+x<\frac{63}{60}\qquad\Longrightarrow\qquad \frac{14}{15}<1+x<\frac{21}{20}$

   ($56$ and $60$ share $4$; $63$ and $60$ share $3$.)

4. 4

   Subtract 1 from all three parts.

   $\frac{14}{15}-\frac{15}{15}<x<\frac{21}{20}-\frac{20}{20}$

   $-\frac{1}{15}<x<\frac{1}{20}$

   One negative and one positive bound, exactly as predicted — and $0$ lies inside ✓.

5. 5

   Compare with the sibling problems. Shifting the window up by one, from $(7,8)$ to $(8,9)$, moves the solution from the entirely negative $\left(-\tfrac{11}{60},-\tfrac{1}{15}\right)$ to this interval containing zero. Each unit of the target range corresponds to a width of $\tfrac{7}{60}$ in $x$, since $x$ changes by $\tfrac{7}{60}$ per unit change of $\tfrac{60(1+x)}{7}$.

6. 6

   Verify both endpoints exactly. At $x=-\tfrac{1}{15}$: $1+x=\tfrac{14}{15}$ and $\tfrac{60}{7}\cdot\tfrac{14}{15}=8$ ✓. At $x=\tfrac{1}{20}$: $1+x=\tfrac{21}{20}$ and $\tfrac{60}{7}\cdot\tfrac{21}{20}=\tfrac{63}{7}=9$ ✓. Both are strict inequalities, so the interval is open at both ends.