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

Step-by-step solution

1. 1

   Predict the sign of the answer. At $x=0$ the middle expression is $\tfrac{60}{7}\approx8.57$, which is above the target window $(7,8)$. So $1+x$ must be shrunk below $1$, meaning $x$ will come out negative — a useful check on the final answer.

2. 2

   Multiply all three parts by 7. Since $7>0$, both inequality signs are preserved:

   $49<60(1+x)<56$

3. 3

   Divide all three parts by 60. Again the sign of the divisor is positive:

   $\frac{49}{60}<1+x<\frac{56}{60}=\frac{14}{15}$

   Reducing $\tfrac{56}{60}$ to $\tfrac{14}{15}$ now keeps the next step tidy.

4. 4

   Subtract 1 from all three parts.

   $\frac{49}{60}-\frac{60}{60}<x<\frac{14}{15}-\frac{15}{15}$

   $-\frac{11}{60}<x<-\frac{1}{15}$

   Both bounds are negative, exactly as predicted in step 1.

5. 5

   Check the ordering of the bounds. On the common denominator $60$: $-\tfrac{11}{60}<-\tfrac{4}{60}$ ✓, so the interval is genuine and non-empty, unlike the inconsistent version with the bounds $8$ and $7$ swapped.

6. 6

   Verify both endpoints exactly. At $x=-\tfrac{11}{60}$: $1+x=\tfrac{49}{60}$ and $\tfrac{60}{7}\cdot\tfrac{49}{60}=\tfrac{49}{7}=7$ — the lower bound exactly ✓. At $x=-\tfrac{1}{15}$: $1+x=\tfrac{14}{15}$ and $\tfrac{60}{7}\cdot\tfrac{14}{15}=\tfrac{56}{7}=8$ ✓. Both inequalities are strict, so the endpoints themselves are excluded.