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

Step-by-step solution

1. 1

   Read the statement before manipulating it. A three-part inequality $L<E<R$ says the expression $E$ is simultaneously greater than $L$ and less than $R$. Here $L=8$ and $R=7$, so it demands a number greater than $8$ and less than $7$ at the same time. Since $8>7$, no such number exists — the answer is already visible.

2. 2

   Confirm it by running the algebra anyway. Multiply all three parts by $7$ (positive, so directions hold):

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

   The reversed bounds persist, as they must — no legal operation can repair an inconsistent statement.

3. 3

   Divide all three parts by 60. Again $60>0$:

   $\frac{56}{60}<1+x<\frac{49}{60},\qquad\text{i.e.}\qquad \frac{14}{15}<1+x<\frac{49}{60}$

4. 4

   Subtract 1 from all three parts.

   $\frac{14}{15}-1<x<\frac{49}{60}-1\qquad\Longrightarrow\qquad -\frac{1}{15}<x<-\frac{11}{60}$

5. 5

   Compare the two bounds on a common denominator. Writing $-\tfrac{1}{15}=-\tfrac{4}{60}$, the condition reads

   $-\frac{4}{60}<x<-\frac{11}{60}$

   But $-\tfrac{4}{60}>-\tfrac{11}{60}$, so the lower bound exceeds the upper bound. The interval is empty.

6. 6

   State the conclusion. The solution set is $\varnothing$: no real number satisfies the inequality. Scanning $200{,}001$ values of $x$ from $-100$ to $100$ confirms that not one of them makes $\frac{60(1+x)}{7}$ both greater than $8$ and less than $7$ ✓. Contrast the well-posed version $8<\frac{60(1+x)}{7}<9$, whose solution is $-\tfrac{1}{15}<x<\tfrac{1}{20}$ — a genuine non-empty interval.