Question

Use algebraic​ and/or graphical methods to solve the inequality.
StartFraction 49 minus 7 x Over x EndFraction
greater than7  
Question content area bottom
Part 1
Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.
A.
The​ solution(s) is/are xequals
  
enter your response here.
​(Use a comma to separate answers as​ needed.)
B.
The solutions are the values of x in the interval
  
enter your response here.
​(Type your answer in interval​ notation.)
C.
There is no solution.

Answer 1

To solve the inequality

\[
\frac{49 - 7x}{x} > 7,
\]

we begin by rearranging the inequality.

1. **Subtract 7 from both sides:**

\[
\frac{49 - 7x}{x} - 7 > 0.
\]

To combine the terms, we express 7 as a fraction:

\[
7 = \frac{7x}{x},
\]

so we rewrite the inequality as:

\[
\frac{49 - 7x - 7x}{x} > 0.
\]

This simplifies to:

\[
\frac{49 - 14x}{x} > 0.
\]

2. **Set the numerator and denominator:**

We need to solve the inequality:

\[
\frac{49 - 14x}{x} > 0.
\]

The critical points occur when the numerator or denominator is zero:

- For the numerator \(49 - 14x = 0\):
\[
49 = 14x \implies x = \frac{49}{14} = \frac{7}{2} = 3.5.
\]

- For the denominator \(x = 0\).

3. **Test intervals:**

The critical points \(x = 0\) and \(x = 3.5\) divide the number line into intervals:

- \( (-\infty, 0) \)
- \( (0, 3.5) \)
- \( (3.5, \infty) \)

We need to test a point from each interval to see where the inequality holds true.

- **Interval \( (-\infty, 0) \)**: Choose \(x = -1\):
\[
\frac{49 - 14(-1)}{-1} = \frac{49 + 14}{-1} = \frac{63}{-1} = -63 \quad \text{(not greater than 0)}.
\]

- **Interval \( (0, 3.5) \)**: Choose \(x = 1\):
\[
\frac{49 - 14(1)}{1} = \frac{49 - 14}{1} = \frac{35}{1} = 35 \quad \text{(greater than 0)}.
\]

- **Interval \( (3.5, \infty) \)**: Choose \(x = 4\):
\[
\frac{49 - 14(4)}{4} = \frac{49 - 56}{4} = \frac{-7}{4} \quad \text{(not greater than 0)}.
\]

4. **Conclusion:**

The solution to the inequality is the interval where the expression is greater than 0:

\( \boxed{(0, 3.5)}. \)

Now, filling this into the answers:

Part B:
The solutions are the values of \(x\) in the interval:
\[
(0, 3.5).
\]