$$ 1

Question

$$ 1 < \quad \frac { 5 x - 10 } { 7 x + 14 } \cdot \frac { 4 x + 8 } { 14 x - 28 } $$

UpStudy AI · Accepted Answer

To solve the inequality

$$
1 < \frac{5x - 10}{7x + 14} \cdot \frac{4x + 8}{14x - 28},
$$

we first simplify the expression on the right side.

1. **Factor the expressions:**

- The first fraction:
     $$
     5x - 10 = 5(x - 2),
     $$
     $$
     7x + 14 = 7(x + 2).
     $$

- The second fraction:
     $$
     4x + 8 = 4(x + 2),
     $$
     $$
     14x - 28 = 14(x - 2).
     $$

2. **Substituting the factored forms:**
   The inequality becomes:
   $$
   1 < \frac{5(x - 2)}{7(x + 2)} \cdot \frac{4(x + 2)}{14(x - 2)}.
   $$

3. **Cancel common factors:**
   The $$ (x - 2) $$ terms cancel out (assuming $$ x \neq 2 $$), and the $$ (x + 2) $$ terms also cancel out (assuming $$ x \neq -2 $$):
   $$
   1 < \frac{5 \cdot 4}{7 \cdot 14} = \frac{20}{98} = \frac{10}{49}.
   $$

4. **Rearranging the inequality:**
   We can rewrite the inequality as:
   $$
   \frac{10}{49} > 1.
   $$
   This is not true, so we need to analyze the conditions under which the original inequality holds.

5. **Finding critical points:**
   The critical points occur when the numerator or denominator is zero:
   - $$ 5x - 10 = 0 $$ gives $$ x = 2 $$.
   - $$ 7x + 14 = 0 $$ gives $$ x = -2 $$.
   - $$ 4x + 8 = 0 $$ gives $$ x = -2 $$ (already found).
   - $$ 14x - 28 = 0 $$ gives $$ x = 2 $$ (already found).

6. **Testing intervals:**
   The critical points divide the number line into intervals: $$ (-\infty, -2) $$, $$ (-2, 2) $$, and $$ (2, \infty) $$.

- **Interval $$ (-\infty, -2) $$:** Choose $$ x = -3 $$:
     $$
     \frac{5(-3) - 10}{7(-3) + 14} \cdot \frac{4(-3) + 8}{14(-3) - 28} = \frac{-15 - 10}{-21 + 14} \cdot \frac{-12 + 8}{-42 - 28} = \frac{-25}{-7} \cdot \frac{-4}{-70} = \frac{25}{7} \cdot \frac{4}{70} = \frac{100}{490} = \frac{10}{49}.
     $$
     Since $$ 1 < \frac{10}{49} $$ is false, this interval does not satisfy the inequality.

- **Interval $$ (-2, 2) $$:** Choose $$ x = 0 $$:
     $$
     \frac{5(0) - 10}{7(0) + 14} \cdot \frac{4(0) + 8}{14(0) - 28} = \frac{-10}{14} \cdot \frac{8}{-28} = \frac{-10}{14} \cdot \frac{-8}{28} = \frac{10 \cdot 8}{14 \cdot 28} = \frac{80}{392} = \frac{10}{49}.
     $$
     Since $$ 1 < \frac{10}{49} $$ is false, this interval does not satisfy the inequality.

- **Interval $$ (2, \infty) $$:** Choose $$ x = 3 $$:
     $$
     \frac{5(3) - 10}{7(3) + 14} \cdot \frac{4(3) + 8}{14(3) - 28} = \frac{15 - 10}{21 + 14} \cdot \frac{12 + 8}{42 - 28} = \frac{5}{35} \cdot \frac{20}{14} = \frac{1}{7} \cdot \frac{10}{7} = \frac{10}{49}.
     $$
     Since $$ 1 < \frac{10}{49} $$ is false, this interval does not satisfy the inequality.

7. **Conclusion:**
   The inequality $$ 1 < \frac{5x - 10}{7x + 14} \cdot \frac{4x + 8}{14x - 28} $$ does not hold for any $$ x $$ in the real numbers, except at the points where the expression is undefined (i.e., $$ x = -2 $$ and $$ x = 2 $$).

Thus, the solution to the inequality is:

$$
\text{No solution.}
$$
The inequality has no solution.

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