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

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: \[ \text{No solution.} \]