1. **State the problem:** We need to find a two-digit number whose digits sum to 12 and when reversed, the new number is \(\frac{4}{7}\) of the original number. 2. **Define variables:** Let the tens digit be \(x\) and the units digit be \(y\). The original number is \(10x + y\). 3. **Write equations from the problem:** - Sum of digits: \(x + y = 12\) - Reversed number: \(10y + x = \frac{4}{7}(10x + y)\) 4. **Express \(y\) from the first equation:** \[ y = 12 - x \] 5. **Substitute \(y\) into the second equation:** \[ 10(12 - x) + x = \frac{4}{7}(10x + 12 - x) \] \[ 120 - 10x + x = \frac{4}{7}(9x + 12) \] \[ 120 - 9x = \frac{4}{7}(9x + 12) \] 6. **Multiply both sides by 7 to clear the denominator:** \[ 7(120 - 9x) = 4(9x + 12) \] \[ 840 - 63x = 36x + 48 \] 7. **Bring all terms to one side:** \[ 840 - 63x - 36x - 48 = 0 \] \[ 792 - 99x = 0 \] 8. **Solve for \(x\):** \[ 792 = 99x \] \[ x = \frac{792}{99} \] \[ x = 8 \] 9. **Find \(y\):** \[ y = 12 - 8 = 4 \] 10. **Final answer:** The original number is \(10x + y = 10 \times 8 + 4 = 84\).