12. If \( D E=4 x+10, E F=2 x-1 \), and \( D F=9 x-15 \), find \( D F \).

To solve for \( D F \), we can use the fact that the segments \( D E \), \( E F \), and \( D F \) are related in the following way: \( D F = D E + E F \). Given: - \( D E = 4x + 10 \) - \( E F = 2x - 1 \) - \( D F = 9x - 15 \) Now, we substitute the expressions for \( D E \) and \( E F \) into the equation for \( D F \): \[ D F = D E + E F \] So we have: \[ 9x - 15 = (4x + 10) + (2x - 1) \] Combining the right side gives: \[ 9x - 15 = 4x + 2x + 10 - 1 \] \[ 9x - 15 = 6x + 9 \] Next, we can solve for \( x \) by moving \( 6x \) to the left side: \[ 9x - 6x - 15 = 9 \] \[ 3x - 15 = 9 \] Now, adding 15 to both sides: \[ 3x = 24 \] \[ x = 8 \] Now that we have \( x \), we can substitute it back to find \( D F \): \[ D F = 9x - 15 \] \[ D F = 9(8) - 15 \] \[ D F = 72 - 15 \] \[ D F = 57 \] Thus, the value of \( D F \) is \( \boxed{57} \).