48. In \( \triangle G H J, A \) is the midpoint of \( \overline{G H}, \overline{C B} \) is a midsegment, and \( \overline{C B} \| \overline{G H} \). What is \( G A \) when \( G H=7 x-5 \) and \( C B=2 x+2 \) ? \( G A=\square \)

In \( \triangle G H J \), point \( A \) is the midpoint of \( \overline{G H} \), and \( \overline{C B} \) is a midsegment parallel to \( \overline{G H} \). By the properties of midsegments in a triangle: 1. The midsegment \( \overline{C B} \) is parallel to \( \overline{G H} \) and is half its length. Therefore, we have the equation: \[ C B = \frac{1}{2} G H \] 2. Substitute the given expressions for \( C B \) and \( G H \): \[ 2x + 2 = \frac{1}{2}(7x - 5) \] 3. Multiply both sides by 2 to eliminate the fraction: \[ 4x + 4 = 7x - 5 \] 4. Rearrange the terms to solve for \( x \): \[ 4 + 5 = 7x - 4x \\ 9 = 3x \\ x = 3 \] 5. Substitute \( x = 3 \) back into \( G H \): \[ G H = 7(3) - 5 = 21 - 5 = 16 \] 6. Since \( A \) is the midpoint of \( \overline{G H} \), the length \( G A \) is half of \( G H \): \[ G A = \frac{G H}{2} = \frac{16}{2} = 8 \] **Answer:** 8