1. **State the problem:** We need to find the length of segment $BM$ in a circle where two chords $ML$ and $JK$ intersect at point $B$ inside the circle. 2. **Recall the chord intersection theorem:** When two chords intersect inside a circle, the products of the lengths of the segments of each chord are equal. That is, $$MB \times BL = JB \times BK$$ 3. **Identify the segments:** From the problem, - On chord $ML$, the segments are $MB = 2x - 6$ and $BL = 15$. - On chord $JK$, the segments are $JB = 12$ and $BK = 10$. 4. **Set up the equation using the theorem:** $$ (2x - 6) \times 15 = 12 \times 10 $$ 5. **Simplify both sides:** $$ 15(2x - 6) = 120 $$ 6. **Distribute 15:** $$ 30x - 90 = 120 $$ 7. **Add 90 to both sides:** $$ 30x - 90 + 90 = 120 + 90 $$ $$ 30x = 210 $$ 8. **Divide both sides by 30:** $$ \cancel{30}x = \frac{210}{\cancel{30}} $$ $$ x = 7 $$ 9. **Find $BM$ by substituting $x=7$ into $2x - 6$:** $$ BM = 2(7) - 6 = 14 - 6 = 8 $$ **Final answer:** $$ BM = 8 $$