1. **Stating the problem:** We are given a trapezoid IJKL with two bases IJ and LK expressed in terms of $x$, and a median MN with length 15. We need to find the value of $x$ and then find the lengths of the bases IJ and LK. 2. **Formula and rules:** The median (or mid-segment) of a trapezoid is the segment that connects the midpoints of the non-parallel sides. Its length is the average of the lengths of the two bases. Mathematically, this is: $$MN = \frac{IJ + LK}{2}$$ 3. **Set up the equation:** Substitute the given expressions: $$15 = \frac{(4x - 6) + (2x + 10)}{2}$$ 4. **Simplify the numerator:** $$15 = \frac{4x - 6 + 2x + 10}{2} = \frac{6x + 4}{2}$$ 5. **Multiply both sides by 2 to clear the denominator:** $$2 \times 15 = \cancel{2} \times \frac{6x + 4}{\cancel{2}}$$ $$30 = 6x + 4$$ 6. **Solve for $x$:** $$30 - 4 = 6x$$ $$26 = 6x$$ $$x = \frac{26}{6}$$ 7. **Simplify the fraction:** $$x = \frac{\cancel{26}}{\cancel{6}} = \frac{13}{3}$$ 8. **Find the lengths of the bases:** - For $IJ$: $$IJ = 4x - 6 = 4 \times \frac{13}{3} - 6 = \frac{52}{3} - 6 = \frac{52}{3} - \frac{18}{3} = \frac{34}{3}$$ - For $LK$: $$LK = 2x + 10 = 2 \times \frac{13}{3} + 10 = \frac{26}{3} + 10 = \frac{26}{3} + \frac{30}{3} = \frac{56}{3}$$ **Final answers:** $$x = \frac{13}{3}$$ $$IJ = \frac{34}{3}$$ $$LK = \frac{56}{3}$$