1. **State the problem:** We have a kite WXYZ with diagonals XZ and WY intersecting at V. Given that $XZ = 18$ and $WY = 52$, and the sides WX, XZ, WY are labeled as follows: $XZ = 23 - 2x$, $WY = 7x - 13$. We need to find the length of side $WX$. 2. **Use the given diagonal lengths:** Since $XZ = 18$, set up the equation: $$23 - 2x = 18$$ 3. **Solve for $x$:** $$23 - 2x = 18$$ $$-2x = 18 - 23$$ $$-2x = -5$$ $$x = \frac{-5}{-2} = 2.5$$ 4. **Find $WY$ using $x$:** $$WY = 7x - 13 = 7(2.5) - 13 = 17.5 - 13 = 4.5$$ 5. **Check the problem statement:** The problem states $WY = 52$, but our calculation gives $4.5$. This suggests a misunderstanding. The problem states $XZ = 18$ and $WY = 52$ as given lengths, but also labels $XZ = 23 - 2x$ and $WY = 7x - 13$. So these expressions must equal the given lengths. 6. **Set up equations for both diagonals:** $$23 - 2x = 18$$ $$7x - 13 = 52$$ 7. **Solve the first equation:** $$23 - 2x = 18$$ $$-2x = -5$$ $$x = 2.5$$ 8. **Solve the second equation:** $$7x - 13 = 52$$ $$7x = 65$$ $$x = \frac{65}{7} \approx 9.2857$$ 9. **Contradiction in $x$ values:** The two equations give different $x$ values, so the problem likely means the expressions $23 - 2x$ and $7x - 13$ represent sides, not diagonals. 10. **Recall properties of a kite:** The diagonals intersect at right angles, and one diagonal is bisected by the other. Since $XZ = 18$ and $WY = 52$, and $V$ is the intersection point, $V$ bisects $WY$. 11. **Calculate half of $WY$:** $$\frac{WY}{2} = \frac{52}{2} = 26$$ 12. **Calculate half of $XZ$:** $$\frac{XZ}{2} = \frac{18}{2} = 9$$ 13. **Use right triangle $W V X$:** Since diagonals intersect at right angles, triangle $W V X$ is right-angled at $V$. 14. **Calculate $WX$ using Pythagoras theorem:** $$WX = \sqrt{(WV)^2 + (VX)^2} = \sqrt{26^2 + 9^2} = \sqrt{676 + 81} = \sqrt{757}$$ 15. **Simplify the square root:** $$\sqrt{757} \approx 27.5$$ **Final answer:** $$WX \approx 27.5$$