1. **State the problem:** We are given three parallel segments AD, BE, and CF with lengths expressed as algebraic expressions: $AD = 3x + 20$, $BE = 2x + 14$, and $CF = x + 25$. Since these segments are parallel and presumably part of a proportional system, their lengths are equal. 2. **Set up the equation:** Because $AD \parallel BE \parallel CF$, the lengths are equal, so: $$3x + 20 = 2x + 14 = x + 25$$ 3. **Solve for $x$ using the first two expressions:** $$3x + 20 = 2x + 14$$ Subtract $2x$ from both sides: $$3x - 2x + 20 = 14$$ Simplify: $$x + 20 = 14$$ Subtract 20 from both sides: $$x = 14 - 20$$ $$x = -6$$ 4. **Check with the third expression:** Check if $2x + 14 = x + 25$ when $x = -6$: $$2(-6) + 14 = -12 + 14 = 2$$ $$-6 + 25 = 19$$ Since $2 \neq 19$, the three lengths are not all equal, so we must check if the problem implies equality between all three or just between pairs. 5. **Check equality between $BE$ and $CF$:** $$2x + 14 = x + 25$$ Subtract $x$ from both sides: $$x + 14 = 25$$ Subtract 14 from both sides: $$x = 11$$ 6. **Check if $x=11$ satisfies $AD = BE$:** Calculate $AD$: $$3(11) + 20 = 33 + 20 = 53$$ Calculate $BE$: $$2(11) + 14 = 22 + 14 = 36$$ Since $53 \neq 36$, $x=11$ does not satisfy $AD = BE$. 7. **Conclusion:** The problem likely means the three segments are equal, so set $AD = BE$ and $BE = CF$ and solve simultaneously: From $AD = BE$: $$3x + 20 = 2x + 14$$ $$x = -6$$ From $BE = CF$: $$2x + 14 = x + 25$$ $$x = 11$$ Since $x$ cannot be both $-6$ and $11$, there is no single $x$ that makes all three segments equal. **Final answer:** There is no value of $x$ that satisfies $AD = BE = CF$ simultaneously. If the problem means the segments are proportional or some other relation, please clarify.