1. The problem asks for the value of $x$ such that quadrilateral ABCD is a parallelogram. 2. In a parallelogram, opposite sides are equal in length. 3. Given that side $AD = 3x + 1$ and side $CB = 6x - 4$, for ABCD to be a parallelogram, these must be equal: $$3x + 1 = 6x - 4$$ 4. Solve for $x$: $$3x + 1 = 6x - 4$$ Subtract $3x$ from both sides: $$\cancel{3x} + 1 = \cancel{3x} + 3x - 4 \implies 1 = 3x - 4$$ Add 4 to both sides: $$1 + 4 = 3x - 4 + 4 \implies 5 = 3x$$ Divide both sides by 3: $$\frac{5}{\cancel{3}} = \frac{3x}{\cancel{3}} \implies x = \frac{5}{3}$$ 5. Therefore, the value of $x$ that makes ABCD a parallelogram is $\boxed{\frac{5}{3}}$.