1. The first problem is to find if there exists an integer $x$ such that $$2x^2 - 4 = 0.$$ 2. Start by solving the equation: $$2x^2 - 4 = 0$$ Add 4 to both sides: $$2x^2 = 4$$ Divide both sides by 2: $$x^2 = 2$$ 3. Now, find $x$ by taking the square root: $$x = \pm \sqrt{2}$$ 4. Since $\sqrt{2}$ is irrational, there is no integer $x$ that satisfies the equation. So, the statement $(\exists x \in \mathbb{Z})(2x^2 - 4 = 0)$ is false. 5. The second problem is to check if there exists a real number $x$ such that for all real numbers $y$, the equation $$2x + y = 5$$ holds. 6. Rearrange the equation to solve for $y$: $$y = 5 - 2x$$ 7. For the statement $(\exists x \in \mathbb{R})(\forall y \in \mathbb{R})(2x + y = 5)$ to be true, the equation must hold for every $y$ in $\mathbb{R}$. 8. But $y = 5 - 2x$ defines a single value of $y$ for each $x$, not all $y$. So, there is no $x$ such that $2x + y = 5$ for all $y$. 9. Therefore, the statement is false. Final answers: - No integer $x$ satisfies $2x^2 - 4 = 0$. - There is no real $x$ such that $2x + y = 5$ for all real $y$.