Recall that:
[tex]a\sqrt{b}+c\sqrt{b}=(a+c)\sqrt{b}.[/tex]Therefore:
[tex]\begin{gathered} 2\sqrt{2}+3\sqrt{2}=(2+3)\sqrt{2}=5\sqrt{2}, \\ 3\sqrt{2}-5\sqrt{2}=(3-5)\sqrt{2}=-2\sqrt{2}. \end{gathered}[/tex]Substituting the above results in the given expression we get:
[tex]5\sqrt{2}-2\sqrt{2}.[/tex]Then:
[tex](2\sqrt{2}+3\sqrt{2})+(3\sqrt{2}-5\sqrt{2})=3\sqrt{2}.[/tex]Now, we know that
[tex]\sqrt{2}[/tex]is an irrational number, then
[tex]3\sqrt{2}[/tex]is also an irrational number.
Answer: The given expression is an irrational number.
