Prove that in the ring $Z[\sqrt2]$: $\langle3+8\sqrt2, 7\rangle$ = $\langle3+\sqrt2\rangle$ So, 
I know that $\langle a\rangle$ = $aR$ and $\langle a,b\rangle$ = $aR + bR$ for any ring R.
I then multiplied each side by $\sqrt2$ and computed them.
These are the steps that I took:
$\langle3+8\sqrt2, 7\rangle$ = $\langle3+\sqrt2\rangle$
$(3+8\sqrt2)\sqrt2 + 7\sqrt2 = (3+\sqrt2)\sqrt2$
$3\sqrt2 + 16 + 7\sqrt2 = 3\sqrt2 + 2$
$16 + 10\sqrt2 = 2+3\sqrt2$
I'm not really sure where to go from here.
Any help would be much appreciated. 

Simply multiplying the generators of the ideals by $\,\sqrt{2}\,$ will not work.
To be more precise, $\,\langle a\rangle = \{\alpha a \mid \alpha \in R\}\,$ and $\,\langle b,c\rangle = \{\beta b + \gamma c\mid \beta, \gamma \in R\}.\,$
To show that two ideals are equal, it is sufficient to show that the each generator of one is an $\,R$-linear combination of the generators of the other, and vice versa. In your specific case, you need to show that:
(i)$\quad\,7=r_1(3+\sqrt2),\,$ for some $\,r_2 \in R\,;\,$ and
(ii) $\quad\,3+8\sqrt2=r_2(3+\sqrt2),\,$ for some $\,r_1 \in R\,;$
(iii)$\quad3+\sqrt2=r_3(3+8\sqrt2)+r_4(7),\,$ for some $\,r_3, r_4 \in R\,.$
For part (i), you might note by inspection that $\,r_1 = 3-\sqrt2,\,$ as $\,(3-\sqrt2)(3+\sqrt2)=9-2=7;\,$ if that's not immediately apparent, you can just do $\,r_1=\frac{7}{3+\sqrt2}\,$ etc.
Then for part (ii), note that $\,3+8\sqrt2=(3+\sqrt2)+7\sqrt{2},\,$ and then you can just substitute the $\,7\,$ with the expression found in part (i) and solve for $\,r_2\,$.
Finally for part (iii), note that $\,3+\sqrt2=(3+8\sqrt2)-(\sqrt{2})(7),\,$ so it should be immediately clear what $\,r_3\,$ and $\,r_4\,$ should be.
The complete solutions are below in case any of the steps are proving elusive.
(i)

By inspection, $\,r_1=3-\sqrt{2};\, \text{or } \,r_1=\dfrac{7}{3+\sqrt2}=\dfrac{7(3-\sqrt2)}{(3+\sqrt2)(3-\sqrt2)}=\dfrac{7(3-\sqrt2)}{9-2}=3-\sqrt2\,$

(ii)

$\,3+8\sqrt2 \, = (3+\sqrt2)+7\sqrt{2} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = (3+\sqrt2) + (3+\sqrt2)(3-\sqrt2)\sqrt{2} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = (3+\sqrt2) \bigl[\,1+(3-\sqrt2)\sqrt{2}\,\bigr] \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad= (3+\sqrt2) \bigl[\,1+3\sqrt2-2\,\bigr] \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = (3+\sqrt2) (3\sqrt2-1)\, \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \text{so} \,r_2 = (3\sqrt2-1).$

(iii)

$\quad3+\sqrt2=(3+8\sqrt2)-(\sqrt{2})(7),\,$ so $\,r_3=1\,$ and $\,r_4=-\sqrt{2}.$

