Question 336626
Since {{{x}}},{{{y}}},{{{z}}},and {{{w}}} are non-negative integers less than {{{3}}}, they can have values of {{{0}}},{{{1}}},or {{{2}}}.



{{{33w+32x+3y+z=34}}}



Let's look at all possible values.

If {{{w=0}}}, then {{{x=1}}}, and {{{z=2}}} and {{{y=0}}}, since {{{32+0+0+2=34}}}, {{{w+z=2}}}

If {{{w=1}}}, then {{{z=1}}} and {{{x=y=0}}} since {{{33+0+0+1=34}}}, {{{w+z=2}}}

If {{{w=2}}}, {{{w}}} can't equal {{{2}}} because {{{33(2)=66>34}}}.
Those are all of the possible solutions.

So,  {{{highlight(w+z=2)}}}.