<p>Solution by starting with choice (C): If we take (C) as our first guess, then 15 students study only Italian, and 18 study only German, so 15+18=33 study both. This is already way too big. So let’s try (A) next. Then 6 study Italian and 9 study German. So 6+9=15 study both. All together we have 30 so that (A) is the answer.</p>
<p>Quickest Solution: Since there are 30 students, and the number that study both subjects equals the number that study only one subject, there are 15 students that study only one subject. So we want to split up 15 as a sum of two numbers such that one is 3 more than the other - we have 9 + 6 = 15 (we can get this using “informal” or “formal” algebra). So 6 students study Italian only, choice (A).</p>
<p>Notes: (1) Here is the “formal” algebra in case you need it: Let x be the number of students that study Italian only. Then the number of students that study German only is x+3. So x + (x+3) = 15. Thus, 2x + 3 = 15. So 2x = 12, and x = 6.</p>
<p>(2) To do this with a Venn diagram, you would draw 2 overlapping circles labelling the first G and the second I. The intersection gets 15 right away, and then we need to put a total of 15 in the other two regions such that region G has 3 more than region I (by G, I actually mean “G only,” and similarly for I) . This should clearly be 9 and 6, respectively.</p>
<p>(3) To solve this algebraically, we can let x be the number of students that study German only, y the number of students that study both, and z the number of students that study Italian only. Then we get the system of equations:</p>
<p>x+y+z=30
x+z=y
x+y=y+z+3</p>
<p>This system simplifies to:</p>
<p>x+y+z=30
x-y+z=0
x-z=3</p>
<p>You can solve this system pretty quickly by hand using the elimination method, or you can do Gauss Jordan Reduction pretty quickly in your graphing calculator by inputting a matrix and using the rref( feature.</p>