Each of $w, x, y$, and $z$ is an integer. If $w+x=45, x+y=51$, and $y+z=28$, what is the value of $w+z$ ?
(A) 28
(B) 22
(C) 17
(D) 23
(E) 15
Since $w+x=45$ and $x+y=51$, then $x+y$ is 6 more than $w+x$ (since 51 is 6 more than 45). Both $x+y$ and $w+x$ contain $x$, and so the difference between these two expressions, 6 , must be the difference between $y$ and $w$.

That is, $y$ is 6 more than $w$ (or $y=6+w$ ).

In the final equation given, $y+z=28, y$ is 6 more than $w$, and so 6 more than $w$ added to $z$ equals 28 .

Since $6+22=28$, then $w+z=22$.

ANsWER: (B)