1998 CEMC Pascal Problems/Problem 10

Problem

If $x + y + z = 25$ and $y + z = 14$ , then $x$ is

$\text{ (A) }\ 8 \qquad\text{ (B) }\ 11 \qquad\text{ (C) }\ 6 \qquad\text{ (D) }\ -6 \qquad\text{ (E) }\ 31$

We can plug in the second equation into the first:

$x + y + z = 25$

$x + 14 = 25$

$x = 25 - 14 = \boxed {\textbf {(B) } 11}$

~anabel.disher