Track / Stage 4 / 12 of 340 #272 of 2000

Problem 272

AMC 12 late, AIME early

Algebra Difficulty 4.4 Find the answer

If $x+y+z=25$ and $y+z=14$ , then $x$ is
(A) 8
(B) 11
(C) 6
(D) -6
(E) 31

Part B: Each question is worth 6 credits.

Your answer

Official solution

If $x+y+z=25$ and $y+z=14$ , then $x$ is
(A) 8
(B) 11
(C) 6
(D) -6
(E) 31

## Solution

We are given that $x+y+z=25$

$\text { and } \quad y+z=14$

Subtract equation (2) from equation (1) to get $x=11$ .

ANSWER: (B)

## PART B:

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.