Let $\displaystyle{x}=\usepackage{cmbright}\Huge\$ the smallest, $\displaystyle{y}=\usepackage{cmbright}\Huge\$ the middle, and $\displaystyle{z}=\usepackage{cmbright}\Huge\$ the largest number. The system is

$\displaystyle{x}+{y}+{z}={25}\usepackage{cmbright}\Huge\$ , $\displaystyle{y}-{x}={1}\usepackage{cmbright}\Huge\$ , and $\displaystyle{2}{y}-{z}={2}\usepackage{cmbright}\Huge\$

Add the first and second equations: $\displaystyle{2}{y}+{z}={26}\usepackage{cmbright}\Huge\$

Add this to the third equation: $\displaystyle{4}{y}={28}\usepackage{cmbright}\Huge\$ or $\displaystyle{y}={7}\usepackage{cmbright}\Huge\$

Substitute into the second equation to get $\displaystyle{7}-{x}={1}\usepackage{cmbright}\Huge\$ , so $\displaystyle{x}={6}\usepackage{cmbright}\Huge\$ .

Substitute into the third equation to get $\displaystyle{14}-{z}={2}\usepackage{cmbright}\Huge\$ , so $\displaystyle{z}={12}\usepackage{cmbright}\Huge\$ .