[tex]\begin{gathered} \text{Let:} \\ x=\text{First number} \\ y=\text{Sencond number} \\ z=\text{Third number} \\ x+y+z=26 \end{gathered}[/tex]
The second number is twice than the first, so:
[tex]y=2x[/tex]
The third number is 6 more than the second, so:
[tex]z=2x+6[/tex]
Replacing the data into the equation:
[tex]\begin{gathered} x+2x+2x+6=26 \\ 5x+6=26 \\ \text{Solving for x:} \\ 5x=26-6 \\ 5x=20 \\ x=\frac{20}{5} \\ x=4 \end{gathered}[/tex]
And:
[tex]\begin{gathered} y=2(4)=8 \\ z=2(4)+6=8+6=14 \\ 4+8+14=26 \\ \text{Which is correct} \end{gathered}[/tex]
