We are asked to determine three consecutive even numbers such that twice the first is 20 more than the second. Let "x" be the first number, therefore, we have:
[tex]\begin{gathered} n_1=x \\ n_2=x+2 \\ n_3=x+4 \end{gathered}[/tex]
Where:
[tex]n_{1,}n_2,n_3=\text{ first, second, and third numbers}[/tex]
According to the condition given we have:
[tex]2n_1=n_2+20[/tex]
Now, we substitute for the numbers in terms of "x":
[tex]2x=x+2+20[/tex]
Adding like terms:
[tex]2x=x+22[/tex]
Now, we subtract "x" from both sides:
[tex]\begin{gathered} 2x-x=x-x+22 \\ x=22 \end{gathered}[/tex]
Therefore, the first number is 22. The second and third numbers are then:
[tex]\begin{gathered} n_1=22 \\ n_2=24 \\ n_3=26 \end{gathered}[/tex]
