Answer:
The three even, consecutive integers are 72, 74, and 76.
Step-by-step explanation:
Let a be the first even integer.
Then the other two consecutive integers, b and c, will be represented by:
[tex]\displaystyle b = a + 2 \text{ and } \\ \\ c = b + 2 = (a + 2) + 2 = a + 4[/tex]
One-sixth of the smallest (that is, a) is three less than one-tenth the sum of the other two even integers (that is, b and c).
Therefore:
[tex]\displaystyle \frac{1}{6} a = \frac{1}{10}\left( b +c\right) - 3[/tex]
Solve for a. Substitute:
[tex]\displaystyle \frac{1}{6} a =\frac{1}{10}\left((a+2)+(a+4)\right) - 3[/tex]
Simplify and solve for a:
[tex]\displaystyle \begin{aligned} \frac{1}{6} a &= \frac{1}{10}(2a + 6) - 3 \\ \\ \frac{1}{6} a &= \frac{1}{5} a + \frac{3}{5} - 3\\ \\ -\frac{1}{30} a &= -\frac{12}{5} \\ \\ a &= 72\end{aligned}[/tex]
Hence, the first even integer is 72.
Therefore, the two other consecutive even integers must be 74, and 76.
In conclusion, the three even, consecutive integers are 72, 74, and 76.
