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hjlf
3 years ago
10

One-sixth of the smallest of three consecutive even

Mathematics
1 answer:
MrRissso [65]3 years ago
8 0

Answer:

The three even, consecutive integers are 72, 74, and 76.

Step-by-step explanation:

Let <em>a</em> be the first even integer.

Then the other two consecutive integers, <em>b</em> and <em>c</em>, will be represented by:

\displaystyle b = a + 2 \text{ and } \\ \\ c = b + 2 = (a + 2) + 2 = a + 4

One-sixth of the smallest (that is, <em>a</em>) is three less than one-tenth the sum of the other two even integers (that is, <em>b</em> and <em>c</em>).

Therefore:

\displaystyle \frac{1}{6} a = \frac{1}{10}\left( b  +c\right) - 3

Solve for <em>a</em>. Substitute:

\displaystyle \frac{1}{6} a =\frac{1}{10}\left((a+2)+(a+4)\right) - 3

Simplify and solve for <em>a: </em>

\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}

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.

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