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Find four consecutive even integers such that if the sum of the first and the third is multiplied by 3, the result is 28 less than 5 times the fourth.

Solución

Solución

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Pasos de solución

Represent the numbers with

x

First number : x

Consecutive number 1 : x + 2

Consecutive number 2 : x + 4

Consecutive number 3 : x + 6

A.T.Q.:

(x + \frac{1}{3}) \cdot 3 = 5 (x + 6) - 28

Resolver

(x + \frac{1}{3}) \cdot 3 = 5 (x + 6) - 28 : x = - \frac{1}{2}

x + 2 = - \frac{1}{2} + 2 = \frac{3}{2}

x + 4 = - \frac{1}{2} + 4 = \frac{7}{2}

x + 6 = - \frac{1}{2} + 6 = \frac{11}{2}

Odd value

\frac{3}{2}

discarded

Sin soluci \overset{o}{ˊ} n

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