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there are five consecutive integers such that three times the largest integer is three more than twice the sum of the two smallest integers

Solución

there are five consecutive integers such that three times the largest integer is three more than twice the sum of the two smallest integers

Solución

The numbers : - 1 \frac{4}{7}, - \frac{4}{7}, \frac{3}{7}, 1 \frac{3}{7}, 2 \frac{3}{7}

Pasos de solución

Represent the numbers with

x

:

First number : x

Consecutive number 1 : x + 1

Consecutive number 2 : x + 2

Consecutive number 3 : x + 3

Consecutive number 4 : x + 4

A.T.Q.:

3 (x + 4) = 3 + 2 (x + x + 1 + x + 2 + x + 3 + x + 4)

Resolver

3 (x + 4) = 3 + 2 (x + x + 1 + x + 2 + x + 3 + x + 4) : x = - \frac{11}{7}

x + 1 = - \frac{11}{7} + 1 = - \frac{4}{7}

x + 2 = - \frac{11}{7} + 2 = \frac{3}{7}

x + 3 = - \frac{11}{7} + 3 = \frac{10}{7}

x + 4 = - \frac{11}{7} + 4 = \frac{17}{7}

Convertir las fracciones impropias a números mixtos:

\frac{11}{7} = 1 \frac{4}{7}

Convertir las fracciones impropias a números mixtos:

\frac{10}{7} = 1 \frac{3}{7}

Convertir las fracciones impropias a números mixtos:

\frac{17}{7} = 2 \frac{3}{7}

The numbers : - 1 \frac{4}{7}, - \frac{4}{7}, \frac{3}{7}, 1 \frac{3}{7}, 2 \frac{3}{7}

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