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Populaire >

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

Solution

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

Solution

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

étapes des solutions

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)

Résoudre

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 des fractions impropres en nombres mixtes:

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

Convertir des fractions impropres en nombres mixtes:

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

Convertir des fractions impropres en nombres mixtes:

\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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