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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

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}

Solution steps

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)

Solve

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}

Convert improper fractions to mixed numbers:

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

Convert improper fractions to mixed numbers:

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

Convert improper fractions to mixed numbers:

\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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