the sum of three consecutive odd integers such that seven times the sum of the first two integers is three more than nine times the third integer

解

t h e s u m o f t h ree co n sec u t i v e o dd in t e g ers s u c h t ha t se v e n t im es t h e s u m o f t h e f i rs t tw o in t e g ers i s t h ree m ore t han nin e t im es t h e t hi r d in t e g er

The numbers : - 5 \frac{2}{7}, - 3 \frac{2}{7}, - 1 \frac{2}{7}

解答ステップ

Represent the numbers with

x

First number : x

Consecutive number 1 : x + 2

Consecutive number 2 : x + 4

A.T.Q.:

x + x + 2 = 3 + 9 (x + 4)

解く

x + x + 2 = 3 + 9 (x + 4) : x = - \frac{37}{7}

x + 2 = - \frac{37}{7} + 2 = - \frac{23}{7}

x + 4 = - \frac{37}{7} + 4 = - \frac{9}{7}

仮分数を帯分数に変換する:

\frac{37}{7} = 5 \frac{2}{7}

仮分数を帯分数に変換する:

\frac{23}{7} = 3 \frac{2}{7}

仮分数を帯分数に変換する:

\frac{9}{7} = 1 \frac{2}{7}

The numbers : - 5 \frac{2}{7}, - 3 \frac{2}{7}, - 1 \frac{2}{7}