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

해법

t h ere a re f i v e co n sec u t i v e in t e g ers s u c h t ha t t h ree t im es t h e l a r g es t in t e g er i s t h ree m ore t han tw i ce t h e s u m o f t h e tw o s ma ll es t in t e g ers

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

솔루션 단계

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)

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

해결 :

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}

부적절한 분수를 혼합 숫자로 변환:

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

부적절한 분수를 혼합 숫자로 변환:

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

부적절한 분수를 혼합 숫자로 변환:

\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}