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Popular Word Problems >

Find four consecutive integers such that twice the sum of the first and third is 11 greater than 3 times the second.

Solution

F in d f o u r co n sec u t i v e in t e g ers s u c h t ha t tw i ce t h e s u m o f t h e f i rs t an d t hi r d i s 11 g re a t er t han 3 t im es t h e seco n d .

Solution

The numbers : - 13 \frac{1}{3}, - 12 \frac{1}{3}, - 11 \frac{1}{3}, - 10 \frac{1}{3}

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

A.T.Q.:

2 (x + \frac{1}{3}) = 11 + 3 (x + 1)

Solve

2 (x + \frac{1}{3}) = 11 + 3 (x + 1) : x = - \frac{40}{3}

x + 1 = - \frac{40}{3} + 1 = - \frac{37}{3}

x + 2 = - \frac{40}{3} + 2 = - \frac{34}{3}

x + 3 = - \frac{40}{3} + 3 = - \frac{31}{3}

Convert improper fractions to mixed numbers:

\frac{40}{3} = 13 \frac{1}{3}

Convert improper fractions to mixed numbers:

\frac{37}{3} = 12 \frac{1}{3}

Convert improper fractions to mixed numbers:

\frac{34}{3} = 11 \frac{1}{3}

Convert improper fractions to mixed numbers:

\frac{31}{3} = 10 \frac{1}{3}

The numbers : - 13 \frac{1}{3}, - 12 \frac{1}{3}, - 11 \frac{1}{3}, - 10 \frac{1}{3}

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