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Popular Word Problems >

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

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

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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