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Phổ biến >

Find four consecutive integers such that 8 times the sum of the first and the third is 40 greater than 10 times the fourth.

Lời Giải

Find four consecutive integers such that 8 times the sum of the first and the third is 40 greater than 10 times the fourth.

Lời Giải

The numbers : - 33 \frac{2}{3}, - 32 \frac{2}{3}, - 31 \frac{2}{3}, - 30 \frac{2}{3}

Các bước giải pháp

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.:

8 (x + \frac{1}{3}) = 40 + 10 (x + 3)

Giải

8 (x + \frac{1}{3}) = 40 + 10 (x + 3) : x = - \frac{101}{3}

x + 1 = - \frac{101}{3} + 1 = - \frac{98}{3}

x + 2 = - \frac{101}{3} + 2 = - \frac{95}{3}

x + 3 = - \frac{101}{3} + 3 = - \frac{92}{3}

Chuyển phân số không thực sự thành hỗn số:

\frac{101}{3} = 33 \frac{2}{3}

Chuyển phân số không thực sự thành hỗn số:

\frac{98}{3} = 32 \frac{2}{3}

Chuyển phân số không thực sự thành hỗn số:

\frac{95}{3} = 31 \frac{2}{3}

Chuyển phân số không thực sự thành hỗn số:

\frac{92}{3} = 30 \frac{2}{3}

The numbers : - 33 \frac{2}{3}, - 32 \frac{2}{3}, - 31 \frac{2}{3}, - 30 \frac{2}{3}

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