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Find two consecutive odd integers such that
4 times the smaller integer is 19 less than 5
times the larger integer.

Sagot :

Answer:

9 and 11

Step-by-step explanation:

let the consecutive odd integers be n and n + 2 , then

4n = 5(n + 2) - 19 , that is

4n = 5n + 10 - 19 ( subtract 5n from both sides )

- n = - 9 ( multiply both sides by - 1 )

n = 9 and n + 2 = 9 + 2 = 11

The 2 integers are 9 and 11

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