prissyn2344
Answered

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find three consecutive odd integers such that the sum of the last two is 15 less than 5 times the first​

Sagot :

hnori22003

2n + 1 , 2n + 3 , 2n + 5

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( 2n + 5 ) + ( 2n + 3 ) = 5 × ( 2n + 1 ) - 15

4n + 8 = 10n + 5 - 15

4n + 8 = 10n - 10

Add both sides 10

4n + 8 + 10 = 10n - 10 + 10

4n + 18 = 10n

Subtract both sides 4n

4n - 4n + 18 = 10n - 4n

18 = 6n

Divide both sides by 6

18 ÷ 6 = 6n ÷ 6

n = 3

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Thus that three numbers are :

2(3) + 1 , 2(3) + 3 , 2(3) + 5

6 + 1 , 6 + 3 , 6 + 5

7 , 9 , 11

And we're done ....

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