Find 3 consecutive odd integers such that 10 times the first is 59 more than the third?

The required three consecutive odd integers are 7, 9 , 11.

Step-by-step explanation:

Let x , x+2 , x+4 be the three consecutive odd integers.

As per given ,  

10(x) = 59+x+4

⇒10x= 59+x+4

⇒ 9x= 63

⇒ x= 7   [ divide both sides by 7]

Hence, the required three consecutive odd integers are 7, 9 , 11.

15

step-by-step explanation:

answer;

the correct answer

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