Find the two consecutive odd integers such that the product of the integers is 77 more than twice the larger integer.

Answer:
Step-by-step explanation:
Consecutive odd integers increase by 2 so:
Let the first integer be y,
The Second will be y+2
Therefore y×y+2=77+2(y+2)
y²+2y=77+2y+4
Move all the terms to one side NB:when the terms move over they change thier sign
y²+2y-81-2y=0
y²+2y-2y-81=0
y²-81=0
y²=81
Take square root
Y=9

Answered by toifbola | 2024-06-12

The first integer is y = 9.
What are the consecutive numbers?
The **consecutive **numbers are those number which follow each other continuously in the order from smallest to largest numbers.
Given tha **Consecutive odd **integers increase by 2 so:
Let the first **integer **be y, then the Second will be y+2
Therefore,
y×y+2=77+2(y+2)
y²+2y=77+2y+4
Now Move all the terms to one side;
y²+2y-81-2y=0
y²+2y-2y-81=0
y²-81=0
y²=81
Y=9
Learn more about the consecutive numbers ;
brainly.com/question/24912446
#SPJ2

Answered by shivishivangi1679 | 2024-06-24

The two consecutive odd integers are 9 and 11. Their product is 99, which is 77 more than twice the larger integer (22 + 77 = 99). Therefore, the condition from the problem is satisfied.
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Answered by shivishivangi1679 | 2025-01-27