Find three consecutive odd integers such that the sum of the
first, two times the second, and three times the third is 70.
Answers
Answer:
9,11,13
Step-by-step explanation:
Given :The sum of the first, twice the second, and three times the third = 70
To find : The three consecutive odd integers.
Proof:
Let the first odd integer = x ,
The second consecutive odd integer = x+2 ,
The third consecutive odd integer = (x+2)+2 = x+4
So,The sum of the first, twice the second, and three times the third will be
x+2(x+2)+3(x+4)
Given the sum of the first, twice the second, and three times the third = 70
∴ x+2(x+2)+3(x+4) = 70
=> x+2x+4+3x+12 = 70
=> 6x+16 = 70
=> 6x = 70-16
=> 6x = 54
=> x = 54÷6
=> x = 9
Now we have;
first odd integer = x
=> x= 9
The second consecutive odd integer = x+2
=> x+2
=> 9+2 = 11
=> x = 11
The third consecutive odd integer = x+4
=> x+4
=> 9+4 = 13
=> x = 13
∴ the three consecutive odd integers x=9, x=11, x= 13.
Hence, the three consecutive odd integers are 9,11,and 13
Answer:
9, 11, 13
Step-by-step explanation:
SOLUTION:
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