Show your own working for this question and i am warning: Only curiousRose will give answer no one will give answer
Eight times the first of three consecutive odd integers is 4 less than thrice the sum of second and third. The second integer is
Share
Given :
Some conditions for three consecutive odd integers..
To find :
Second odd integer
Solution :
Let the three consecutive odd integers be (x + 1), (x + 3) and (x + 5)
So,
▪Eight times the first odd integer
=> 8(x + 1)
▪Sum of second and third integer
=> (x + 3) + (x + 5)
=> (x + 3 + x + 5)
=> (2x + 8)
▪ Thrice of the sum of second and third integer
=> 3(2x + 8)
▪ 4 less than thrice the sum of second and third. The second integer
=> 3(2x + 8) - 4
According to the condition given,
[tex] = > 8(x + 1) = 3(2x + 8) - 4[/tex]
[tex] = > 8x + 8 = 6x + 24 - 4[/tex]
[tex] = > 8x - 6x = 24 - 4 - 8[/tex]
[tex] = > 2x = 12[/tex]
[tex] = > x = \frac{12}{2} = 6 \\ [/tex]
If x = 6 then,
Second odd integer = (x + 3)
=> (6 + 3) = 9
Therefore, the second odd integer is 9.