split - Consecutive Odd Integers

[TamaraHill]

Determine three consecutive odd integers such that the twice the largest plus three timed the smallest is 58 more than the "middle" number

 

[Denis]

Re: Consecutive Odd Integers

Determine three consecutive odd integers such that the twice the largest plus three timed the smallest is 58 more than the "middle" number

Let 2x + 1 = smallest, then middle = 2x + 3 and highest = 2x + 5

2(2x + 5) + 3(2x + 1) = 2x + 3 + 58
Solve for x

 

[masters]

Re: Consecutive Odd Integers

Determine three consecutive odd integers such that the twice the largest plus three timed the smallest is 58 more than the "middle" number

Hi TamaraHill,

Let x = smallest odd integer

Let x + 2 = middle odd integer

Let x + 4 = largest odd integer

\(\displaystyle 2(x+4)+3x=(x+2)+58\)

 

[masters]

Re: Consecutive Odd Integers

Determine three consecutive odd integers such that the twice the largest plus three timed the smallest is 58 more than the "middle" number

Let 2x + 1 = smallest, then middle = 2x + 3 and highest = 2x + 5

2(2x + 5) + 3(2x + 1) = 2x + 3 + 58
Solve for x

Denis, did you solve this equation?

 

[Deleted member 4993]

Re: Consecutive Odd Integers

Determine three consecutive odd integers such that the twice the largest plus three timed the smallest is 58 more than the "middle" number

Let 2x + 1 = smallest, then middle = 2x + 3 and highest = 2x + 5

2(2x + 5) + 3(2x + 1) = 2x + 3 + 58
Solve for x

Denis, did you solve this equation?

x = 6

and the odd numbers are 13, 15 & 17

 

[TamaraHill]

Re: Consecutive Odd Integers

Determine three consecutive odd integers such that the twice the largest plus three timed the smallest is 58 more than the "middle" number

Hi TamaraHill,

Let x = smallest odd integer

Let x + 2 = middle odd integer

Let x + 4 = largest odd integer

[2x+8+3x]2(x+4)+3x=(x+2)+58[x+2+58]

5x+8=x+60
4x+8=60
4x=52
x=13

 

[masters]

Re: Consecutive Odd Integers

Determine three consecutive odd integers such that the twice the largest plus three timed the smallest is 58 more than the "middle" number

Let 2x + 1 = smallest, then middle = 2x + 3 and highest = 2x + 5

2(2x + 5) + 3(2x + 1) = 2x + 3 + 58
Solve for x

Denis, did you solve this equation?

x = 6

and the odd numbers are 13, 15 & 17

Yeah, thanks. I ignored the initial setup. Sorry Denis. Whoopsie Daisy!

 

[Denis]

Re: Consecutive Odd Integers

Yeah, thanks. I ignored the initial setup. Sorry Denis. Whoopsie Daisy!

OKAY Masters....go stand in the corner ...

Thanks for coming to my rescue, Subhotosh : I presume you want to borrow some money :shock:

Btw, I still maintain that 2x+1 is better than x as an "odd number" :idea: