Question

If 6 is subtracted from the third of three consecutive odd integers and the result is multiplied by 2, the answer is 31 less than the sum of the first and twice the second of the integers. Find the integers.
the first integer is

Answer 1

(x+2y)-31 = 2(z-6)

Answer 2

consecutive....how
ie next odd number ascending from 0 ... so y = x+2
and z = y + 2
...?

Answer 3

@tryder ?

Answer 4

ok, so lets assume that it is consecutive in they're ascending odd numbers
so intergers are x, y, z
y = x+2
z = y + 2
so
(x+2y)-31 = 2(z-6)
(x+2[x+2])-31 = 2([y+2]-6)
(x+2[x+2])-31 = 2([{x+2}+2]-6)
(x+2x+4-31 = 2(x+2+2-6)
3x-27 = 2x+4+4-12
3x-27 = 2x-4
x = 31

Answer 5

so x = 31, y = 33, z = 35...?

Answer 6

(x+1)+2(x+3)-31 = 2(x+5-6)
x+1+2x+6-31 = 2x-2
x = 22

x+1 = 23
x+3 = 25
x+5 = 27

Answer 7

To check:
23+2(25)-31 = 2(27-6)
73-31 = 2(21)
42 = 42
Therefore, correct is 23, 25, 27. :)

Answer 8

yep, hes right... where's my math wrong tho...?

Answer 9

no, found it, last line
was:
3x-27 = 2x+4+4-12
3x-27 = 2x-4
x = 31
should be
3x-27 = 2x+4+4-12
3x-27 = 2x-4
x = -23

Answer 10

duh, typo x = 23

Answer 11

it's no typo, mental calculation error it is

Answer 12

aye, cheers @Yttrium
slaters

Answer 13

can be done a bit more simply: let numbers be \(x,x+2,x+4\) and just verify that result is odd

\[(x+4-6)*2+31 = x+2(x+2)\]\[2x+27=x+2x+6\]\[27-6=x+2x-2x\]\[x=23\]which is odd, as required, giving us our answer of \(23,25,27\)

@jack1 your error was in solving

\[3x-27=2x-4\]You didn't change the sign on the -27 when you shifted it to the other side of the = sign.
\[3x-27=2x-4\]subtract \(2x\) from each side giving\[x-27=-4\]add 27 to each side giving\[x=-4+27 = 23\]

Answer 14

to true, cheers man @whpalmer4

Answer 15

as i say, mental calculation error it is