Question

Help please!

Find three consecutive odd integers whose sum is 10 less than 4 times the last integer

Answer 1

let the middle odd integer be 'x'
then 3 integers are x-2, x,x+2
can u form the equation using given info ?

Answer 2

Whats the equation? I think im wrong with x-2+x+x+2=10-4(x-2)

Answer 3

If your numbers are x-2, x, and x+2, then the right side would be "10 less than 4 times the last integer", or 4(x+2) - 10

Answer 4

However the solution does not yield an odd integer.

Answer 5

Are you sure with the right side? I thought itwas 10-4(x-2)

Answer 6

If the last integer is "x+2", then 4 times the last integer is 4(x+2). Ten less than that would be 4(x+2) less 10 = 4(x+2) - 10

Answer 7

Oh sorry i mean least integer not last

Answer 8

sum will be just 3x
it is 10 less than 4 times (x-2)
so.
3x+10 = 4(x-2)
solve this to find 'x'

Answer 9

That still results in an even integer though.

Answer 10

yes.

Answer 11

But i need odd integer

Answer 12

@Kittykatkat14 When you solve the equation 3x + 10 = 4(x-2) as @hartnn suggested, what value do you obtain for the middle integer, x?

Answer 13

12

Answer 14

Ahm it says here that if the problem has no solution, explain why

Answer 15

Double check your work because I get the middle integer x = 18. In any event, if you set up the equation properly you find that there is only one number that satisfies the given conditions and that number happens to be even.

Answer 16

Oh thanks, so it has no solution,

Answer 17

the integers can't be odd, because the sum of odd integers is odd number, addin 10 will give you odd number, but 4 times the least will give you even number....and odd number cannot equal even number, hence, no solution....