MarziaSparta
MarziaSparta MarziaSparta
  • 21-10-2016
  • Mathematics
contestada

Find four consecutive odd integers such that the sum of the first three exceeds the fourth by 18?

Respuesta :

SJ2006
SJ2006 SJ2006
  • 22-10-2016

Let, numbers = x,(x+2),(x+4),(x+6)

A.T.Q,

x+x+2+x+4 = x+6+18

3x+6 = x+24

3x-x= 24-6

2x = 18

x = 9

Numbers would be 9, 11, 13, 15

Answer Link
lena2712
lena2712 lena2712
  • 06-01-2022

Answer:

9,11,13,15

Step-by-step explanation:

Let our 4 numbers be

-n

-n+2

-n+4

-n+6

first, we have to write an expression and we gradually simplify.

n+n+n+2+4=n+6+18

combine like terms:

3n+6==n+24

subtract from both parts of the equation.

3n-n=24-6

now are finally finishing our problem.

2n=18

n=9

Now that we know our first integer, we need to consecute.

9,11,13,15

Hope this helps!    

Answer Link

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