algebra problem please help?

max-h

the sum of four consecutive odd intergers is 64. what are the integers thanks

viktor

your 4 odd integers are:

2n+1, 2n+3, 2n+5, 2n+7

they are such that

(2n+1)+(2n+3)+(2n+5)+(2n+7)=64

8n+16=64

8n=64-16=48

n=48/8=6

then your numbers are

2*6+1, 2*6+3, 2*6+5, 2*6+7

13, 15, 17 and 19

mickeyfinn

x+x+2+x+4+x+6=64

so

4x+12=64

subtract 12 from both sides

4x=52

divide both sides by 4

x=13

so the four consecutive odd integers are:

13, 15, 17, 19

jun-agruda

1st integer—x:

x + x + 2 + x + 4 + x + 6 = 64

4x = 52

x = 13

2nd integer:

= 13 + 2

= 15

3rd integer:

= 13 + 4

= 17

4th integer:

= 13 + 6

= 19

Answer: 13, 15, 17 & 19 are the integers. Their sum is 64.

dove

Let the odd numbers be

2k-3, 2k-1, 2k+1 and 2k+3

So sum= 8k= 64

k=8

Numbers are

13,15,17 and 19

animefan

the equation is set up like this:

x+(x+2)+(x+4)+(x+6) =64

4x+12=64

4x=52

x=13

integers are 13,15,17,19