Find three consecutive even integers, such that 7 times the middle integer is 10 more than the sum of first and third. Please an explanation would be nice.

Question

anonymous · Answer

Nope sorry

anonymous · Answer

7y=10+(x+z)


$$7*4=10+(2+6)=18$$
$$28\neq 18$$

anonymous · Answer

so let your integers be x,(x+1),(x+2).. 
then form an equation.. 
here the middle integer is (x+1) so 7 times of it wud be...&(x+1)...
it is now given that 7 times the mid integer is 10 more than  sum of first and third.. so 
equation is 7(x+1)-(x+(x+2)=10

anonymous · Answer

Can you solve it now?? @ccardenas2322 ??

anonymous · Answer

Okay let me try :)

anonymous · Answer

keep me posted..

anonymous · Answer

if x =2 , the middle integer is then odd

anonymous · Answer

i believe it's on the right track but you need to make it x (x+2) (x+4)

anonymous · Answer

yeah sorry it shud be x,(x+2),(x+4)

anonymous · Answer

so its 7(x+2)-x(x+4)=10

anonymous · Answer

no its 7(x+2+-{x+(x+4)}=10

anonymous · Answer

That equation looks confusing,but i'll try.

anonymous · Answer

$$7(x+2)=10+((x+(x+4))$$
$$7x+14=10+(2x+4)$$
$$7x+14=10+2x+4$$
$$5x=-4+4$$

$$x=0$$

anonymous · Answer

does this make sense?

anonymous · Answer

however im not sure if 0 is even? lol

anonymous · Answer

okay that one makes more sense so the integers are 0,2,4

anonymous · Answer

yeah i guess you're correct.

anonymous · Answer

well check to see =]

using the equality

if you let x =0 ,y=2, z=4
$$7y=10+(x+z)$$

anonymous · Answer

well it does satisfy your condition..

anonymous · Answer

It does thank you guys ! :)

anonymous · Answer

sure thing..

anonymous · Answer

thanks for the medal @Outkast3r09 ..