Two consecutive even integers?

Question

Given that (x+37)/5 and (7x-1)/2 are two consecutive even integers, find the value of x.

Ans: x =3


Please show me the workings. Thanks a lot. :)

Answer 1

If they're both even integers, then their difference is 2, so
(x+37)/5 - (7x-1)/2 = 2 or
(7x-1)/2 - (x+37)/5 = 2
Now just solve the equations for x.

HTH

Doug

Answer 2

Consecutive integers are two integers whose difference is 1. Define the first integer as = x then, the second integer can be x + 1, or it can be x - 1 Either definition will give the same answers The product of the two integers is: x (x + 1) = x^2 + x = 182 Set equal to zero and factor (x + 14)(x - 13) = 182 x = -14 and x + 1 = -13 or, x = 13 and x + 1 = 14 Check The second problem is done the same way. x^2 + x - 272 = 0 (x + 17)(x - 16) = 0 You can get both sets of answers. Check.

Answer 3

If they are consecutive even integers, they differ by 2. Then (x+37)/5 + or - 2 = (7x-1)/2
First, we note that for (7x-1) to be even, x has to be odd. If x is odd, its unit digit has to equal 3 for x+37 to be divisible by 5. You can either solve the relation above or note that 3 is the only value that will yield the consecutive even numbers.

Answer 4

If they are consecutive even integers, then the second expression must be equivalent to the first expression plus 2, or the first expression is equal to the second expression minus 2.

So, (x+37)/5 = (7x-1)/2 - 2

solving for x:
x+37=(7x-1)5/2 - 10
x=(7x-1)5/2 - 47
x=17.5x - 49.5
16.5x = 49.5
x= 3

Answer 5

okay, let's start fom the beginning
since they are even, means that they are 2 units apart..
so one of them = the other + 2; right?
so substituting,
(x+37)/5 = (7x-1)/2 + 2
transpose,
(x+37)/5 - (7x-1)/2 = 2
look for the GCF = 10; then multiplying by 10
you get:
2(x+37) - 5(7x-1) = 20
Distributing:
2x - 74 - 35x + 5 = 20
33x = 69 + 20
33x = 99
x = 3

hope this helps!

Answer 6

(x+37)/5=n
(7x-1)/2=n+2

(7x-1)/2=(x+37)/5 + 2...multiply by 10 to eliminate the fractions

5(7x-1)=2(x+37) + 20...distribute

35x-5=2x+74+20...subtract 2x from and add 5 to both sides and combine like terms

33x=99...divide both sides by 33

x=3

hope this helps

Answer 7

if
n1 = (x+37)/5
and
n2 = (7x-1)/2

we also know that n1 + 2 = n2

so

(x+37)/5 + 2 = (7x-1)/2
(x+37)/5 + 10/5 = (7x-1)/2
(x+47)/5 = (7x-1)/2
cross multiply
2(x+47) = 5(7x-1)
2x + 94 = 35x - 5
99 = 33x
x = 99/33
x = 3

Answer 8

(7x-1)/2 - (x+37) /5 = 2 . .. solving . .
5 (7x-1) - 2 (x+37) = 20
35x - 5 - 2x - 74 = 20
33x = 99
x = 3

Answer 9

Jusat set these two equations EQUAL to each other and solve for x. EX:

(x+37)/5 = (7x- 1)/2
> 1/5x + 37/5 = 7/2x - 1/2
> just keep going, you'll get it.

Actually, I like the way the first two guys did it because it's easier. The trick is to add them and realize they must equal two. But my method will also work, if you take the time to do it.