find two consecutive even integers such that 5 more than the first is divided by 2 less than the second, the result is 13/12.
let the two integers be x and x+2
<<5 more than the first is divided by 2 less than the second, the result is 13/12>
---> (x+5)/x = 13/12
I will let you solve it.
(n+5)/(n+2-2)=13/12
is it 60 im confused i missed class today so, i really didn't get the whole of the lesson. =)
yes, the two integers are 60 and 62
thanks that's what i got just making sure =]
To solve this problem, we can translate the given information into mathematical equations.
Let's assume the first even integer is x.
Then, the second even integer would be x + 2 (since the integers are consecutive and even integers have a difference of 2).
According to the problem, "5 more than the first is divided by 2 less than the second, the result is 13/12."
This can be written as:
(x + 5) / (x + 2 - 2) = 13/12
Simplifying the equation, we get:
(x + 5) / x = 13/12
12(x + 5) = 13x
Now, let's solve the equation for x.
12x + 60 = 13x
60 = 13x - 12x
60 = x
So, the first even integer is x = 60.
The second even integer can be found by adding 2 to the first even integer:
x + 2 = 60 + 2 = 62
Therefore, the two consecutive even integers that satisfy the given conditions are 60 and 62.