Question

Hey there,

Here is a test to how good your math is...
Four times the first of three consecutive even integers is six more than the product of two and the third integer. Find the integers.

Answer 1

I decided to do it again.
Being consecutive even numbers, we need to add 2 to the previous number.
Assign variables :
Let x = first consecutive even
Let x + 2 = second consecutive even
Let x + 4 = third consecutive even
4x = 6 + 2(x+4)
4x = 6 + 2x + 8
4x = 2x + 14
4x - 2x = 14

2x = 14
x = 14/2
x = 7
The numbers are 7, 9, and 11.
As you can see, the numbers are odd not even. There could be a typo in your original question. There is no way we can get three consecutive even integers from this question. We get three consecutive odd integers.

Answer 2

Hi ,

Let x , ( x + 2 ) and ( x + 4 ) are three

consecutive even numbers ,

According to the problem given ,

4 times of first number = product of 2

and third number + 6

4x = 2(x + 4 ) + 6

4x = 2x + 8 +6

4x - 2x = 14

2x = 14

x = 14/2

x = 7

Therefore ,

Required even numbers are ,

x = 7 ;

x + 2 = 7 + 2 = 9 ;

x + 4 = 7 + 4 = 11 :

I hope this helps you.

:)