Algebra word problem

Question

Three consecutive even integers have the property that the sum of 10 more than the first, 22 less than the second, and 4 times the third is 62. What are these numbers? 

Answer 1

n represents the first even

n + 2 represents the second even

n + 4 represents the third even

 

n + 10 represents 10 more than the first

n + 2 - 22 represents 22 less than the second

4(n + 4) represents 4 times the third

 

n + 10 + n + 2 - 22 + 4(n + 4) = 62

2n - 10 + 4n + 16 = 62

6n + 6 = 62

6n = 56

n = 9 1/3

 

This is a solution for the algebra. It is not an integer. Either no integer solution exists or some information is incorrect.

Answer 2

So there is something we need to realize from the beginning of the problem where it states that these are 3 consecutive even integers. This means something like 2,4,6, 8, etc. If we analyze the difference between 3 consecutive even integers at any point along the continuum we see that each number can be represented in this way: n, n+2, n+4. When n=2, you see you get 2,4,6. If n=24 for example, you'd get 24, 26, 28. Also note that the information given for this problem point towards altered versions of our integers leading to a sum, so we must structure our work towards this sum. There are 3 numbers that go towards this sum, but they aren't immediately obvious, so we can refer to them as 'a', 'b', and 'c'.

Now back to the problem. Let's start with the first bit of information given. "10 more than the first". Let's call the 'first' consecutive even integer x then. So we have x+10 as 'a'. To get the second consecutive even integer we have x+2 because of that rule I was explaining regarding the consecutive even integers before. Then the problem states "22 less than the second" for 'b', thus 'b' is x+2-22=x-20. Our third consecutive even integers will by default be x+4. Finally, our last bit of information states that adding 'a' and 'b' with "4 times the third is 62". So we can formulate this into an expression: x+10+(x-20)+4(x+4)=62, where 4(x+4)='c'. Simplifying the expression we get 6x+6=62. So now we solve for x. Step by step==>6x=56==>x=28/3

 

The number doesn't come out to be an integer, so there are likely to be some incorrect numbers, or whoever constructed the problem was not careful to inspect that the resulting 'integers' were in fact integers. However, if we were to ignore the limitation of the numbers having to be integers our answer would be: 28/3, 34/3, and 40/3 respectively AKA x, x+2, and x+4.

Answer 3

First, let's consider the nature of integers.  The integers are ... , -3, -2, -1, 0, 1, 2, 3, ....  Even and odd numbers have been discussed since antiquity.  To be even, a number must be divisible by 2, and the term odd refers to those numbers that are not divisible by two.  So, a number dichotomy is created by even and odd that can be used to separate the integers into two sets: the evens and the odds.  The even integers are ... -4, -2, 0, 2, 4, ....  By definition, all of the even integers can be represented as multiples of 2.  Starting with any even integer, we may generate a list of consecutive even integers beyond it by adding or subtracting two, repeatedly.  For example, starting with 18, we get a list of consecutive even integers by writing

18, 18 + 2, 18 + 2 + 2, 18 + 2 + 2 + 2, ... and so on.  Using this information, and a little ingenuity, the problem of finding the set of consecutive even integers fitting given constraints can now be solved.  However, using algebraic techniques can help you avoid trial and error.  Looking back at the list we generated from 18, we see that the list may also be written 2(9), 2(9 + 1), 2(9 + 2), 2(9 + 3), ....  Now, start an unknown integer, n.  How do you generate a list of unknown, but consecutive even integers with n?  Once you do that, all that remains is to construct an equation that represents the constraints in the given problem.  The solution to that equation will the integer you need to find the three consecutive integers in question.