Find five consecutive integers such that: "the sum of the first and 4 times the third is equal to 60 less than 3 times the sum of the second, fourth, and fifth."

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Mathematics

6 March, 19:21

Find five consecutive integers such that: "the sum of the first and 4 times the third is equal to 60 less than 3 times the sum of the second, fourth, and fifth."

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Zaniyah Montgomery · Answer 1

Consecutive integers are integers that come one after another in order. For example, 2, 3, and 4 are all consecutive integers, as well as 100, 101, and 102. In general, if we have some integer n, the consecutive integers following it will be n + 1, n + 2, n + 3, etc. Here, we have five. If we

Call the first integer n, Then the second integer will be n + 1, The third will be n + 2, The fourth will be n + 3, And the fifth will be n + 4.

Now, let's break down the equation in English. We'll use #1, #2, #3, #4, and #5 as stand-ins for the five integers just so we can get things set up, then we'll replace them with the expressions we came up with in the first part. Starting from the beginning of the statement, let's translate words to symbols:

"The sum of the first and 4 times the third" → #1 + 4 * #3 "is equal to" → = (maybe obvious, but still!) "60 less than 3 times the sum of the second, fourth, and fifth" → 3 * (#2 + #4 + #5) - 60

Putting that all together, we get the equation

#1 + 4 * #3 = 3 * (#2 + #4 + #5) - 60.

Replacing our stand-ins for the expressions we came up with, it becomes

n + 4 (n + 2) = 3[ (n + 1) + (n + 3) + (n + 4) ] - 60

We can now start to simplify our equation to find n (the first integer) and the rest will follow easily from that. I'll start by focusing on the left side. We can distribute and combine like terms like so:

n + 4 (n+2) = n + 4n + 8 = 5n + 8

Dealing with the right, we can get rid of some redundant parentheses and carry out the same process:

3[ (n + 1) + (n + 3) + (n + 4) ] - 60 = 3 (n + 1 + n + 3 + n + 4) - 60

= 3 (3n + 8) - 60 = 9n + 24 - 60 = 9n - 36

Our simplified equation now becomes

5n + 8 = 9n - 36.

Subtracting 5n from either side:

8 = 4n - 36

Adding 36 to either side:

44 = 4n

And dividing either side by 4:

11 = n.

We now know that our first integer is 11, which means that our 5 consecutive integers are 11, 12, 13, 14, and 15.