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find the three consecutive even intergers such that the sum of twice the smallest number and three times the largest is 42

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3 votes

Answer:

6, 8, 10.

Explanation:

Let n be any integer.

Then our first even integer will be
2n

And the consecutive integers will be
2n+2 and
2n+4

We want to find the integers such that the sum of twice the smallest number (2n) and three times the largest number (2n+4) is 42. In other words:


2(2n)+3(2n+4)=42

Solve for n. Distribute:


4n+6n+12=42

Combine like terms:


10n+12=42

Subtract 12 from both sides:


10n=30

Divide both sides by 10:


n=3

So, the value of n is 3.

This means that our first even integer is 2(3) or 6.

So, our sequence of integers is: 6, 8, 10.

And we're done!

Notes:

We use 2n because anything integer multiplied by 2 ensures that the resulting number is even. Starting out, n can be either even or odd, but by multiplying it by 2, we will get an even number.

User Gfullam
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