Question

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

Answer 1

Answer:

6, 8, 10.

Step-by-step 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.