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

6, 8, 10.

Step-by-step explanation:

Let n be any integer.

Then our first even integer will be

And the consecutive integers will be and

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:

Solve for n. Distribute:

Combine like terms:

Subtract 12 from both sides:

Divide both sides by 10:

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.

30,40,50 is correct answer