Three numbers have an average of 14. The smallest number is six less than the middle number and the largest number is four less than twice the middle number. Use an algebraic model to find all three numbers.

Let x (largest), y (middle) and z (smallest) be the 3 numbers.

According to your question: z=y-3 and * = 2y-4

Because we know the average we'd do the following: (x+y+z) / 3=14 (add all the three numbers and divide them by 3)

Now just substitute

[ (2y-4) + y + (y-3) ]/3=14

Now simplify

2y+y+y=4y

-4-3=-7

You get [4y-7]/3=14

Now multiply both sides by 3

4y-7=42

Now add 7 to both sides

4y=49

Now divide both sides by 4

y=49/4 (middle number)

Substitute y in z and x

z = (49/4) - 3 = 37/4 (smallest)

x=2 (49/4) - 4 = 41/2 (largest)

Those are the three numbers

To check your work substitute x, y and z in the first equation, you will get 14