Question

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.

Answer 1

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

Answer 2

Answer: The required three numbers are 7, 13 and 22.

Step-by-step explanation: Given that the average of three numbers is 14. The smallest number is six less than the middle number and the largest number is four less than twice the middle number.

We are to find all the three numbers.

Let the middle number be represented by x.

Then, according to the given information, we have

`((x-6)+x+(2x-4))/(3)=14\\\\\Rightarrow 4x-10=14*3\\\\\Rightarrow 4x=42+10\\\\\Rightarrow 4x=52\\\\\Rightarrow x=(52)/(4)\\\\\Rightarrow x=13.`

Therefore, middle number = 13,

smallest number = 13-6 = 7

and

the largest number = 2×13 - 4=26 - 4 = 22.

Thus, the required three numbers are 7, 13 and 22.