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# Additional Algebra I and II topics

Joined: 07/22/2017Posts: 24
Question #16 from the online student center lesson 4 additional practice.

Let the function f be defined by f(x)=x+6. If 3f(n) = 39. what is the value of f(2n)?

I've asked about a problem like this before...this question is wording it differently because it's asking for the value of f(2n). I tried backplugging but then realized that I can't divide the f(2n) by 2 because it is inside the parentheses. Also figured out that f(n)=13.

Thanks!

## Posts

• PowerScore Staff Joined: 10/31/2016Posts: 200
Hi, Diane,

This problem requires some algebra but can also be solved using problem solving techniques, in this case Backplugging.

Let's take a look. Remember that the first thing to do with functions is to get the function by itself. You can manipulate equations with functions in the same manner in which you'd handle any other algebraic function.

Since you're starting with 3f(n) = 39, you want to start by isolating f(n). Divide both sides by 3. You end up with f(n) = 13.

You now have two possible paths to the solution.

## Pure Algebra

Record What You Know™

f(x) = x + 6
f(n) = 13
13 = n + 6
n = 7
f(2n) = f(14) = 14 + 6 = 20

## Backplugging

Record What You Know™

f(x) = x + 6
f(n) = 13 = n + 6
n = 7

… f(2n) = … ??? = 2n + 6 …… Solve for n … Does n = 7?
(A)
(B)
(C) 20 ……… 20 = 2n + 6 …… n = 7 ………… YES! we've found the answer
(D)
(E)

I hope this helps!