Algebra...Solve

One eighth of the square root of 7 less than a number is 2. What is the number?

What i get is the equation 1/8 sqrt7 - x = 2 but I get stuck on the solving part. Can anyone please help me? Thanks in advance!

Or,

Nothing but basic Algebra.

n - (1/8)[sqrt(7)] = 2
n - (1/8)[sqrt(7)] + (1/8)[sqrt(7)] = 2 + (1/8)[sqrt(7)]
n = 2 + (1/8)[sqrt(7)]
The number is 2 + (1/8)[sqrt(7)]

That's what I got.

That's not simplest form. I got the same thing, but simplified.

[Sillysidley]

Hmmm... Some people could read this and think it as:
=2
It's not exactly clear which one it is.

I would see this as

(1/8)sqrt(7) - n = 2
...

how do you guys get the forula to come up here?

[Scion]

Originally Posted by soccerstar6997

how do you guys get the forula to come up here?

latex

[hannah7]

Originally Posted by somedude

Hmmm... Some people could read this and think it as:
=2
It's not exactly clear which one it is.

how do you even make those things?oohhhhhhhhhhhhhh riiight...latex....is this algebra?

Look at the title.

http://www.artofproblemsolving.com/L...PS_L_About.php - Good LaTeX tutorial.