Algebra...Solve - XP Math - Forums

 XP Math - Forums Algebra...Solve

 05-15-2007 #1 alwaysalillost Guest   Posts: n/a 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!
 05-15-2007 #2 Temperal Guest   Posts: n/a $x-\frac {1}{8}\sqrt{7}=2$ $x=\frac {1}{8}\sqrt{7}+2$ $x \approx 2.33$ Or, $x=2+\sqrt{0.4375}$ Nothing but basic Algebra.
 05-15-2007 #3 alwaysalillost Guest   Posts: n/a 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.
 05-15-2007 #4 Temperal Guest   Posts: n/a That's not simplest form. I got the same thing, but simplified.
 05-17-2007 #5 Sillysidley   Join Date: Oct 2006 Posts: 822 Hmmm... Some people could read this and think it as: $\frac{1}{8}\sqrt{x-7}$=2 It's not exactly clear which one it is. __________________ .
 06-15-2007 #6 YvesBenoit Guest   Posts: n/a I would see this as (1/8)sqrt(7) - n = 2 ...
06-22-2007   #7
hannah7

Join Date: Mar 2007
Posts: 265

Quote:
 Originally Posted by somedude Hmmm... Some people could read this and think it as: $\frac{1}{8}\sqrt{x-7}$=2 It's not exactly clear which one it is.

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