SOLUTION: I would appreciate it if someone would help me solve this equation: sqrt3x-1/2=4 All of 3x-1/2 is supposed to be square rooted. So I tried squaring both sides of the equation to

Algebra ->  Radicals -> SOLUTION: I would appreciate it if someone would help me solve this equation: sqrt3x-1/2=4 All of 3x-1/2 is supposed to be square rooted. So I tried squaring both sides of the equation to       Log On


   

Question 7103: I would appreciate it if someone would help me solve this equation:
sqrt3x-1/2=4
All of 3x-1/2 is supposed to be square rooted. So I tried squaring both sides of the equation to get rid of the square root. So I got:
3x-1/2=16
Then I cross multiplied and now I have:
3x-1/32
It's been a year since I took Algebra1, so even if I've been correct so far, I don't know where to go from here. Thank you in advance to the tutor who helps me with this.
Answer by prince_abubu(198) About Me  (Show Source):
You can put this solution on YOUR website!
Let's take it from the point where you got +3x+-+1%2F2+=+16+. You don't need to do any cross-multiplying because the 2 is NOT a denominator of the entire expression on the left hand side of the equals sign. the -1/2 really is -0.5. So, your equation would be +3x+-+0.5+=+16+ - just like any normal linear equation.

So, going on further, you'll get +3x+=+16.5+ <---- 0.5 added to both sides. When you finally solve for x, x = 5.5.