equation question

G

good girl

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Mar 28, 2013
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I've been stuck on this question. Can someone help me?

it says, solve algebraically for x: 3/4 = -(x+11)/4x+1/2x
 
What you have written is this:

\(\displaystyle \dfrac{3}{4} = -\dfrac{x+11}{4}x + \dfrac{1}{2}x\)

Is this what you intended?

Maybe this is what you intended, but have not written?

\(\displaystyle \dfrac{3}{4} = -\dfrac{x+11}{4x} + \dfrac{1}{2x}\)
 
Yes, but the x and 11 have parentheses around them. Does that make a difference?
 
the second question but the negative sign is outside the parenthesis
-(x+11)
 
Last edited:
good girl,

then I believe you mean the following:


\(\displaystyle \dfrac{3}{4} \ = \ \dfrac{-(x+11)}{4x} \ + \ \dfrac{1}{2x}\)


Is that correct?
 
\(\displaystyle \dfrac{3}{4} \ = \ \dfrac{-(x+11)}{4x} \ + \ \dfrac{1}{2x}\)


I would suggest that you start by multiplying both sides of the equation by 4x. That will "get rid" of the fractions.
 
Denis said:
OK then. Just rearrange:
(x + 11) / (4x) = 1 / (2x) - 3/4

Let's see your stuff!
Click to expand...

So then I get

4x = 2x^2 +22x - 3/4X - 8 1/4

4x = 2x^2 + 21 1/4x - 8 1/4

0 = 2x^2 + 17 1/4x - 8 1/4

Is this correct? What do I do next?
 
\(\displaystyle \dfrac{3}{4} \ = \ \dfrac{-(x+11)}{4x} \ + \ \dfrac{1}{2x}\)
You are not doing what was suggested. If you multiply both sides by 4x you get
\(\displaystyle \frac{3}{4}(4x)= \frac{-x- 11}{4x}(4x)+ \frac{1}{2x}(4x)\)
3x= -x- 11+ 2
 
HallsofIvy said:
\(\displaystyle \dfrac{3}{4} \ = \ \dfrac{-(x+11)}{4x} \ + \ \dfrac{1}{2x}\)
You are not doing what was suggested. If you multiply both sides by 4x you get
\(\displaystyle \frac{3}{4}(4x)= \frac{-x- 11}{4x}(4x)+ \frac{1}{2x}(4x)\)
3x= -x- 11+ 2
Click to expand...

Thank you
 
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