# Math Help - Need to know intermediate steps for this equation..

1. ## Need to know intermediate steps for this equation..

I haven't touched algebra in years, so bear with me.
The start of the equation is..

(plus or minus) +/-.76 = x/0.0200-x

the two it splits into are..

x= 0.0200 X 0.76 / 1.76 = 0.0086
and
x= -0.0200 X 0.76 / 0.24 = -0.063

I understand you're solving for both positive and negative numbers, but without intermediate steps I'm not sure how the first part jumps to the next two equations.

Help would be very appreciated. Thanks.

2. you're trying to solve:

$\pm 0.76 = \frac{x}{0.0200}-x$

right?

If so... I think you'll wanna factorise the right hand side of the equation to give:

$\pm 0.76 = x(\frac{1}{0.0200} - 1)$

and then divide both sides of the equation by:

$(\frac{1}{0.0200} - 1)$

to give:

$\frac{\pm 0.76}{50 - 1} = x$

whether that gives you the same answers I don't know, maybe I have misunderstood your question.

3. Originally Posted by Subcydian
you're trying to solve:

$\pm 0.76 = \frac{x}{0.0200}-x$

right?
Well, it's actually +/-76 = x / (0.0200 - x )^2

Sorry, that was my mistake.

and it turns into

x = (0.0200 x 0.76) / 1.76

and

x = (-0.0200 x 0.76) / 0.24

Why are these numbers...there. I'm just all sorts of confused haha. Thanks for trying to help. My mistake in posting a wrongish problem before.

4. Originally Posted by climbo
Well, it's actually +/-76 = x / (0.0200 - x )^2
Multiply both sides by [tex](0.02- x)^2[tex] to get $\pm 76(.02- x)^2= x$. [tex](.02- x)^2= x^2- .04x+ .0004[tex] so $\pm 76(.02- x)^2= \pm(0.0304- 3.04x+ 76x^2)$.

That is, your equation becomes $\pm(0.0304- 3.04x+ 76x^2)= x$. With the plus sign, $76x^2- 4.04x+ 0.0304= 0$. With the minus sign, $76x^2-2.04x+ 0.0304= 0$. You can solve those with the quadratic formula. There should actually be four different solutions.

Sorry, that was my mistake.

and it turns into

x = (0.0200 x 0.76) / 1.76

and

x = (-0.0200 x 0.76) / 0.24

Why are these numbers...there. I'm just all sorts of confused haha. Thanks for trying to help. My mistake in posting a wrongish problem before.