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How do I solve for $x$ from this equation?

$$ -\frac{1}{x^2} + \frac{9}{(4-x-y)^2} = 0.$$

I need to get this into $x=$"blah"?

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  • $\begingroup$ If $x$ and $y$ are real you don't since the left side is positive. If they are complex, rearrange to get $[(4-x-y)/x]^2=-9$. $\endgroup$ – Robin Chapman Oct 31 '10 at 16:43
  • $\begingroup$ fixed a typo. should have been -1/x^2 $\endgroup$ – PhilCK Oct 31 '10 at 16:48
  • $\begingroup$ funky formatting is standard LaTex. See en.wikipedia.org/wiki/LaTeX for more information. $\endgroup$ – Djaian Oct 31 '10 at 16:57
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    $\begingroup$ Hint for LaTex formatting, if you are not used to it, you can copy/paste what other write if you look at the source code of the page. $\endgroup$ – Djaian Oct 31 '10 at 17:19
  • $\begingroup$ @PhilCK: It's called "solving for $x$", not "making $x$ subject of", or "making $x=$ out of". $\endgroup$ – Arturo Magidin Nov 7 '10 at 19:07
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If you add $\frac{1}{x^{2}}$ to both sides of your equation

$$-\frac{1}{x^{2}}+\frac{9}{\left( 4-x-y\right) ^{2}}=0\qquad (1)$$

you get this equivalent one (provided that $\frac{1}{x^{2}}$ is finite, i.e $x\neq 0$)

$$\frac{1}{x^{2}}=\frac{9}{\left( 4-x-y\right) ^{2}}.\qquad (2)$$

It is satisfied if the square root of one side is equal or symmetric to the square root of the other side:

$$\frac{1}{x}=\pm \frac{3}{4-x-y}.\qquad (3)$$

Equation $(3)$ is equivalent to

$$3x=\pm \left( 4-x-y\right) \qquad (4)$$

provided that $x\neq 0$ and $4-x-y\neq 0$.

The equation $(4)$ represents two equations. One is

$$3x=4-x-y,\qquad (5)$$

which is equivalent to

$$4x=4-y\Leftrightarrow x=1-\frac{1}{4}y\qquad (6)$$

and the other

$$3x=-\left( 4-x-y\right), \qquad (7)$$

is equivalent to

$$3x=-4+x+y\Leftrightarrow x=-2+\frac{1}{2}y.\qquad (9)$$

Thus $(1)$ is equivalent to

$$x=1-\frac{1}{4}y\qquad (10)$$

or

$$x=-2+\frac{1}{2}y,\qquad (11)$$

provided that $y\neq 4$ because the conditions $x\neq 0$ and $4-x-y\neq 0$ correspond to

$$x\neq 0\Leftrightarrow 1-\frac{1}{4}y\neq 0\Leftrightarrow y\neq 4$$

$$x\neq 0\Leftrightarrow -2+\frac{1}{2}y\neq 0\Leftrightarrow y\neq 4$$

and

$$4-x-y\neq 0\Leftrightarrow 4-\left( 1-\frac{1}{4}y\right) -y\neq 0\Leftrightarrow y\neq 4$$

$$4-x-y\neq 0\Leftrightarrow 4-\left( -2+\frac{1}{2}y\right) -y\neq 0\Leftrightarrow y\neq 4.$$

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Put $z = 4-x-y\:.\ $ Multiplying by $(xz)^2$ yields $$0 = 9x^2-z^2 = (3x-z)\:(3x+z) = (4x+y-4)(2x-y+4)$$

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When you have fractions involved, the usual trick is to put everything on the same side, so you have "= 0" on the other side (This step is already ok in your case). Then, you put all the fractions on the same denominator and you multiply by this denominator (which vanishes since you have 0 on the other side). Usually it's then easier, since you don't have fractions anymore (well, it depends on the situation, but this is a good idea in general).

So, in your case:

$$\frac{-1}{x^2} + \frac{9}{(4-x-y)^2} = 0$$

becomes

$$\frac{-(4-x-y)^2}{x^2(4-x-y)^2} + \frac{9x^2}{x^2(4-x-y)^2} = 0$$

and then

$$-(4-x-y)^2 + 9x^2 = 0$$

Now you can extand $(4-x-y)^2$:

$$(4-x-y)^2 = ((4-y)-x)^2 = (4-y)^2 - 2(4-y)x + x^2$$

And thus your equation becomes:

$$-(4-y)^2 + 2(4-y)x + 8x^2 = 0$$

Which is a standard equation, the same as $ax^2 + bx + c = 0$, that can be solved for $x$ (consider $y$ as if it was any number).

At the end, don't forget to check if the solution you obtain is ok with your beginning equation (no division by 0 for example).

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If you're working on the same problem as asked about in this question, the answer is that you don't need to right away, since you can first use the other equation to get a relation between $x$ and $y$ that will simplify things a little. See my comment there.

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  • $\begingroup$ lol, thats my flat mate! Yes same question. $\endgroup$ – PhilCK Oct 31 '10 at 18:36

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