# How do you solve rational equations 2/(3x+1) = 1/x - (6x)/(3x+1)?

Dec 14, 2015

$x = \frac{1}{2}$

#### Explanation:

1) Check when the equation is defined

First of all, bear in mind that your equation is not defined if any denominator is equal to zero.

Thus, let's first compute which values for $x$ are possible:

• impossible values for the first and third demoninator: $3 x + 1 = 0 \iff x = - \frac{1}{3}$
• impossible values for the second denominator: $x = 0$

Thus, for your equation, $x \ne 0$ and $x \ne - \frac{1}{3}$.

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Now, let's solve the equation!

$\frac{2}{3 x + 1} = \frac{1}{x} - \frac{6 x}{3 x + 1}$

... to eliminate the first and the third fraction, multiply both sides of the equation with $\left(3 x + 1\right)$...

$\iff 2 = \frac{3 x + 1}{x} - 6 x$

... to eliminate the last remaining fraction, multiply both sides of the equation with $x$...

$\iff 2 x = 3 x + 1 - 6 {x}^{2}$

... bring all terms to the left side of the equation...
(to do so, add $- 3 x - 1 + 6 {x}^{2}$ on both sides of the equation)

$\iff 6 {x}^{2} - x - 1 = 0$

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At this point, you have a quadratic equation which can be solved e.g. with the quadratic formula.

In your case, $a = 6$, $b = - 1$ and $c = - 1$. Thus,

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{1 \pm \sqrt{1 + 24}}{12} = \frac{1 \pm 5}{12}$

The two solutions for this quadratic equation are

$x = \frac{1}{2}$ and $x = - \frac{1}{3}$.

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4) Check which solutions are possible

Since we have noted before that $x \ne - \frac{1}{3}$ needs to hold, we need to discard the second solution.

Thus, the only solution for this equation is $x = \frac{1}{2}$.

Dec 14, 2015

Another way of approaching the initial stage. It has to be the same way as given but just looks different!

#### Explanation:

To make all the denominators the same:

Let some value be k then

$x \times k = \left(3 x + 1\right)$

So $k = \left(3 + \frac{1}{x}\right)$

Thus $\frac{1}{x}$ may be written as $\frac{3 + \frac{1}{x}}{3 x + 1} \to \frac{1 \times k}{x \times k} = \frac{1}{x}$

Now we have:

$\frac{2}{3 x + 1} = \frac{3 + \frac{1}{x}}{3 x + 1} - \frac{6 x}{3 x + 1}$

As they all have the same denominator the resulting equation would still be a true if we totally removed it.

$2 = 3 + \frac{1}{x} - 6 x$

$6 x - 1 = \frac{1}{x}$

$6 {x}^{2} - x - 1 = 0$

Then solve in the normal way!

Dec 14, 2015

Just another way

#### Explanation:

$\frac{2}{3 x + 1} = \frac{1}{x} - \frac{6 x}{3 x + 1}$

Take common terms to same side

$\frac{6 x + 2}{3 x + 1} = \frac{1}{x}$

Cross multiply both sides

$6 {x}^{2} + 2 x = 3 x + 1$
$6 {x}^{2} - 3 x + 2 x - 1 = 0$
$3 x \left(2 x - 1\right) + 1 \left(2 x - 1\right) = 0$
$\left(3 x + 1\right) \left(2 x - 1\right) = 0$
So we have two solutions $x = - \frac{1}{3} , \frac{1}{2}$

x =-1/3 leads to $\infty$. Hence the solution is $x = \frac{1}{2}$