Rational Equations Practice

First, we note that ‍$x=1$ and $x=3$ make at least one of the denominators zero and so neither of them is a solution.

Now, we multiply both sides of the equation by the least common multiple of the denominators, which is $(x-1)(x-3)$.

$\begin{aligned}\dfrac{x}{x-1}&=-\dfrac{2}{x-3}\\\\(x-1)(x-3)\cdot\dfrac{x}{x-1}&=(x-1)(x-3)\cdot\left(-\dfrac{2}{x-3}\right)\\\\x^2-3x&=-2x+2\\\\x^2-x-2&=0\\\\(x+1)(x-2)&=0\end{aligned}$

We get that $x=-1$ and $x=2$ are potential solutions. Neither of them makes the denominator zero, so they are both solutions.

In conclusion, the solutions are $x=-1$ and ‍$x=2$.

First, we note that ‍$x=3$ and $x=4$ make at least one of the denominators zero and so neither of them is a solution.

Now, we multiply both sides of the equation by the least common multiple of the denominators, which is $(x-3)(x-4)$.

$\begin{aligned}\dfrac{17-3x}{(x-3)(x-4)}&=\dfrac{x+1}{x-4}\\\\(x-3)(x-4)\cdot\dfrac{17-3x}{(x-3)(x-4)}&=(x-3)(x-4)\cdot\dfrac{x+1}{x-4}\\\\17-3x&=(x-3)(x+1)\\\\17-3x&=x^2-2x-3\\\\0&=x^2+x-20\\\\0&=(x+5)(x-4)\end{aligned}$

We get that $x=-5$ and $x=4$ are potential solutions. Recall that $x=4$ cannot be a solution.

Therefore, the only solution is $x=-5$.

$x=4$ is an extraneous solution, because we got it as a potential solution although it isn't actually a solution.

First, we note that ‍$x=3$ and $x=6$ make at least one of the denominators zero and so neither of them is a solution.

Now, we multiply both sides of the equation by the least common multiple of the denominators, which is $(x-3)(x-6)$.

$\begin{aligned}-\dfrac{2}{x-3}&=\dfrac{x}{x-6}\\\\(x-3)(x-6)\cdot\left(-\dfrac{2}{x-3}\right)&=(x-3)(x-6)\cdot\dfrac{x}{x-6}\\\\-2x+12&=x^2-3x\\\\0&=x^2-x-12\\\\0&=(x+3)(x-4)\end{aligned}$

We get that $x=-3$ and $x=4$ are potential solutions. Neither of them makes the denominator zero, so they are both solutions.

In conclusion, the solutions are $x=-3$ and $x=4$.

First, we note that ‍$x=-2$ and $x=0$ make at least one of the denominators zero and so neither of them is a solution.

Now, we multiply both sides of the equation by the least common multiple of the denominators, which is $x(x+2)$.

$\begin{aligned}-\dfrac{2}{x(x+2)}&=\dfrac{x+3}{x+2}\\\\x(x+2)\cdot\left(-\dfrac{2}{x(x+2)}\right)&=x(x+2)\cdot\dfrac{x+3}{x+2}\\\\-2&=x^2+3x\\\\0&=x^2+3x+2\\\\0&=(x+1)(x+2)\end{aligned}$

We get that ‍$x=-2$ and $x=-1$ are potential solutions. Recall that $x=-2$ cannot be a solution.

Therefore, the only solution is $x=-1$.

$x=-2$ is an extraneous solution, because we got it as a potential solution although it isn't actually a solution.