Question

How do you solve $\dfrac{1}{2x} + \dfrac{3}{x+7}=-\dfrac{1}{x}$ and find any extraneous solutions?

Answer 1

Hint: First to solve this equation, you should multiply all the terms with both the terms 2x and x+7 . Then you can add or subtract the terms in the numerator and then solve the equation accordingly. After you get the answer, you can simplify it if possible.

Complete step by step solution:
Here is the complete step by step solution.
The first step to do is to multiply all the terms with terms 2x and x + 7. We multiply 2x and x + 7 because they are the lcm of the equation. Therefore, we get the equation as:
$\Rightarrow \dfrac{2x\left(x+7\right)}{2x} + \dfrac{6x\left(x+7\right)}{x+7}=-\dfrac{2x\left(x+7\right)}{x}$
Now, we can cancel out the terms. Then, we get
$\Rightarrow x+7+6x=-2x-14$
Now we simplify the equation.
$\Rightarrow 9x = -21$
Now we take the number 9 to the right hand side, to get the answer.
$\Rightarrow x=-\dfrac{21}{9}$
Now, we can see that, we can also simplify the answer as 3 is common in both the terms. So, we get the answer as
$\Rightarrow x=-\dfrac{7}{3}$
Therefore, we get the final answer of the question, how do you solve $\dfrac{1}{2x} + \dfrac{3}{x+7}=-\dfrac{1}{x}$ and find any extraneous solutions as $ x=-\dfrac{7}{3}$.

Note: Extraneous solutions are the solutions which you get while solving the equations, but they are not the actual answers as when they are put in the original equation, it does not satisfy the equation. Here we do not have any extraneous solutions. Extraneous solutions usually appear when there are square terms or square root terms in the equation.