Suppose we’re given an equation to solve like this

$$ \begin{align*} \frac{5}{x+1} = - 4 - \frac{3}{x+2} \end{align*} $$

The easiest way to solve a rational equation like this is just to multiply both sides by the LCD (least common multiple) of all the denominators. We can see above that the LCD is \((x+1)(x+2)\) so let’s multiply both sides of the equation by \((x+1)(x+2)\).

$$ \begin{align*} (x+1)(x+2)\frac{5}{x+1} &= (x+1)(x+2) \big(- 4 - \frac{3}{x+2}\big) \\ 5(x+2) &= -4(x+1)(x+2) - 3(x+1) \\ 5x + 10 &= -4(x^2 + 3x + 2) - 3x - 3 \\ 5x + 10 &= -4x^2 - 12x - 8 - 3x - 3 \\ 5x + 10 &= -4x^2 - 15x - 11 \\ 4x^2 + 15x + 11 + 5x + 10 &= 0 \\ 4x^2 + 20x + 21 &= 0 \\ (2x + 3)(2x + 7) &= 0 \\ \end{align*} $$

so we have two solutions \(\{\frac{2}{3}, \frac{7}{2} \}\). Note that we need to make sure that neither of the solutions would make the denominator zero. If $x=-1$ was a solution for example, then we exclude that term from the final list of solutions.


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