In this method, we will use the LCM of the denominators to clear (get rid of) all the the denominators in a rational equation. Make sure to use the Lowest common multiple: Your work will always be easier if you clear the denominators in the equation by multiplying by the LCM of all the denominators. So practice finding the LCM if necessary. Here is a Khan Academy video example. I did one example in class, and here is another:
Example: (this is one version of problem 9 from Test 4)
Solve $\frac{2}{x^{2} – 1} – \frac{1}{x-1} = \frac{1}{2}$
The denominators are
$x^{2} – 1 = (x+1)(x-1)$
$x-1$
$2$
The Lowest Common Multiple of the denominators is $2(x+1)(x-1)$ and remember to leave it in its factored form!
Now multiply each term on both sides of the equation by this LCM. Note: we are not changing to a common denominator. The goal is to get rid of the denominators.
$\left(\frac{2}{(x+1)(x-1)}\right) 2(x+1)(x-1) – \left(\frac{1}{x-1}\right) 2(x+1)(x-1) = \left(\frac{1}{2}\right) 2(x+1)(x-1)$
Cancel the common factors to clear the denominators:
$\left(2\right) 2 – 2(x+1) = (x+1)(x-1)$
$4 – 2x – 2 = x^{2} -1$
$0 = x^{2} + 2x – 3$
Solve by factoring and using the zero product property:
$(x+3)(x-1) = 0$
$x+3 = 0$ or $x-1 = 0$
$x = -3$ or $x = 1$
These solve the quadratic equation, but do they solve the original rational equation? By substitution we can see that the first one works but the second one gives a zero denominator, so there is one solution to the equation: $x = -3$
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See also: what NOT to do when solving a rational equation
Whenever you solve any type of equation, it is important NOT to divide out or “cancel” any factor which contains a variable (and so could possibly be equal to 0). If you do this you risk losing a solution to the equation!
For example, here is a thing I have witnessed: in trying to solve
$5x^{2} = 3x$
one person thought to divide both sides by $x$: (WRONG!)
$\frac{5x^{2}}{x} = \frac{3x}{x}$
$5x = 3$
$x = \frac{3}{5}$
This certainly is a solution of the original equation, as you can see by substituting it back in: the problem is that there is another solution, namely $x = 0$, which we lost when we divided out that factor. This being a quadratic equation, a safe way to solve it is to use the Zero Product Property:
$5x^{2} = 3x$
$5x^{2} – 3x = 0$
$x(5x – 3) = 0$
$
x = 0$ or $5x – 3 = 0 \implies x = \frac{
3}{5}$
In any case, avoid at all costs dividing both sides of any equation by an expression which might possibly be 0.
