To solve rational equations is quite simple. Every rational equation comes down to a polynomial equation with one important exception. Since we have denominators, and possibly in some points those denominators may be equal to zero. These are the cases we have to eliminate. This means that we have to be careful, if the solution of our equation is equal to the zero of the denominator, given equation will have no solutions. Also, it may happen that we get a numerical identity. This will mean that the set of solutions is the whole set of real numbers.
First thing that you must pay attention is whether this equation has problems in denominator. Since we only have real numbers, we don’t have to worry about solutions we get.
The easiest way to solve this equation is the method of cross product.
$ 2(x – 1) = 3(2x + 1)$
$ 2x – 2 = 6x + 3$
$ 4x = – 5$
$ x = -\frac{5}{4}$
Example 2. Solve following rational equation $\frac{3x + 8}{2x – 1}$.
First thing you should pay attention to is the denominator. Since we have unknown in the denominator we should isolate cases where it is equal to zero, because these are the cases we know that cannot be valid. If we get the zero of the denominator as the solution of the equation we should simply cross it out as a solution. If that is the only solution, then that equation has no solution.
When we found values of $x$ that cannot happen, we can proceed with solving the equation.
$ 2x – 1 \not= 0 -> 2x \not= 1 -> x \not= \frac{1}{2}$
$ 3x + 8 = 3 \cdot (2x – 1)$
$ 3x + 8 = 6x – 3$
$ -3x = -11$
$ x = \frac{11}{3}$
Example 3. Solve following rational equation $\frac{2}{x – 3} + \frac{4}{x}$
This kind of equation can be solved in two ways. First way is to add these two fractions and then just equalize numerator with zero. Second way is to “transfer one fraction to the other side” and again use cross product. Of course, first thing we have to do is to find the points that can’t be solutions of this equation.
$ x\not= 0$, $ x\not= 3$
$\frac{2}{x – 3} + \frac{4}{x} = 0$
$\frac{2x + 4x – 12}{x(x – 3)} = 0$
$ 6x = 12$
$ x = 2$
Example 4. Solve following rational equation.
$\frac{1}{x + 1} + \frac{2}{x + 2} = \frac{1}{x^2 + 3x + 2}$
If we can’t see the zeros of the polynomial in the denominator, we should factorize it first.
$ x^2 + 3x + 2 = (x + 1)(x + 2)$
$ x\not= – 1$, $ x\not= – 2$
We can notice that the factorized form of the denominator on the right side of the equation is equal to the product of denominators on the left. This is great because when cross multiply them one big part of this equation will be canceled out.
$\frac{(x + 2) + 2(x + 1)}{(x + 1)(x + 2)} = \frac{1}{(x + 1)(x + 2)}$
$ x + 2 + 2x + 2 = 1$
$ 3x = – 3$
$ x = – 1$
We got that $ x = -1$ but we have a condition which says that $ x\not=- 1$ which means that this equation has no solutions.
