# Solving Rational Equations

How to solve rational equations: example and its solution.

## Example: Solve 3/*x* + *x*/(*x* - 1) = 1/[*x*(*x* - 1)]

Find the LCM of the denominators.

LCM (Least common multiple)

Write [*x* = *x*], [*x* - 1 = (*x* - 1)], and [*x*(*x* - 1) = *x*(*x* - 1)]

like this.

Arrange the same factors vertically.

Draw a horizontal line below.

Write the factors

that are in either numbers

below the horizontal line:*x*(*x* - 1).

This is the LCM.

Write the excluded values.

Excluded value

The excluded values can be found

by setting (each factor) = 0.

So *x* ≠ 0.

And *x* - 1 ≠ 0.

So *x* ≠ 1.

So *x* cannot be 0 and 1.

Multiply the LCM, *x*(*x* - 1),

to each fraction of the given equation.

Cancel the common factor for each fraction.

Then 3(*x* - 1) + *x*⋅*x* = 1.

Solve this quadratic equation.

3(*x* - 1) = 3*x* - 3*x*⋅*x* = *x*^{2}

So 3*x* - 3 + *x*^{2} = 1.

Move 1 to the left side.

And arrange the terms in descending order.

Then *x*^{2} + 3*x* - 4 = 0.

Factor *x*^{2} [+ 3]*x* [- 4].

Factor a quadratic trinomial

Find a pair of numbers

whose product is the constant term [-4]

and whose sum is the middle term's coefficient [+3].

The constant term is (-).

So the signs of the numbers are different:

one is (+), and the other is (-).

[-4] = -1⋅4

-1 + 4 = 3 = [+3]

So -1 and 4 are the right numbers.

Use -1 and +4

to write a factored form:

(*x* - 1)(*x* + 4).

Solve (*x* - 1)(*x* + 4) = 0.

1) *x* - 1 = 0

So *x* = 1.

But *x* cannot be 1,

because it's the excluded value.

(See the excluded values above.)

So *x* = 1 is not the solution.

In other words,*x* = 1 is the [extraneous solution].

The [extraneous solution] is the solution

which is found by solving the equation

but cannot be the answer.

2) *x* + 4 = 0

So *x* = -4.

Solving a quadratic equation by factoring

So *x* = -4 is the answer.