Solving Rational Equations

Solving Rational Equations

How to solve rational equations: example and its solution.

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

Solve the given expression. 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) + xx = 1.

Solve this quadratic equation.

3(x - 1) = 3x - 3
xx = x2

So 3x - 3 + x2 = 1.

Move 1 to the left side.
And arrange the terms in descending order.

Then x2 + 3x - 4 = 0.

Factor x2 [+ 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.