# Solving Rational Inequalities

How to solve a rational inequality: examples and their solutions.

## Example 1: Solve 1/x - 1/2x ≥ 3

Before solving the inequality,
first find the excluded values.

Excluded value

Set the denominators with the variable as 0.

Then, from 1/x term,
x ≠ 0.

And, from -1/2x term,
2x ≠ 0.

So [x ≠ 0].

Then solve the given inequality.

Move 3 to the left side.

To add and subtract the fractions,
change the denominator of each term
to the LCM of the denominators:
2x.

For each term,
multiply the missing factors
to both of the numerator and the denominator.

LCM (Least common multiple)

Adding and subtracting rational expressions

The denominators are all the same: 2x.

So add and subtract the fractions.

To make the -6x term (+),
Multiply -1 on both sides.

Then (6x - 1)/2x ≤ 0.

A minus sign number is multiplies on both sides.
So the order of the inequality sign changes.

Divide both sides by 2.

Multiply the square of the denominator, x2,
on both sides.

Before solving the inequality,
you found [x ≠ 0].

So x2 is (+),
neither 0 nor (-).

So the order of the inequality sign
does not change.

x2⋅(6x - 1)/x = x(6x - 1)

0⋅x2 = 0

Then x(6x - 1) ≤ 0.

Find the zeros of x(6x - 1) ≤ 0.

1) x = 0

2) 6x - 1 = 0

Move -1 to the right side
and divide both sides by 6.

Then x = 1/6.

So the zeros are x = 0, 1/6.

Quadratic function - Finding zeros

Roughly draw y = x(6x - 1)
on the x-axis.

Use the zeros 0 and 1/6.

Because of the excluded value,
[x ≠ 0].

So draw an empty circle
at [x = 0].

The inequality sign shows that
y = x(6x - 1) ≤ 0.

So color the function below the x-axis,
including the zero 1/6.

Solving quadratic inequalities

Then 0 < x ≤ 1/6.

## Example 2: Solve 4/(x - 1) + 1 ≤ 1/x

Before solving the inequality,
first find the excluded values.

Set the denominators with the variable as 0.

Then, from 4/(x - 1) term,
x - 1 ≠ 0.
So x ≠ 1.

And, from 1/x term,
x ≠ 0.

So [x ≠ 0] and [x ≠ 1].

Then solve the given inequality.

Move 1/x to the left side.

To add and subtract the fractions,
change the denominator of each term
to the LCM of the denominators:
x(x - 1).

For each term,
multiply the missing factors
to both of the numerator and the denominator.

The denominators are all the same:
x(x - 1).

So add and subtract the fractions.

-(x - 1) = -x + 1

4x - x - x = 2x

So [x2 + 2x + 1]/[x(x - 1)] ≤ 0.

x2 + 2x + 1 = (x + 1)2

Factor a perfect square trinomial (a2 ± 2ab + b2)

Multiply the square of the denominator, [x(x - 1)]2,
on both sides.

Before solving the inequality,
you found [x ≠ 0] and [x - 1 ≠ 0].

So x(x - 1) ≠ 0.

So [x(x - 1)]2 is (+),
neither 0 nor (-).

So the order of the inequality sign
does not change.

[x(x - 1)]2⋅(x + 1)2/[x(x - 1)] = x(x - 1)(x + 1)2

0⋅x2 = 0

So x(x - 1)(x + 1)2 ≤ 0.

The zeros of x(x - 1)(x + 1)2 ≤ 0 are
x = 0, 1, and -1.

Roughly draw y = x(x - 1)(x + 1)2
on the x-axis.

Use the zeros 0, 1, and -1.

Solving polynomial inequalities

Because of the excluded values,
[x ≠ 0] and [x ≠ 1].

So draw empty circles
at [x = 0] and [x = 1].

The inequality sign shows that
y = x(x - 1)(x + 1)2 ≤ 0.

So color the function below the x-axis,
including the zero -1.

Then x = -1, 0 < x < 1.