# Solving Rational Inequalities

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

## Example 1: Solve 1/*x* - 1/2*x* ≥ 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 -1/2*x* term,

2*x* ≠ 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:

2*x*.

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: 2*x*.

So add and subtract the fractions.

To make the -6*x* term (+),

Multiply -1 on both sides.

Then (6*x* - 1)/2*x* ≤ 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, *x*^{2},

on both sides.

Before solving the inequality,

you found [*x* ≠ 0].

So *x*^{2} is (+),

neither 0 nor (-).

So the order of the inequality sign

does not change.

*x*^{2}⋅(6*x* - 1)/*x* = *x*(6*x* - 1)

0⋅*x*^{2} = 0

Then *x*(6*x* - 1) ≤ 0.

Find the zeros of *x*(6*x* - 1) ≤ 0.

1) *x* = 0

2) 6*x* - 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*(6*x* - 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*(6*x* - 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

4*x* - *x* - *x* = 2*x*

So [*x*^{2} + 2*x* + 1]/[*x*(*x* - 1)] ≤ 0.

*x*^{2} + 2*x* + 1 = (*x* + 1)^{2}

Factor a perfect square trinomial (*a*^{2} ± 2*ab* + *b*^{2})

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⋅*x*^{2} = 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.