Rational Inequality

How to solve a rational inequality: 2 examples and their solutions.

Example 1

Example

Solution

Before solving the inequality,
first set (denominator) ≠ 0.

Excluded Value

See 1/x.

The denominator is x.

So x ≠ 0.

Next, see 1/2x.

The denominator is 2x.

So 2x ≠ 0.
Then x ≠ 0.

From the denominators,
you found that
x ≠ 0.

The x values should satisfy this condition.

Next, solve the rational inequality.

First, find the least common multiple, LCM,
of the denominators.

The denominators are
x and 2x.

So the LCM is
2x.

Change the denominators of the rational expression
to the LCM: 2x.

Add and Subtract Rational Expressions

See 1/x.

The denominator is x.

The factor 2 is missing.

So multiply 2
to both of the numerator and the denominator.

1/x
= [1/x]⋅[2/2]

See -1/2x.

The denominator is 2x:
the LCM.

So you don't have to change -1/2x.

Write the inequality sign ≥.

See the right side 3.

The denominator is 1.

2x is missing.

So multiply 2x
to both of the numerator and the denominator.

3
= 3⋅[2x/2x]

So
1/x - 1/2x ≥ 3
becomes
[1/x]⋅[2/2] - 1/2x ≥ 3⋅[2x/2x].

[1/x]⋅[2/2]
= 2/2x

3⋅[2x/2x] = 6x/2x

2/2x - 1/2x = 1/2x

Move 6x/2x to the right side.

Multiply -1 to both sides.

Then (6x - 1)/2x ≤ 0.

Multiplying (-) on both sides
changes the order of the inequality sign:
≥ → ≤.

Linear Inequality (One Variable)

Divide both sides by 2.

Then (6x - 1)/x ≤ 0.

Move the denominator x
to the numerator.

(6x - 1)/x → x(6x - 1)

This is
multiplying the square of the denominator, x2,
to both sides.

x ≠ 0
So you can multiply x2 on both sides.

And x2 is (+).
So the order of the inequality sign
doesn't change.

And write ≠ 0
under x.

Find the zeros of x(6x - 1).

Quadratic Equation: by Factoring

Case 1) x = 0

(We know that x ≠ 0.
This is the zero you're going to use
when graphing the inequality on the x-axis.)

Case 2) 6x - 1 = 0

Then x = 1/6.

Case 1) x = 0
Case 2) x = 1/6

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

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

First point the zeros x = 0 and 1/6.
And draw a parabola
that passes through x = 0 and 1/6.

Quadratic Function: Find Zeros

x ≠ 0

So draw an empty circle on x = 0.

See x(6x - 1) ≤ 0.

The left side is less than or equal to 0.

So color the region
where the graph is below the x-axis (y = 0).

The inequality sign includes equal to [=].

So draw a full circle on x = 1/6.

(Don't fill the empty circle on x = 0.
This is the excluded value.)

The colored region is
0 < x ≤ 1/6.

So
0 < x ≤ 1/6
is the answer.

Example 2

Example

Solution

Before solving the inequality,
first set (denominator) ≠ 0.

See 4/(x - 1).

The denominator is (x - 1).

So x - 1 ≠ 0.
Then x ≠ 1.

Next, see 1/x.

The denominator is x.

So x ≠ 0.

From the denominators,
you found that
x - 1 ≠ 0, x ≠ 0.

The x values should satisfy these conditions.

Next, solve the rational inequality.

First, find the least common multiple, LCM,
of the denominators.

The denominators are
(x - 1) and x.

So the LCM is
x(x - 1).

Change the denominators of the rational expression
to the LCM: x(x - 1).

See 4/(x - 1).

The denominator is (x - 1).

The factor x is missing.

So multiply x
to both of the numerator and the denominator.

4/(x - 1)
= [4/(x - 1)]⋅[x/x]

See the next term +1.

The denominator is 1.

x(x - 1) is missing.

So multiply x(x - 1)
to both of the numerator and the denominator.

+1
= +1⋅[[x(x - 1)]/[x(x - 1)]]

Write the inequality sign ≤.

See the right side 1/x.

The denominator is x.

The factor (x - 1) is missing.

So multiply (x - 1)
to both of the numerator and the denominator.

1/x
= [1/x]⋅[(x - 1)/(x - 1)]

So
4/(x - 1) + 1 ≤ 1/x
becomes
[4/(x - 1)]⋅[x/x] + 1⋅[[x(x - 1)]/[[x(x - 1)]] ≤ [1/x]⋅[(x - 1)/(x - 1)].

4⋅x = 4x

+1⋅x(x - 1) = +x2 - x

Multiply a Monomial and a Polynomial

1⋅(x - 1) = x - 1

4x + x2 - x
= x2 + 3x

Move (x - 1)/[x(x - 1)] to the left side.

Then (x2 + 3x - x + 1)/[x(x - 1)].

+3x - x = +2x

x2 + 2x + 1
= x2 + 2⋅x⋅1 + 12
= (x + 1)2

Factor a Perfect Square Trinomial

Move the denominator x(x - 1)
to the numerator.

(x + 1)2/[x(x - 1)] → (x + 1)2x(x - 1)

This is
multiplying the square of the denominator,
[x(x - 1)]2,
to both sides.

(x - 1) ≠ 0, x ≠ 0
So x(x - 1) ≠ 0.
So you can multiply [x(x - 1)]2 on both sides.

And [x(x - 1)]2 is (+).
So the order of the inequality sign
doesn't change.

And write ≠ 0
under x and (x - 1).

Write the zeros.
x = -1, 0, 1

(We know that x ≠ 0, 1.
These are the zeros you're going to use
when graphing the inequality on the x-axis.)

Draw the x-axis.

Point the zeros x = -1, 0, 1.

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

Polynomial Inequality

The highest degree term of y = (x + 1)2x(x - 1)
is x4.
The coefficient is (+).

So starting from the top right of the x-axis,
draw the graph
that goes toward the nearest zero:
x = 1.

See the factor (x - 1).
(x - 1) = (x - 1)1

The exponent is 1.
It's odd.

Then draw the graph
that passes through the x-axis
at x = 1.

Next, see the factor x.
x = x1

The exponent is 1.
It's odd.

Then draw the graph
that passes through the x-axis
at x = 0.

See the factor (x + 1)2.

The exponent is 2.
It's even.

Then draw the graph
that bounces off the x-axis
at x = -1.

So this is the graph of the polynomial
y = (x + 1)2x(x - 1)
on the x-axis.

x ≠ 0, (x - 1) ≠ 0

So draw empty circles on x = 0 and x = 1.

See (x + 1)2x(x - 1) ≤ 0.
The left side is less than 0.

So color the region
where the graph is below the x-axis (y = 0).

The inequality sign does not include equal to [=].
So draw a full circle on x = -1.

(Don't fill the empty circles on x = 0 and x = 1.
These are the excluded values.)

The colored regions are
x = -1, 0 < x < 1.

So
x = -1, 0 < x < 1
is the answer.