# Rational (Math)

See how to simplify a rational expression/equation/inequality.

17 examples and their solutions.

## Simplifying a Rational Expression

### Example

x - 5x

Solution ^{2}- 7x + 10 x

= (x - 2)(x - 5) - [1]

x - 5x

= x - 5(x - 2)(x - 5)

= 1x - 2

^{2}- 7x + 10= (x - 2)(x - 5) - [1]

x - 5x

^{2}- 7x + 10= x - 5(x - 2)(x - 5)

= 1x - 2

Close

### Example

2x

Solution ^{2}+ x - 1x^{2}- 2x - 3 2x

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

x

= (x - 3)(x + 1) - [1]

2x

= (2x - 1)(x + 1)(x - 3)(x + 1)

= 2x - 1x - 3

^{2}+ x - 1= (2x - 1)(x + 1) - [1]

x

^{2}- 2x - 3= (x - 3)(x + 1) - [1]

2x

^{2}+ x - 1x^{2}- 2x - 3= (2x - 1)(x + 1)(x - 3)(x + 1)

= 2x - 1x - 3

Close

## Dividing Rational Expressions

### Example

x3x - 1 ÷ x + 72x

Solution x3x - 1 ÷ x + 72x

= x3x - 1 ⋅ 2xx + 7

= 2x

= x3x - 1 ⋅ 2xx + 7

= 2x

^{2}(3x - 1)(x - 7)Close

### Example

x

Solution ^{2}- 4x - 1 ÷ x + 25 x

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

x

= (x + 2)(x - 2)x - 1 ÷ x + 25

= (x + 2)(x - 2)x - 1 ⋅ 5x + 2

= 5(x - 2)x - 1

^{2}- 4= (x + 2)(x - 2) - [1]

x

^{2}- 4x - 1 ÷ x + 25= (x + 2)(x - 2)x - 1 ÷ x + 25

= (x + 2)(x - 2)x - 1 ⋅ 5x + 2

= 5(x - 2)x - 1

Close

## Complex Fraction

### Formula

ab cd = adbc

### Example

x + 2x 6x - 1

Solution x + 2x 6x - 1 = (x + 2)(x - 1)6x

Close

### Example

4x - 1x (2x - 1)

Solution ^{2} 4x - 1x

= 4x⋅xx - 1x

= 4x

= 4x

= (2x)

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

4x - 1x (2x - 1)

= (2x + 1)(2x - 1)x (2x - 1)

= (2x + 1)(2x - 1)⋅1x(2x - 1)

= (2x + 1)x(2x - 1)

= 4x⋅xx - 1x

= 4x

^{2}x - 1x= 4x

^{2}- 1x= (2x)

^{2}- 1^{2}x= (2x + 1)(2x - 1)x - [1]

4x - 1x (2x - 1)

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

^{2}1= (2x + 1)(2x - 1)⋅1x(2x - 1)

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

Close

## Adding and Subtracting Rational Expressions

### Example

3x + xx + 2

Solution (LCM) = x(x + 2) - [1]

3x + xx + 2

= 3(x + 2)x(x + 2) + x⋅xx(x + 2) - [2]

= 3x + 6x(x + 2) + x

= 3x + 6 + x

= x

3x + xx + 2

= 3(x + 2)x(x + 2) + x⋅xx(x + 2) - [2]

= 3x + 6x(x + 2) + x

^{2}x(x + 2)= 3x + 6 + x

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

^{2}+ 3x + 6x(x + 2)[1]

LCM of the denominators

[2]

Change the denominators to x(x + 2)

by multiplying the missing factors.

by multiplying the missing factors.

Close

### Example

2xx

Solution ^{2}- 1 - 5x^{2}+ x x

x

(LCM) = x(x + 1)(x - 1)

2xx

= 2x(x + 1)(x - 1) - 5x(x + 1)

= 2x⋅xx(x + 1)(x - 1) - 5(x - 1)x(x + 1)(x - 1)

= 2x

= 2x

= 2x

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

^{2}+ x = x(x + 1)(LCM) = x(x + 1)(x - 1)

2xx

^{2}- 1 - 5x^{2}+ x= 2x(x + 1)(x - 1) - 5x(x + 1)

= 2x⋅xx(x + 1)(x - 1) - 5(x - 1)x(x + 1)(x - 1)

= 2x

^{2}x(x + 1)(x - 1) - 5x - 5x(x + 1)(x - 1)= 2x

^{2}- (5x - 5)x(x + 1)(x - 1)= 2x

^{2}- 5x + 5x(x + 1)(x - 1)Close

## Excluded Value

### Formula

ab

Excluded value: b = 0

The excluded value of a fractionExcluded value: b = 0

is the x value

that makes the denominator 0.

The denominator cannot be 0.

So the excluded value

cannot be the solution.

To find the excluded value,

set (denominator) = 0.

### Example

x

Solution ^{2}+ 5x - 13x - 2 x

3x - 2 = 0

3x = 2

x = 23

^{2}+ 5x - 13x - 23x - 2 = 0

3x = 2

x = 23

Close

## Rational Equation

### Example

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

Solution 3x + xx - 1 = 1x(x - 1)

x ≠ 0 - [1]

x - 1 ≠ 0 - [1]

x ≠ 1

(LCM) = x(x - 1) - [2]

3x⋅x(x - 1) + xx - 1⋅x(x - 1) = 1x(x - 1)⋅x(x - 1) - [3]

3(x - 1) + x⋅x = 1

3x - 3 + x

x

x

(x + 4)(x - 1) = 0 - [4] [5]

1) x + 4 = 0

x = -4 ( o ) - [6]

2) x - 1 = 0

x = 1 ( x ) - [7]

x = -4

x ≠ 0 - [1]

x - 1 ≠ 0 - [1]

x ≠ 1

(LCM) = x(x - 1) - [2]

3x⋅x(x - 1) + xx - 1⋅x(x - 1) = 1x(x - 1)⋅x(x - 1) - [3]

3(x - 1) + x⋅x = 1

3x - 3 + x

^{2}= 1x

^{2}+ 3x - 3 - 1 = 0x

^{2}+ 3x - 4 = 0(x + 4)(x - 1) = 0 - [4] [5]

1) x + 4 = 0

x = -4 ( o ) - [6]

2) x - 1 = 0

x = 1 ( x ) - [7]

x = -4

[1]

[2]

LCM of the denominators

[3]

× x(x - 1) to both sides.

[6]

x = -4 satisfies x ≠ 0, x ≠ 1.

So x = -4 is the root.

So x = -4 is the root.

[7]

x = 1 does not satisfy x ≠ 0, x ≠ 1.

So x = 1 is not the root.

So x = 1 is not the root.

Close

## Rational Inequality

### Example

4x - 1 + 1 ≤ 1x

Solution 4x - 1 + 1 ≤ 1x

x - 1 ≠ 0 - [1]

x ≠ 1

x ≠ 0 - [1]

(LCM) = x(x - 1) - [2]

4⋅xx(x - 1) + x(x - 1)x(x - 1) ≤ 1⋅(x - 1)x(x - 1) - [3]

4xx(x - 1) + x

4xx(x - 1) + x

4x + x

x

x

(x + 1)

(x + 1)

(x + 1)

≠ 0≠ 0 - [6]

- [7] [8]

x = -1, 0 < x < 1

x - 1 ≠ 0 - [1]

x ≠ 1

x ≠ 0 - [1]

(LCM) = x(x - 1) - [2]

4⋅xx(x - 1) + x(x - 1)x(x - 1) ≤ 1⋅(x - 1)x(x - 1) - [3]

4xx(x - 1) + x

^{2}- xx(x - 1) ≤ x - 1x(x - 1)4xx(x - 1) + x

^{2}- xx(x - 1) - x - 1x(x - 1) ≤ 04x + x

^{2}- x - (x - 1)x(x - 1) ≤ 0x

^{2}+ 3x - x + 1x(x - 1) ≤ 0x

^{2}+ 2x + 1x(x - 1) ≤ 0(x + 1)

^{2}x(x - 1) ≤ 0 - [4](x + 1)

^{2}x(x - 1)⋅(x(x - 1))^{2}≤ 0⋅(x(x - 1))^{2}- [5](x + 1)

^{2}x(x - 1) ≤ 0≠ 0≠ 0 - [6]

x = -1, 0 < x < 1

[1]

[2]

LCM of the denominators

[3]

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

[5]

× (denominator)

(denominator)

So the order of the inequality sign

doesn't change.

^{2}to both sides.(denominator)

^{2}is (+).So the order of the inequality sign

doesn't change.

[6]

Write ≠ 0 below x and (x - 1).

[7]

[8]

(x + 1)

So color the region

where the graph is below the x-axis.

x ≠ 0, x ≠ 1

So draw empty circles on x = 0, 1.

The inequality sign, ≤, does include '='.

So draw a full circle on x = -1.

^{2}x(x - 1) ≤ 0So color the region

where the graph is below the x-axis.

x ≠ 0, x ≠ 1

So draw empty circles on x = 0, 1.

The inequality sign, ≤, does include '='.

So draw a full circle on x = -1.

Close

## Partial Fraction Decomposition

Partial fraction decomposition is a way

to split a rational expression

to its partial fractions.

(= fractions with reduced denominators)

### Example

3x - 2x(x - 1)

Solution 3x - 2x(x - 1) = Ax + Bx - 1 - [1]

= A(x - 1)x(x - 1) + Bxx(x - 1)

= A(x - 1) + Bxx(x - 1)

= Ax - A + Bxx(x - 1)

= (A + B)x - Ax(x - 1)

A + B = 3 - [2]

-A = -2 - [3]

A = 2 - [4]

2 + B = 3 - [5]

B = 1

(given) = 2x + 1x - 1 - [6]

= A(x - 1)x(x - 1) + Bxx(x - 1)

= A(x - 1) + Bxx(x - 1)

= Ax - A + Bxx(x - 1)

= (A + B)x - Ax(x - 1)

A + B = 3 - [2]

-A = -2 - [3]

A = 2 - [4]

2 + B = 3 - [5]

B = 1

(given) = 2x + 1x - 1 - [6]

[1]

Denominator: x(x - 1)

Reduced factors: x, (x - 1)

So set

(given) = A/x + B/(x - 1).

The goal is to find A and B.

Reduced factors: x, (x - 1)

So set

(given) = A/x + B/(x - 1).

The goal is to find A and B.

[2]

Numerators of both sides: 3x - 2 = (A + B)x - A

x coefficients: A + B = 3

x coefficients: A + B = 3

[3]

Constants: -A = -2

[4]

-A = -2

A = 2

A = 2

[5]

A = 2 → A + B = 3

System of Linear Equations

System of Linear Equations

[6]

A = 2, B = 1 → A/x + B/(x - 1)

Close

### Example

5x

Solution ^{2}- 1x^{2}(x + 1) 5x

= Ax(x + 1)x

= Ax

= Ax

= (A + C)x

A + C = 5 - [2]

A + B = 0 - [3]

B = -1 - [4]

A + (-1) = 0 - [5]

A = 1

1 + C = 5 - [6]

C = 4

(given) = 1x - 1x

^{2}- 1x^{2}(x + 1) = Ax + Bx^{2}+ Cx + 1 - [1]= Ax(x + 1)x

^{2}(x + 1) + B(x + 1)x^{2}(x + 1) + Cx^{2}x^{2}(x + 1)= Ax

^{2}+ Axx^{2}(x + 1) + Bx + Bx^{2}(x + 1) + Cx^{2}x^{2}(x + 1)= Ax

^{2}+ Ax + Bx + B + Cx^{2}x^{2}(x + 1)= (A + C)x

^{2}+ (A + B)x + Bx^{2}(x + 1)A + C = 5 - [2]

A + B = 0 - [3]

B = -1 - [4]

A + (-1) = 0 - [5]

A = 1

1 + C = 5 - [6]

C = 4

(given) = 1x - 1x

^{2}+ 4x + 1 - [7][1]

Denominator: x

Reduced factors: x, x

So set

(given) = A/x + B/x

The goal is to find A, B, and C.

^{2}(x + 1)Reduced factors: x, x

^{2}, (x + 1)So set

(given) = A/x + B/x

^{2}+ C/(x + 1).The goal is to find A, B, and C.

[2]

Numerators of both sides:

5x

x

5x

^{2}- 1 = (A + C)x^{2}+ (A + B)x + Bx

^{2}coefficients: A + C = 5[3]

x coefficients: A + B = 0

[4]

Constants: B = -1

[5]

B = -1 → A + B = 0

[6]

A = 1 → A + C = 5

[7]

A = 1, B = -1, C = 4

→ A/x + B/x

→ A/x + B/x

^{2}+ C/(x + 1)Close

### Formula

1AB = 1B - A(1A - 1B)

When B - A = (constant),use this formula.

### Example

1x(x + 1)

Solution 1x(x + 1)

= 1(x + 1) - x(1x - 1x + 1)

= 11(1x - 1x + 1)

= 1x - 1x + 1

= 1(x + 1) - x(1x - 1x + 1)

= 11(1x - 1x + 1)

= 1x - 1x + 1

[1]

(x + 1) - x = 1 (constant)

So use the formula.

So use the formula.

Close

## Rational Function: Graph

### Graph: y = a/x (a > 0)

Then the graph is

on the right top and the left bottom

of the axes.

### Graph: y = a/x (a < 0)

Then the graph is

on the left top and the right bottom

of the axes.

### Asymptotes

y = ax

Asymptotes: x = 0

y = 0

a vertical asymptote

and a horizontal asymptote.

(asymptote: a line the graph follows)

To find the asymptotes of y = ax,

1. Set (denominator) = 0.

→ x = 0

2. Set (fraction) = 0.

→ y = 0

### Example

Graph y = 4x - 1 + 2.

Solution y = 4x - 1 + 2

x - 1 = 0

x = 1

y = 0 + 2

y = 2

[1]

x - 1 = 0

x = 1

y = 0 + 2

y = 2

[1]

First draw the asymptotes

x = 1, y = 2.

(dashed lines)

y = 4/(x - 1) + 2

The numerator, 4, is (+).

So draw the graph

on the right top and the left bottom

of the asymptotes.

x = 1, y = 2.

(dashed lines)

y = 4/(x - 1) + 2

The numerator, 4, is (+).

So draw the graph

on the right top and the left bottom

of the asymptotes.

Close

### Example

Graph y = 13 - x - 1.

Solution y = 13 - x - 1

y = -1x - 3 - 1

x - 3 = 0

x = 3

y = 0 - 1

y = -1

[1]

y = -1x - 3 - 1

x - 3 = 0

x = 3

y = 0 - 1

y = -1

[1]

First draw the asymptotes

x = 3, y = -1.

(dashed lines)

y = -1/(x - 3) - 1

The numerator, -1, is (-).

So draw the graph

on the left top and the right bottom

of the asymptotes.

x = 3, y = -1.

(dashed lines)

y = -1/(x - 3) - 1

The numerator, -1, is (-).

So draw the graph

on the left top and the right bottom

of the asymptotes.

Close

### Example

Graph y = 3x - 5x - 2.

Solution y = 3x - 5x - 2

= 3(x - 2) + 3⋅2 - 5x - 2- [1]

= 3(x - 2) + 6 - 5x - 2

= 3(x - 2) + 1x - 2

= 3(x - 2)x - 2 + 1x - 2

= 3 + 1x - 2

y = 1x - 2 + 3

x - 2 = 0

x = 2

y = 0 + 3

y = 3

[2]

= 3(x - 2) + 3⋅2 - 5x - 2- [1]

= 3(x - 2) + 6 - 5x - 2

= 3(x - 2) + 1x - 2

= 3(x - 2)x - 2 + 1x - 2

= 3 + 1x - 2

y = 1x - 2 + 3

x - 2 = 0

x = 2

y = 0 + 3

y = 3

[1]

The numerator, 3x - 5, has a variable: x.

To find the asymptotes,

the numerator should be a constant.

So, to remove x in the numerator,

change 3x to 3(x - 2).

(x - 2): denominator

3⋅(-2) is added.

So, to undo the change,

write + 3⋅2.

To find the asymptotes,

the numerator should be a constant.

So, to remove x in the numerator,

change 3x to 3(x - 2).

(x - 2): denominator

3⋅(-2) is added.

So, to undo the change,

write + 3⋅2.

[2]

First draw the asymptotes

x = 2, y = 3.

(dashed lines)

y = 1/(x - 2) + 3

The numerator, 1, is (+).

So draw the graph

on the right top and the left bottom

of the asymptotes.

x = 2, y = 3.

(dashed lines)

y = 1/(x - 2) + 3

The numerator, 1, is (+).

So draw the graph

on the right top and the left bottom

of the asymptotes.

Close