Rational (Math)
See how to simplify a rational expression/equation/inequality.
17 examples and their solutions.
Simplifying a Rational Expression
Example
x - 5x2 - 7x + 10
Solution x2 - 7x + 10
= (x - 2)(x - 5) - [1]
x - 5x2 - 7x + 10
= x - 5(x - 2)(x - 5)
= 1x - 2
= (x - 2)(x - 5) - [1]
x - 5x2 - 7x + 10
= x - 5(x - 2)(x - 5)
= 1x - 2
Close
Example
2x2 + x - 1x2 - 2x - 3
Solution 2x2 + x - 1
= (2x - 1)(x + 1) - [1]
x2 - 2x - 3
= (x - 3)(x + 1) - [1]
2x2 + x - 1x2 - 2x - 3
= (2x - 1)(x + 1)(x - 3)(x + 1)
= 2x - 1x - 3
= (2x - 1)(x + 1) - [1]
x2 - 2x - 3
= (x - 3)(x + 1) - [1]
2x2 + x - 1x2 - 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
= 2x2(3x - 1)(x - 7)
= x3x - 1 ⋅ 2xx + 7
= 2x2(3x - 1)(x - 7)
Close
Example
x2 - 4x - 1 ÷ x + 25
Solution x2 - 4
= (x + 2)(x - 2) - [1]
x2 - 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
= (x + 2)(x - 2) - [1]
x2 - 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)2
Solution 4x - 1x
= 4x⋅xx - 1x
= 4x2x - 1x
= 4x2 - 1x
= (2x)2 - 12x
= (2x + 1)(2x - 1)x - [1]
4x - 1x (2x - 1)2
= (2x + 1)(2x - 1)x (2x - 1)21
= (2x + 1)(2x - 1)⋅1x(2x - 1)2
= (2x + 1)x(2x - 1)
= 4x⋅xx - 1x
= 4x2x - 1x
= 4x2 - 1x
= (2x)2 - 12x
= (2x + 1)(2x - 1)x - [1]
4x - 1x (2x - 1)2
= (2x + 1)(2x - 1)x (2x - 1)21
= (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) + x2x(x + 2)
= 3x + 6 + x2x(x + 2)
= x2 + 3x + 6x(x + 2)
3x + xx + 2
= 3(x + 2)x(x + 2) + x⋅xx(x + 2) - [2]
= 3x + 6x(x + 2) + x2x(x + 2)
= 3x + 6 + x2x(x + 2)
= x2 + 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
2xx2 - 1 - 5x2 + x
Solution x2 - 1 = (x + 1)(x - 1)
x2 + x = x(x + 1)
(LCM) = x(x + 1)(x - 1)
2xx2 - 1 - 5x2 + x
= 2x(x + 1)(x - 1) - 5x(x + 1)
= 2x⋅xx(x + 1)(x - 1) - 5(x - 1)x(x + 1)(x - 1)
= 2x2x(x + 1)(x - 1) - 5x - 5x(x + 1)(x - 1)
= 2x2 - (5x - 5)x(x + 1)(x - 1)
= 2x2 - 5x + 5x(x + 1)(x - 1)
x2 + x = x(x + 1)
(LCM) = x(x + 1)(x - 1)
2xx2 - 1 - 5x2 + x
= 2x(x + 1)(x - 1) - 5x(x + 1)
= 2x⋅xx(x + 1)(x - 1) - 5(x - 1)x(x + 1)(x - 1)
= 2x2x(x + 1)(x - 1) - 5x - 5x(x + 1)(x - 1)
= 2x2 - (5x - 5)x(x + 1)(x - 1)
= 2x2 - 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
x2 + 5x - 13x - 2
Solution x2 + 5x - 13x - 2
3x - 2 = 0
3x = 2
x = 23
3x - 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 + x2 = 1
x2 + 3x - 3 - 1 = 0
x2 + 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
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 + x2 = 1
x2 + 3x - 3 - 1 = 0
x2 + 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) + x2 - xx(x - 1) ≤ x - 1x(x - 1)
4xx(x - 1) + x2 - xx(x - 1) - x - 1x(x - 1) ≤ 0
4x + x2 - x - (x - 1)x(x - 1) ≤ 0
x2 + 3x - x + 1x(x - 1) ≤ 0
x2 + 2x + 1x(x - 1) ≤ 0
(x + 1)2x(x - 1) ≤ 0 - [4]
(x + 1)2x(x - 1)⋅(x(x - 1))2 ≤ 0⋅(x(x - 1))2 - [5]
(x + 1)2x(x - 1) ≤ 0
≠ 0≠ 0 - [6]
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) + x2 - xx(x - 1) ≤ x - 1x(x - 1)
4xx(x - 1) + x2 - xx(x - 1) - x - 1x(x - 1) ≤ 0
4x + x2 - x - (x - 1)x(x - 1) ≤ 0
x2 + 3x - x + 1x(x - 1) ≤ 0
x2 + 2x + 1x(x - 1) ≤ 0
(x + 1)2x(x - 1) ≤ 0 - [4]
(x + 1)2x(x - 1)⋅(x(x - 1))2 ≤ 0⋅(x(x - 1))2 - [5]
(x + 1)2x(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)2 to both sides.
(denominator)2 is (+).
So the order of the inequality sign
doesn't change.
(denominator)2 is (+).
So the order of the inequality sign
doesn't change.
[6]
Write ≠ 0 below x and (x - 1).
[7]
Graph y = (x + 1)2x(x - 1) on a coordinate plane.
Polynomial Inequality
Polynomial Inequality
[8]
(x + 1)2x(x - 1) ≤ 0
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.
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.
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
5x2 - 1x2(x + 1)
Solution 5x2 - 1x2(x + 1) = Ax + Bx2 + Cx + 1 - [1]
= Ax(x + 1)x2(x + 1) + B(x + 1)x2(x + 1) + Cx2x2(x + 1)
= Ax2 + Axx2(x + 1) + Bx + Bx2(x + 1) + Cx2x2(x + 1)
= Ax2 + Ax + Bx + B + Cx2x2(x + 1)
= (A + C)x2 + (A + B)x + Bx2(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 - 1x2 + 4x + 1 - [7]
= Ax(x + 1)x2(x + 1) + B(x + 1)x2(x + 1) + Cx2x2(x + 1)
= Ax2 + Axx2(x + 1) + Bx + Bx2(x + 1) + Cx2x2(x + 1)
= Ax2 + Ax + Bx + B + Cx2x2(x + 1)
= (A + C)x2 + (A + B)x + Bx2(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 - 1x2 + 4x + 1 - [7]
[1]
Denominator: x2(x + 1)
Reduced factors: x, x2, (x + 1)
So set
(given) = A/x + B/x2 + C/(x + 1).
The goal is to find A, B, and C.
Reduced factors: x, x2, (x + 1)
So set
(given) = A/x + B/x2 + C/(x + 1).
The goal is to find A, B, and C.
[2]
Numerators of both sides:
5x2 - 1 = (A + C)x2 + (A + B)x + B
x2 coefficients: A + C = 5
5x2 - 1 = (A + C)x2 + (A + B)x + B
x2 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/x2 + C/(x + 1)
→ A/x + B/x2 + 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
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
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
= 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