The Central Idea
Rational expressions behave like fractions because they are fractions. If
\[ \frac{a}{b} \quad\text{and}\quad \frac{c}{d} \]are defined, then the familiar fraction rules still apply:
\[ \frac{a}{b}\cdot\frac{c}{d} = \frac{ac}{bd}, \] \[ \frac{a}{b}\div\frac{c}{d} = \frac{a}{b}\cdot\frac{d}{c}, \] \[ \frac{a}{b}+\frac{c}{d} = \frac{ad+bc}{bd}, \]and
\[ \frac{a}{b}-\frac{c}{d} = \frac{ad-bc}{bd}. \]Algebra adds one major responsibility: we must track all values that make an original denominator zero.
Begin with Domain Restrictions
Before operating on rational expressions, identify the values that make any original denominator zero.
For example, in
\[ \frac{x+1}{x-2} + \frac{3}{x+5}, \]the restrictions are
\[ x\neq2 \]and
\[ x\neq-5. \]These restrictions remain part of the final result even if later simplification removes one of the corresponding factors.
Multiplying Rational Expressions
Multiply numerators together and multiply denominators together:
\[ \boxed{ \frac{a}{b}\cdot\frac{c}{d} = \frac{ac}{bd} }. \]With rational expressions, it is usually best to factor before carrying out all of the multiplication.
Factoring exposes common factors that may cancel.
Example: Direct Multiplication
Multiply
\[ \frac{2x}{3}\cdot\frac{5}{x+1}. \]Multiply the numerators:
\[ 2x\cdot5=10x. \]Multiply the denominators:
\[ 3(x+1). \]Therefore,
\[ \boxed{ \frac{10x}{3(x+1)}, \qquad x\neq-1 }. \]Factor Before Multiplying
Consider
\[ \frac{x^2-9}{x^2+5x+6} \cdot \frac{x+2}{x-3}. \]Expanding products would make the expression harder to simplify. Instead, factor first.
\[ x^2-9=(x-3)(x+3), \] \[ x^2+5x+6=(x+2)(x+3). \]Example: Multiply and Cancel
Using the factorizations,
\[ \frac{(x-3)(x+3)}{(x+2)(x+3)} \cdot \frac{x+2}{x-3}. \]The original restrictions are
\[ x\neq-2,\qquad x\neq-3,\qquad x\neq3. \]Every visible factor cancels:
\[ \boxed{ 1, \qquad x\neq-3,\ -2,\ 3 }. \]The result is numerically \(1\) everywhere the original product is defined.
Cross-Canceling Is Still Factor Cancellation
Students often describe cancellation between opposite corners of two multiplied fractions as cross-canceling.
For example,
\[ \frac{6x}{5} \cdot \frac{10}{9x}. \]Since multiplication is associative and commutative, all numerator factors form one product and all denominator factors form another:
\[ \frac{6x\cdot10}{5\cdot9x}. \]Canceling is therefore ordinary common-factor reduction:
\[ \boxed{ \frac43, \qquad x\neq0 }. \]Example: Multiplication with Several Factors
Simplify
\[ \frac{3(x-2)^2}{4(x+5)} \cdot \frac{8(x+5)^2}{9(x-2)}. \]The original restrictions are
\[ x\neq-5,\qquad x\neq2. \]Reduce the numerical factors and cancel common polynomial factors:
\[ \frac{3\cdot8}{4\cdot9} \cdot \frac{(x-2)^2(x+5)^2}{(x+5)(x-2)}. \] \[ = \frac23(x-2)(x+5). \]Therefore,
\[ \boxed{ \frac{2(x-2)(x+5)}{3}, \qquad x\neq-5,\ 2 }. \]Division Uses the Reciprocal
Dividing by a nonzero fraction is equivalent to multiplying by its reciprocal:
\[ \boxed{ \frac{a}{b}\div\frac{c}{d} = \frac{a}{b}\cdot\frac{d}{c} }. \]The second fraction is inverted because
\[ \frac{c}{d}\cdot\frac{d}{c}=1 \]whenever both \(c\) and \(d\) are nonzero.
Keep, Change, Flip
A common memory aid for fraction division is:
- keep the first fraction;
- change division to multiplication;
- flip the second fraction.
The phrase is useful only when its algebraic meaning is understood: division by a quantity is multiplication by its reciprocal.
Example: Dividing Rational Expressions
Simplify
\[ \frac{x+2}{x-4} \div \frac{x+5}{x+1}. \]The original denominator restrictions are
\[ x\neq4,\qquad x\neq-1. \]The divisor must also be nonzero, so its numerator cannot be zero:
\[ x+5\neq0. \]Therefore,
\[ x\neq-5 \]as well.
Division Introduces an Additional Restriction
Continue with
\[ \frac{x+2}{x-4} \div \frac{x+5}{x+1}. \]Rewrite as multiplication by the reciprocal:
\[ \frac{x+2}{x-4} \cdot \frac{x+1}{x+5}. \]Therefore,
\[ \boxed{ \frac{(x+2)(x+1)} {(x-4)(x+5)}, \qquad x\neq4,\ -1,\ -5 }. \]The restriction \(x\neq-5\) comes from the requirement that the original divisor not equal zero.
Why a Divisor Cannot Equal Zero
Consider
\[ \frac{A}{B}\div\frac{C}{D}. \]The expression is defined only if
\[ B\neq0,\qquad D\neq0, \]and also
\[ C\neq0, \]because
\[ \frac{C}{D}=0 \]when \(C=0\), and division by zero is undefined.
Example: Divide, Factor, and Cancel
Simplify
\[ \frac{x^2-4}{x^2+3x+2} \div \frac{x-2}{x+1}. \]Factor:
\[ x^2-4=(x-2)(x+2), \] \[ x^2+3x+2=(x+1)(x+2). \]The original restrictions are
\[ x\neq-1,\qquad x\neq-2, \]and the divisor cannot equal zero, so
\[ x\neq2. \]Complete the Division Example
Rewrite division as multiplication:
\[ \frac{(x-2)(x+2)} {(x+1)(x+2)} \cdot \frac{x+1}{x-2}. \]Cancel all common factors:
\[ \boxed{ 1, \qquad x\neq-2,\ -1,\ 2 }. \]Again, the constant simplified result does not erase the original restrictions.
Addition Requires a Common Denominator
Fractions can be added directly only when they represent parts of the same size.
If two rational expressions already have the same denominator,
\[ \frac{A}{D}+\frac{B}{D}, \]then
\[ \boxed{ \frac{A}{D}+\frac{B}{D} = \frac{A+B}{D} }. \]The denominator is retained, and the numerators are combined.
Example: Add with a Common Denominator
Add
\[ \frac{2x}{x+3} + \frac{5}{x+3}. \]The denominators already match:
\[ \frac{2x+5}{x+3}. \]Therefore,
\[ \boxed{ \frac{2x+5}{x+3}, \qquad x\neq-3 }. \]Subtraction with a Common Denominator
If the denominators match,
\[ \frac{A}{D}-\frac{B}{D} = \frac{A-B}{D}. \]For example,
\[ \frac{3x+4}{x-2} - \frac{x+1}{x-2} \]becomes
\[ \frac{(3x+4)-(x+1)}{x-2}. \]Simplify the numerator:
\[ 3x+4-x-1=2x+3. \]Thus,
\[ \boxed{ \frac{2x+3}{x-2}, \qquad x\neq2 }. \]Use Parentheses When Subtracting
Parentheses protect the entire numerator being subtracted.
In
\[ \frac{A}{D}-\frac{B+C}{D}, \]write
\[ \frac{A-(B+C)}{D}. \]The negative sign must distribute to every term:
\[ A-(B+C)=A-B-C. \]Omitting parentheses is a common source of sign errors.
Different Denominators Require an LCD
To add or subtract rational expressions with different denominators, first find a least common denominator, or LCD.
The LCD is built from the factors needed so that every original denominator divides it.
Once the expressions have equivalent common-denominator forms, their numerators can be combined.
Finding an LCD from Distinct Linear Factors
Consider the denominators
\[ x-2 \]and
\[ x+5. \]These factors are distinct, so the LCD is
\[ \boxed{(x-2)(x+5)}. \]The corresponding restrictions are
\[ x\neq2,\qquad x\neq-5. \]Example: Add with Different Linear Denominators
Add
\[ \frac{2}{x-2} + \frac{3}{x+5}. \]The LCD is
\[ (x-2)(x+5). \]Rewrite each fraction:
\[ \frac{2(x+5)}{(x-2)(x+5)} + \frac{3(x-2)}{(x+5)(x-2)}. \]Complete the Addition Example
Combine the numerators:
\[ \frac{2(x+5)+3(x-2)} {(x-2)(x+5)}. \]Expand only the numerator:
\[ 2x+10+3x-6=5x+4. \]Therefore,
\[ \boxed{ \frac{5x+4}{(x-2)(x+5)}, \qquad x\neq2,\ -5 }. \]Example: Subtract with Different Linear Denominators
Subtract
\[ \frac{4}{x+1} - \frac{1}{x-3}. \]The LCD is
\[ (x+1)(x-3). \]Rewrite:
\[ \frac{4(x-3)} {(x+1)(x-3)} - \frac{x+1} {(x-3)(x+1)}. \]Combine:
\[ \frac{4(x-3)-(x+1)} {(x+1)(x-3)}. \]Simplify:
\[ 4x-12-x-1=3x-13. \]Thus,
\[ \boxed{ \frac{3x-13}{(x+1)(x-3)}, \qquad x\neq-1,\ 3 }. \]The LCD Uses Each Factor at Its Greatest Needed Power
Suppose the denominators are
\[ (x-1)^2(x+3) \]and
\[ (x-1)(x+3)^3. \]The LCD must contain enough copies of every factor to accommodate both:
\[ \boxed{(x-1)^2(x+3)^3}. \]We use the greatest exponent appearing for each distinct factor.
Example: Denominators with Shared Factors
Add
\[ \frac{1}{x(x+2)} + \frac{3}{x+2}. \]The denominators are
\[ x(x+2) \]and
\[ x+2. \]The LCD is simply
\[ x(x+2). \]Only the second fraction needs an additional factor of \(x\).
Complete the Shared-Factor Example
Rewrite
\[ \frac{3}{x+2} \]as
\[ \frac{3x}{x(x+2)}. \]Then
\[ \frac{1}{x(x+2)} + \frac{3x}{x(x+2)} = \frac{3x+1}{x(x+2)}. \]Therefore,
\[ \boxed{ \frac{3x+1}{x(x+2)}, \qquad x\neq0,\ -2 }. \]Factor Denominators Before Finding the LCD
Consider
\[ \frac{1}{x^2-4} + \frac{2}{x^2+4x+4}. \]Factor:
\[ x^2-4=(x-2)(x+2), \] \[ x^2+4x+4=(x+2)^2. \]The LCD is
\[ \boxed{(x-2)(x+2)^2}. \]Factoring first prevents unnecessary duplication of factors.
Example: Add Using a Factored LCD
Continue with
\[ \frac{1}{(x-2)(x+2)} + \frac{2}{(x+2)^2}. \]The restrictions are
\[ x\neq2,\qquad x\neq-2. \]Rewrite using the LCD:
\[ \frac{x+2}{(x-2)(x+2)^2} + \frac{2(x-2)}{(x-2)(x+2)^2}. \]Combine:
\[ \frac{x+2+2x-4} {(x-2)(x+2)^2}. \]Therefore,
\[ \boxed{ \frac{3x-2}{(x-2)(x+2)^2}, \qquad x\neq-2,\ 2 }. \]Equivalent Fractions Are Created by Multiplying by One
When we change
\[ \frac{A}{B} \]into an equivalent fraction with a larger denominator, we multiply by a carefully chosen form of \(1\).
For example,
\[ \frac{2}{x-3} \cdot \frac{x+1}{x+1} = \frac{2(x+1)}{(x-3)(x+1)}. \]The value is unchanged wherever \(x+1\neq0\).
Example: A Numerical and Variable LCD
Add
\[ \frac{1}{6x} + \frac{5}{4x^2}. \]The least common multiple of \(6\) and \(4\) is \(12\).
The greatest power of \(x\) needed is \(x^2\).
Therefore, the LCD is
\[ 12x^2. \]Rewrite:
\[ \frac{2x}{12x^2} + \frac{15}{12x^2}. \]Thus,
\[ \boxed{ \frac{2x+15}{12x^2}, \qquad x\neq0 }. \]Adding More Than Two Rational Expressions
The same LCD method works with three or more terms.
Consider
\[ \frac1x+\frac1{x+1}+\frac1{x-1}. \]The LCD is
\[ x(x+1)(x-1). \]The restrictions are
\[ x\neq0,\qquad x\neq-1,\qquad x\neq1. \]Example: Three Rational Terms
Rewrite each term:
\[ \frac{(x+1)(x-1)} {x(x+1)(x-1)} + \frac{x(x-1)} {x(x+1)(x-1)} + \frac{x(x+1)} {x(x+1)(x-1)}. \]Combine numerators:
\[ (x+1)(x-1)+x(x-1)+x(x+1). \]Expand:
\[ x^2-1+x^2-x+x^2+x = 3x^2-1. \]Therefore,
\[ \boxed{ \frac{3x^2-1}{x(x+1)(x-1)}, \qquad x\neq-1,\ 0,\ 1 }. \]Simplify After Combining
After adding or subtracting rational expressions, inspect the resulting numerator and denominator for common factors.
For example,
\[ \frac{x}{x+2} + \frac{2}{x+2} = \frac{x+2}{x+2}. \]Therefore,
\[ \boxed{1,\qquad x\neq-2}. \]The original restriction remains even though the final formula is a constant.
Example: Addition Followed by Cancellation
Add
\[ \frac{x}{x-3} - \frac{3}{x-3}. \]Combine the numerators:
\[ \frac{x-3}{x-3}. \]For \(x\neq3\),
\[ \frac{x-3}{x-3}=1. \]Thus,
\[ \boxed{1,\qquad x\neq3}. \]Do Not Cancel Before Combining Addition
Consider
\[ \frac{x}{x+2} + \frac{2}{x+2}. \]We cannot cancel \(x\) with anything in the denominator of the first term, nor can we cancel \(2\) with the denominator of the second term.
First combine:
\[ \frac{x+2}{x+2}. \]Only then does the entire numerator become a factor matching the denominator.
Opposite Denominators
Sometimes denominators differ only by a negative sign.
Since
\[ a-b=-(b-a), \]we can rewrite one denominator so the factors match.
For example,
\[ \frac{1}{x-4} + \frac{2}{4-x} \]becomes
\[ \frac{1}{x-4} - \frac{2}{x-4}. \]Example: Add with Opposite Denominators
Continue:
\[ \frac{1}{x-4} - \frac{2}{x-4} = \frac{-1}{x-4}. \]Therefore,
\[ \boxed{ -\frac{1}{x-4}, \qquad x\neq4 }. \]Rewriting opposite factors first can turn an apparently different- denominator problem into a same-denominator problem.
Multiplication and Addition Behave Differently
In a product,
\[ \frac{a}{b}\cdot\frac{c}{d}, \]factors can often cancel before multiplication is completed.
In a sum,
\[ \frac{a}{b}+\frac{c}{d}, \]cross-cancellation is not valid because addition separates the terms.
A common denominator must be created first.
Common Error: Adding Denominators
A frequent mistake is
\[ \frac{a}{b}+\frac{c}{d} \not= \frac{a+c}{b+d}. \]For example,
\[ \frac12+\frac13 \]is not
\[ \frac25. \]The correct calculation is
\[ \frac36+\frac26=\frac56. \]Rational expressions follow the same rule.
Common Error: Cross-Canceling Across Addition
Consider
\[ \frac{x}{x+1} + \frac{1}{x}. \]The \(x\) in the first numerator cannot cancel with the \(x\) in the second denominator because the fractions are being added, not multiplied.
The correct LCD is
\[ x(x+1). \]Common Error: Forgetting the Reciprocal in Division
Dividing
\[ \frac{a}{b}\div\frac{c}{d} \]does not give
\[ \frac{ac}{bd}. \]That would be multiplication.
Division requires the reciprocal:
\[ \boxed{ \frac{a}{b}\div\frac{c}{d} = \frac{a}{b}\cdot\frac{d}{c} }. \]Common Error: Forgetting That the Divisor Cannot Be Zero
In
\[ \frac{x+1}{x-2} \div \frac{x-4}{x+3}, \]the original denominators require
\[ x\neq2,\qquad x\neq-3. \]But the divisor also cannot equal zero, so
\[ x-4\neq0. \]Therefore,
\[ \boxed{x\neq2,\ -3,\ 4}. \]Common Error: Using an Unnecessarily Large Denominator
Suppose the denominators are
\[ x(x+2) \]and
\[ x+2. \]Multiplying them mechanically gives
\[ x(x+2)^2, \]which is a common denominator but not the least common denominator.
The LCD is
\[ \boxed{x(x+2)}. \]Factoring denominators first makes the least common structure visible.
Common Error: Distributing a Negative Sign Incorrectly
Consider
\[ \frac{3x+2}{x+1} - \frac{x-4}{x+1}. \]The numerator becomes
\[ (3x+2)-(x-4). \]Distribute the negative sign:
\[ 3x+2-x+4 = 2x+6. \]Therefore,
\[ \boxed{ \frac{2x+6}{x+1}, \qquad x\neq-1 }. \]Common Error: Expanding Too Early
In multiplication and division, expanded polynomials often hide common factors.
For
\[ \frac{x^2-4}{x^2+5x+6} \cdot \frac{x+3}{x-2}, \]factoring immediately gives
\[ \frac{(x-2)(x+2)} {(x+2)(x+3)} \cdot \frac{x+3}{x-2}, \]making the cancellation obvious.
Expanding first would create unnecessary work.
A Reliable Procedure for Multiplication
- Record restrictions from every original denominator.
- Factor every numerator and denominator completely when useful.
- Combine the fractions into one product.
- Cancel common factors from numerator and denominator.
- Reduce numerical coefficients.
- Multiply any remaining factors only if expanded form is useful.
- Carry all original restrictions into the final answer.
A Reliable Procedure for Division
- Record restrictions from every original denominator.
- Require the entire divisor to be nonzero.
- Rewrite division as multiplication by the reciprocal.
- Factor all numerators and denominators.
- Cancel common factors.
- Simplify numerical coefficients.
- Preserve all restrictions from the original problem.
A Reliable Procedure for Addition and Subtraction
- Record restrictions from every original denominator.
- Factor the denominators.
- Identify the least common denominator.
- Rewrite each rational expression with the LCD.
- Combine the numerators while keeping the common denominator.
- Use parentheses carefully when subtracting.
- Simplify the resulting numerator.
- Factor and cancel any common factors that appear afterward.
- Carry all original restrictions into the final answer.
Example: Full Multiplication Workflow
Simplify
\[ \frac{x^2-4x}{x^2-9} \cdot \frac{x+3}{2x}. \]The restrictions are
\[ x\neq-3,\qquad x\neq3,\qquad x\neq0. \]Factor:
\[ x^2-4x=x(x-4), \] \[ x^2-9=(x-3)(x+3). \]Then
\[ \frac{x(x-4)}{(x-3)(x+3)} \cdot \frac{x+3}{2x}. \]Cancel \(x\) and \(x+3\):
\[ \boxed{ \frac{x-4}{2(x-3)}, \qquad x\neq-3,\ 0,\ 3 }. \]Example: Full Division Workflow
Simplify
\[ \frac{x^2-1}{x^2+4x+3} \div \frac{x-1}{x+3}. \]Factor:
\[ x^2-1=(x-1)(x+1), \] \[ x^2+4x+3=(x+1)(x+3). \]Original denominators require
\[ x\neq-1,\qquad x\neq-3. \]The divisor cannot equal zero, so
\[ x\neq1. \]Rewrite:
\[ \frac{(x-1)(x+1)} {(x+1)(x+3)} \cdot \frac{x+3}{x-1}. \]Everything cancels:
\[ \boxed{ 1, \qquad x\neq-3,\ -1,\ 1 }. \]Example: Full Addition Workflow
Add
\[ \frac{2}{x-1} + \frac{3}{x+2}. \]Restrictions:
\[ x\neq1,\qquad x\neq-2. \]LCD:
\[ (x-1)(x+2). \]Rewrite:
\[ \frac{2(x+2)}{(x-1)(x+2)} + \frac{3(x-1)}{(x+2)(x-1)}. \]Combine:
\[ \frac{2x+4+3x-3}{(x-1)(x+2)} = \frac{5x+1}{(x-1)(x+2)}. \]Therefore,
\[ \boxed{ \frac{5x+1}{(x-1)(x+2)}, \qquad x\neq1,\ -2 }. \]Example: Full Subtraction Workflow
Subtract
\[ \frac{x}{x+3} - \frac{2}{x-1}. \]Restrictions:
\[ x\neq-3,\qquad x\neq1. \]The LCD is
\[ (x+3)(x-1). \]Rewrite:
\[ \frac{x(x-1)}{(x+3)(x-1)} - \frac{2(x+3)}{(x-1)(x+3)}. \]Combine:
\[ \frac{x(x-1)-2(x+3)} {(x+3)(x-1)}. \]Simplify the numerator:
\[ x^2-x-2x-6 = x^2-3x-6. \]Therefore,
\[ \boxed{ \frac{x^2-3x-6}{(x+3)(x-1)}, \qquad x\neq-3,\ 1 }. \]Practice 1: Multiplication
Simplify and state the restrictions:
\[ \frac{3x}{x+2} \cdot \frac{x+2}{6}. \]Practice 2: Multiply After Factoring
Simplify and state the restrictions:
\[ \frac{x^2-9}{x^2+x-6} \cdot \frac{x-2}{x+3}. \]Practice 3: Division
Simplify and state all restrictions:
\[ \frac{x+4}{x-1} \div \frac{x+2}{x+5}. \]Practice 4: Divide After Factoring
Simplify and state all restrictions:
\[ \frac{x^2-4}{x^2+5x+6} \div \frac{x-2}{x+3}. \]Practice 5: Same Denominator
Add:
\[ \frac{4x}{x-5} + \frac{7}{x-5}. \]Practice 6: Different Denominators
Add:
\[ \frac{1}{x} + \frac{2}{x+3}. \]Practice 7: Subtraction
Simplify:
\[ \frac{3}{x+2} - \frac{1}{x-1}. \]Practice 8: Shared Denominator Factor
Add:
\[ \frac{2}{x(x+1)} + \frac{5}{x+1}. \]Practice 9: Factor Before Finding the LCD
Add:
\[ \frac{1}{x^2-4} + \frac{1}{x+2}. \]Practice 10: Identify the Error
A student writes
\[ \frac{x}{x+1} + \frac{1}{x} = \frac{x+1}{2x+1}. \]Explain the error and find the correct result.
Practice Answers
Practice 1
Cancel \(x+2\) and reduce \(3/6\):
\[ \boxed{ \frac{x}{2}, \qquad x\neq-2 }. \]Practice 2
Factor:
\[ x^2-9=(x-3)(x+3), \] \[ x^2+x-6=(x+3)(x-2). \]The original restrictions are
\[ x\neq-3,\qquad x\neq2. \]Then
\[ \frac{(x-3)(x+3)} {(x+3)(x-2)} \cdot \frac{x-2}{x+3}. \]Simplifying gives
\[ \boxed{ \frac{x-3}{x+3}, \qquad x\neq-3,\ 2 }. \]Practice 3
Original denominators require
\[ x\neq1,\qquad x\neq-5. \]The divisor cannot equal zero, so
\[ x\neq-2. \]Rewrite:
\[ \frac{x+4}{x-1} \cdot \frac{x+5}{x+2}. \]Therefore,
\[ \boxed{ \frac{(x+4)(x+5)} {(x-1)(x+2)}, \qquad x\neq1,\ -5,\ -2 }. \]Practice 4
Factor:
\[ x^2-4=(x-2)(x+2), \] \[ x^2+5x+6=(x+2)(x+3). \]The original denominator restrictions are
\[ x\neq-2,\qquad x\neq-3. \]The divisor cannot equal zero, so
\[ x\neq2. \]Rewrite:
\[ \frac{(x-2)(x+2)} {(x+2)(x+3)} \cdot \frac{x+3}{x-2}. \]Everything cancels:
\[ \boxed{ 1, \qquad x\neq-3,\ -2,\ 2 }. \]Practice 5
The denominator already matches:
\[ \boxed{ \frac{4x+7}{x-5}, \qquad x\neq5 }. \]Practice 6
The LCD is
\[ x(x+3). \]Rewrite:
\[ \frac{x+3}{x(x+3)} + \frac{2x}{x(x+3)}. \]Therefore,
\[ \boxed{ \frac{3x+3}{x(x+3)} = \frac{3(x+1)}{x(x+3)}, \qquad x\neq0,\ -3 }. \]Practice 7
The LCD is
\[ (x+2)(x-1). \]Combine:
\[ \frac{3(x-1)-(x+2)} {(x+2)(x-1)} = \frac{2x-5} {(x+2)(x-1)}. \]Thus,
\[ \boxed{ \frac{2x-5}{(x+2)(x-1)}, \qquad x\neq-2,\ 1 }. \]Practice 8
The LCD is already
\[ x(x+1). \]Rewrite the second fraction:
\[ \frac{5x}{x(x+1)}. \]Then
\[ \boxed{ \frac{5x+2}{x(x+1)}, \qquad x\neq0,\ -1 }. \]Practice 9
Factor
\[ x^2-4=(x-2)(x+2). \]The LCD is
\[ (x-2)(x+2). \]Rewrite:
\[ \frac{1}{(x-2)(x+2)} + \frac{x-2}{(x+2)(x-2)}. \]Combine:
\[ \frac{1+x-2}{(x-2)(x+2)} = \frac{x-1}{(x-2)(x+2)}. \]Therefore,
\[ \boxed{ \frac{x-1}{(x-2)(x+2)}, \qquad x\neq-2,\ 2 }. \]Practice 10
The student incorrectly added numerators and denominators directly.
The LCD is
\[ x(x+1). \]Rewrite:
\[ \frac{x^2}{x(x+1)} + \frac{x+1}{x(x+1)}. \]Combine:
\[ \frac{x^2+x+1}{x(x+1)}. \]Therefore,
\[ \boxed{ \frac{x^2+x+1}{x(x+1)}, \qquad x\neq0,\ -1 }. \]What to Remember
Rational expressions use the ordinary laws of fraction arithmetic, but algebraic factoring and domain restrictions determine how those laws are applied.
For multiplication:
\[ \boxed{ \frac{a}{b}\cdot\frac{c}{d} = \frac{ac}{bd} }. \]For division:
\[ \boxed{ \frac{a}{b}\div\frac{c}{d} = \frac{a}{b}\cdot\frac{d}{c} }. \]For addition and subtraction:
\[ \boxed{ \frac{a}{b}\pm\frac{c}{d} = \frac{ad\pm bc}{bd} }. \]The essential habits are:
- record restrictions before operating;
- factor before multiplying, dividing, or finding an LCD;
- cancel only common factors;
- multiply by the reciprocal when dividing;
- use a common denominator before adding or subtracting;
- protect subtraction with parentheses;
- simplify the final result;
- preserve every restriction inherited from the original expressions.
Where We Go Next
Ordinary rational-expression operations can be nested inside one another. A fraction may have rational expressions in its numerator, denominator, or both, creating what is called a complex fraction.
The next lesson, Complex Fractions, develops several methods for simplifying these layered expressions, including division by reciprocals and clearing internal denominators with a least common denominator.