Rational Expressions

Study Sheet

Rational Expressions

Excluded values, simplifying, multiplying, dividing, adding, subtracting, complex fractions, equations, and work problems

What Is a Rational Expression?

Concept
Definition

A rational expression is a ratio of two polynomials—one polynomial on top (the numerator) divided by another polynomial on the bottom (the denominator):

polynomialpolynomial,such as  2x+1x−3,x2−9x2+5x+6,5x.\frac{\text{polynomial}}{\text{polynomial}},\qquad\text{such as }\ \frac{2x+1}{x-3},\quad \frac{x^2-9}{x^2+5x+6},\quad \frac{5}{x}.

It is just like a numerical fraction (such as 34\tfrac{3}{4}), except the top and bottom can contain variables.

A rational expression is a fraction of polynomials, and it follows fraction rules exactly: factor, cancel common factors, use a common denominator to add. The new ingredient is the domain — values that make a denominator zero are excluded, even if they cancel out on paper.

Rational expressions relate to polynomials exactly as fractions relate to integers: a polynomial over a polynomial. That is why every fraction skill transfers: simplify by cancelling common factors, multiply straight across, divide by flipping, add over a common denominator. The one new danger is that the denominator can be zero for some values of the variable.

Example
Example: Which are rational expressions?
  • 2x+1x−3\dfrac{2x+1}{x-3} Yes — a polynomial over a polynomial.
  • 5x2\dfrac{5}{x^2} Yes — the constant 55 and x2x^2 are both polynomials.
  • x+1\sqrt{x}+1 No — x\sqrt{x} is not a polynomial (a variable under a root).
Side note
Where it goes wrong

Cancelling terms instead of factors: x2−9x−3\tfrac{x^2 - 9}{x - 3} is x+3x + 3 only after factoring the top as (x−3)(x+3)(x - 3)(x + 3), and x+9x+3\tfrac{x + 9}{x + 3} does not simplify at all.

Tip

Remember: A constant like 55 is a polynomial, so 5x\dfrac{5}{x} counts. But roots of a variable, or a variable in an exponent, are not allowed in a polynomial.

Try it
Try it: rational or not?

Which are rational expressions: 2x+1x−3\dfrac{2x + 1}{x - 3}, x+1\sqrt{x} + 1, 7x\dfrac{7}{x}, x22x\dfrac{x^2}{2^x}?

Show the answer

2x+1x−3\tfrac{2x + 1}{x - 3}: yes, polynomial over polynomial. x+1\sqrt x + 1: no, x\sqrt x is not a polynomial. 7x\tfrac7x: yes, 77 is a (constant) polynomial. x22x\tfrac{x^2}{2^x}: no, the variable in an exponent makes 2x2^x not a polynomial.

Excluded Values (Domain Restrictions)

Concept
Why the Denominator Matters

Division by zero is undefined. So a rational expression is undefined for any value that makes the denominator equal to 00. Those values are called excluded values.

Method: Set the denominator equal to 00, solve, and exclude those values.

Division by zero has no answer: 50\tfrac{5}{0} would have to be a number that, times 00, gives 55, and no number does. So an expression with x2−x−6x^2 - x - 6 in the denominator simply has no value at x=3x = 3 or x=−2x = -2. The graph shows it: near those inputs the values blow up, and at them the graph is missing.

Example
Example: Find the excluded values of x+4x2−x−6\dfrac{x+4}{x^2-x-6}
xy−6−5−4−3−2−1123456−6−5−4−3−2−112345x = 3x = −2
The graph breaks at the excluded values x=−2x = -2 and x=3x = 3.

Set the denominator to 00 and factor:

x2−x−6=0(x−3)(x+2)=0x=3orx=−2.\begin{aligned} x^2 - x - 6 &= 0 \\ (x-3)(x+2) &= 0 \\ x = 3 \quad&\text{or}\quad x = -2. \end{aligned}

So the excluded values are x≠3x \neq 3 and x≠−2x \neq -2.

Side note
Where it goes wrong

Setting the numerator equal to zero: "x+4=0x + 4 = 0, so −4-4 is excluded." At x=−4x = -4 the expression is 014=0\tfrac{0}{14} = 0, a perfectly good value. Only zeros of the denominator are excluded.

Tip

Tip: Look at the original denominator to find excluded values, before you cancel anything. Canceling can hide a restriction, but the restriction is still there.

Try it
Try it: excluded values

Find the excluded values of x−1x2−16\dfrac{x - 1}{x^2 - 16} and 3x2+4\dfrac{3}{x^2 + 4}.

Show the answer

x2−16=(x−4)(x+4)=0x^2 - 16 = (x - 4)(x + 4) = 0 at x=±4x = \pm 4, so exclude 44 and −4-4. x2+4x^2 + 4 is at least 44 and never zero, so 3x2+4\tfrac{3}{x^2 + 4} has no excluded values.

Simplifying Rational Expressions

Concept
Factor, Then Cancel Common Factors

To simplify a rational expression:

  • Factor the numerator and denominator completely.
  • Cancel any factor that appears in both the top and the bottom.
  • State the excluded values (from the original denominator).

Cancelling works because (x−3)(x+3)(x+2)(x+3)\tfrac{(x - 3)(x + 3)}{(x + 2)(x + 3)} is x−3x+2\tfrac{x - 3}{x + 2} multiplied by x+3x+3\tfrac{x + 3}{x + 3}, and that factor equals 11, except at x=−3x = -3, where it is 00\tfrac00. So the simplified form agrees with the original everywhere except x=−3x = -3. That is why the restriction must be written down: the graph of the original has a hole there.

Example
Example: Simplify x2−9x2+5x+6\dfrac{x^2-9}{x^2+5x+6}
xy−6−5−4−3−2−1123−4−22468hole at (−3, 6)
The graph of x−3x+2\tfrac{x - 3}{x + 2} with the point at x=−3x = -3 removed.

Factor top and bottom:

x2−9x2+5x+6=(x−3)(x+3)(x+2)(x+3).\frac{x^2-9}{x^2+5x+6} = \frac{(x-3)(x+3)}{(x+2)(x+3)}.

Cancel the common factor (x+3)(x+3):

=x−3x+2,x≠−3, x≠−2.= \frac{x-3}{x+2},\qquad x\neq -3,\ x\neq -2.
Side note
Where it goes wrong

Cancelling across a sum: x+6x+2=62=3\tfrac{x + 6}{x + 2} = \tfrac{6}{2} = 3. Test x=2x = 2: 84=2\tfrac84 = 2, not 33. Only factors of the whole numerator and whole denominator cancel; x+6x+2\tfrac{x + 6}{x + 2} is already simplified.

Example
Example: Simplify x2−xx\dfrac{x^2-x}{x}

Factor the top: x2−x=x(x−1)x^2 - x = x(x-1). Then cancel the xx:

x(x−1)x=x−1,x≠0.\frac{x(x-1)}{x} = x-1,\qquad x\neq 0.
Tip

You may only cancel FACTORS, never terms!   In x+2x\dfrac{x+2}{x} you cannot cancel the xx's, because x+2x+2 is a sum, not a product. Only cancel things that are multiplied. For instance x+3x+2\dfrac{x+3}{x+2} is already simplified—nothing cancels.

Try it
Try it: simplify

Simplify 2x2−8x2+x−6\dfrac{2x^2 - 8}{x^2 + x - 6} and state the excluded values.

Show the answer

Factor: 2(x−2)(x+2)(x+3)(x−2)\tfrac{2(x - 2)(x + 2)}{(x + 3)(x - 2)}. Cancel (x−2)(x - 2): 2(x+2)x+3\tfrac{2(x + 2)}{x + 3}, with x≠−3x \ne -3 and x≠2x \ne 2 (the second from the original denominator).

Multiplying Rational Expressions

Concept
Factor, Cancel, Multiply

To multiply, factor everything, cancel any factor common to a top and a bottom, then multiply the remaining tops together and the remaining bottoms together. (You do not need a common denominator to multiply.)

Multiplying fractions multiplies tops and bottoms, so every factor ends up in one big fraction; cancelling beforehand is just cancelling in that big fraction early. Factoring first is what makes the common factors visible.

Example
Example: x2−4x2+6x+9⋅x+3x−2\dfrac{x^2-4}{x^2+6x+9}\cdot\dfrac{x+3}{x-2}

Factor each part:

(x−2)(x+2)(x+3)(x+3)⋅x+3x−2.\frac{(x-2)(x+2)}{(x+3)(x+3)}\cdot\frac{x+3}{x-2}.

Cancel one (x−2)(x-2) and one (x+3)(x+3):

=x+2x+3,x≠−3, x≠2.= \frac{x+2}{x+3},\qquad x\neq -3,\ x\neq 2.
Side note
Where it goes wrong

Multiplying out before cancelling and then being unable to factor: x2−4x2+6x+9⋅x+3x−2\tfrac{x^2 - 4}{x^2 + 6x + 9} \cdot \tfrac{x + 3}{x - 2} expanded to x3+3x2−4x−12x3+4x2−3x−18\tfrac{x^3 + 3x^2 - 4x - 12}{x^3 + 4x^2 - 3x - 18} hides the common factors. Factor first, cancel, then multiply.

Tip

Tip: Cancel before you multiply out—it keeps the numbers small. Collect the excluded values from every original denominator.

Try it
Try it: multiply

Multiply x2−253x⋅6x2x+5\dfrac{x^2 - 25}{3x} \cdot \dfrac{6x^2}{x + 5}.

Show the answer

Factor: (x−5)(x+5)3x⋅6x2x+5\tfrac{(x - 5)(x + 5)}{3x} \cdot \tfrac{6x^2}{x + 5}. Cancel (x+5)(x + 5) and 3x3x (leaving 2x2x from 6x26x^2): 2x(x−5)2x(x - 5), with x≠0x \ne 0, x≠−5x \ne -5.

Dividing Rational Expressions

Concept
Multiply by the Reciprocal

To divide, flip the second fraction (take its reciprocal) and multiply. Then factor and cancel as usual.

AB÷CD=AB⋅DC.\frac{A}{B}\div\frac{C}{D} = \frac{A}{B}\cdot\frac{D}{C}.

Dividing by CD\tfrac{C}{D} means multiplying by whatever undoes CD\tfrac CD, and that is DC\tfrac DC, because CD⋅DC=1\tfrac CD \cdot \tfrac DC = 1. The flip also moves CC into a denominator, which is why its zeros join the excluded values.

Example
Example: x2−1x2−4÷x+1x−2\dfrac{x^2-1}{x^2-4}\div\dfrac{x+1}{x-2}

Flip the divisor and multiply:

x2−1x2−4⋅x−2x+1=(x−1)(x+1)(x−2)(x+2)⋅x−2x+1.\frac{x^2-1}{x^2-4}\cdot\frac{x-2}{x+1} = \frac{(x-1)(x+1)}{(x-2)(x+2)}\cdot\frac{x-2}{x+1}.

Cancel (x+1)(x+1) and (x−2)(x-2):

=x−1x+2,x≠2, x≠−2, x≠−1.= \frac{x-1}{x+2},\qquad x\neq 2,\ x\neq -2,\ x\neq -1.
Side note
Where it goes wrong

Flipping the first fraction instead of the second, or forgetting the restriction from the divisor's numerator. In AB÷x+1x−2\tfrac{A}{B} \div \tfrac{x + 1}{x - 2}, x=−1x = -1 is excluded: at x=−1x = -1 you would be dividing by 0−3=0\tfrac{0}{-3} = 0.

Tip

Careful with restrictions when dividing. The factor you flip (x+1x+1 here) was a denominator after flipping, so x≠−1x\neq -1 too. Track excluded values from both the original bottoms and the divisor's top.

Try it
Try it: divide

Divide x2−9x+4÷x−32x+8\dfrac{x^2 - 9}{x + 4} \div \dfrac{x - 3}{2x + 8}.

Show the answer

Flip and factor: (x−3)(x+3)x+4⋅2(x+4)x−3=2(x+3)\tfrac{(x - 3)(x + 3)}{x + 4} \cdot \tfrac{2(x + 4)}{x - 3} = 2(x + 3), with x≠−4x \ne -4 and x≠3x \ne 3.

Adding & Subtracting with LIKE Denominators

Concept
Same Bottom? Combine the Tops

When the denominators are already the same, add or subtract the numerators and keep the common denominator. Then simplify if possible.

AD±BD=A±BD.\frac{A}{D}\pm\frac{B}{D} = \frac{A\pm B}{D}.

With a common denominator, the fractions count the same kind of piece, so adding them adds the counts, exactly as 38+18=48\tfrac38 + \tfrac18 = \tfrac48. The subtraction warning comes from the distributive property: subtracting a whole numerator subtracts every term of it.

Example
Example: x2x−2−4x−2\dfrac{x^2}{x-2}-\dfrac{4}{x-2}

Subtract the numerators over the common denominator, then factor and cancel:

x2−4x−2=(x−2)(x+2)x−2=x+2,x≠2.\frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{x-2} = x+2,\qquad x\neq 2.
Side note
Where it goes wrong

Subtracting only the first term of the second numerator: 5xx+1−2x−3x+1=5x−2x−3x+1\tfrac{5x}{x + 1} - \tfrac{2x - 3}{x + 1} = \tfrac{5x - 2x - 3}{x + 1}. The whole numerator is subtracted: 5x−2x+3x+1=3x+3x+1=3\tfrac{5x - 2x + 3}{x + 1} = \tfrac{3x + 3}{x + 1} = 3 (for x≠−1x \ne -1).

Tip

Subtracting? Subtract the whole numerator—use parentheses. AD−BD=A−BD\dfrac{A}{D}-\dfrac{B}{D}=\dfrac{A-B}{D}, and the minus sign hits every term of BB.

Try it
Try it: combine

Simplify x2+5xx+2+6x+2\dfrac{x^2 + 5x}{x + 2} + \dfrac{6}{x + 2}.

Show the answer

Add the numerators over the common denominator: x2+5x+6x+2=(x+2)(x+3)x+2=x+3\tfrac{x^2 + 5x + 6}{x + 2} = \tfrac{(x + 2)(x + 3)}{x + 2} = x + 3, with x≠−2x \ne -2.

Adding & Subtracting with UNLIKE Denominators

Concept
Find the LCD

When denominators differ, build a least common denominator (LCD):

  • Factor each denominator.
  • The LCD contains each different factor the greatest number of times it appears in any one denominator.
  • Rewrite each fraction over the LCD, then combine the numerators.

Each fraction is multiplied by a form of 11, such as x+1x+1\tfrac{x + 1}{x + 1}, to rewrite it over the LCD; that changes its appearance, not its value. The least common denominator is used because any larger common denominator adds factors that must be cancelled later.

Example
Example: 2x+3x+1\dfrac{2}{x}+\dfrac{3}{x+1}

The denominators xx and x+1x+1 share no factors, so the LCD is x(x+1)x(x+1). Rewrite each fraction:

2x⋅x+1x+1+3x+1⋅xx=2(x+1)x(x+1)+3xx(x+1)=2x+2+3xx(x+1)=5x+2x(x+1),x≠0, x≠−1.\begin{aligned} \frac{2}{x}\cdot\frac{x+1}{x+1} + \frac{3}{x+1}\cdot\frac{x}{x} &= \frac{2(x+1)}{x(x+1)} + \frac{3x}{x(x+1)} \\ &= \frac{2x+2+3x}{x(x+1)} = \frac{5x+2}{x(x+1)},\qquad x\neq 0,\ x\neq -1. \end{aligned}
Side note
Where it goes wrong

Adding tops and bottoms: 2x+3x+1=52x+1\tfrac2x + \tfrac{3}{x + 1} = \tfrac{5}{2x + 1}. Test x=1x = 1: the left side is 2+1.5=3.52 + 1.5 = 3.5, the right side 53\tfrac53. Rewrite over x(x+1)x(x + 1): 5x+2x(x+1)\tfrac{5x + 2}{x(x + 1)}, which at x=1x = 1 is 72=3.5\tfrac72 = 3.5.

Example
Example with factoring: 3x+2+xx2−4\dfrac{3}{x+2}+\dfrac{x}{x^2-4}

Factor x2−4=(x−2)(x+2)x^2-4=(x-2)(x+2), so the LCD is (x−2)(x+2)(x-2)(x+2):

3(x−2)(x−2)(x+2)+x(x−2)(x+2)=3x−6+x(x−2)(x+2)=4x−6(x−2)(x+2)=2(2x−3)(x−2)(x+2),x≠2, x≠−2.\begin{aligned} \frac{3(x-2)}{(x-2)(x+2)} + \frac{x}{(x-2)(x+2)} &= \frac{3x-6+x}{(x-2)(x+2)} \\ &= \frac{4x-6}{(x-2)(x+2)} = \frac{2(2x-3)}{(x-2)(x+2)},\qquad x\neq 2,\ x\neq -2. \end{aligned}
Tip

Do not just multiply the denominators blindly. Factor first so the LCD is as small as possible, and only multiply each numerator by the factors its denominator is missing.

Try it
Try it: LCD

Simplify xx−2−8x2−4\dfrac{x}{x - 2} - \dfrac{8}{x^2 - 4}.

Show the answer

LCD (x−2)(x+2)(x - 2)(x + 2): x(x+2)−8(x−2)(x+2)=x2+2x−8(x−2)(x+2)=(x+4)(x−2)(x−2)(x+2)=x+4x+2\tfrac{x(x + 2) - 8}{(x - 2)(x + 2)} = \tfrac{x^2 + 2x - 8}{(x - 2)(x + 2)} = \tfrac{(x + 4)(x - 2)}{(x - 2)(x + 2)} = \tfrac{x + 4}{x + 2}, x≠±2x \ne \pm 2.

Complex Fractions (Intro)

Concept
A Fraction Within a Fraction

A complex fraction has fractions in its numerator, its denominator, or both. Two ways to simplify:

  • Multiply through by the LCD of every little fraction, or
  • Combine the top into one fraction and the bottom into one fraction, then divide (multiply by the reciprocal).

A complex fraction is one division written on top of others. Multiplying its top and bottom by the LCD of all the little fractions multiplies by a form of 11, so the value is unchanged, and every small denominator cancels at once.

Example
Example: 1+1x1−1x\dfrac{1+\frac{1}{x}}{1-\frac{1}{x}}

The only small denominator is xx, so multiply the top and bottom by xx:

(1+1x)⋅x(1−1x)⋅x=x+1x−1,x≠0, x≠1.\frac{\left(1+\frac{1}{x}\right)\cdot x}{\left(1-\frac{1}{x}\right)\cdot x} = \frac{x+1}{x-1},\qquad x\neq 0,\ x\neq 1.
Side note
Where it goes wrong

Multiplying only the fraction pieces by the LCD: 1+1x1−1x\tfrac{1 + \frac1x}{1 - \frac1x} times xx\tfrac xx written as 1+11−1\tfrac{1 + 1}{1 - 1}. Every term of the top and bottom must be multiplied, the 11s included: x+1x−1\tfrac{x + 1}{x - 1}.

Tip

Tip: Multiplying every part by the LCD clears all the little fractions in one clean step. Watch the excluded values: x≠0x\neq 0 (the inner denominator) and x≠1x\neq 1 (the new bottom).

Try it
Try it: simplify

Simplify 2x−124−xx\dfrac{\frac{2}{x} - \frac{1}{2}}{\frac{4 - x}{x}}.

Show the answer

Multiply top and bottom by 2x2x: top 4−x4 - x, bottom 2(4−x)2(4 - x). So the value is 4−x2(4−x)=12\tfrac{4 - x}{2(4 - x)} = \tfrac12, for x≠0x \ne 0, x≠4x \ne 4.

Solving Rational Equations

Concept
Clear the Denominators

To solve an equation with rational expressions:

  • Note the excluded values (denominators ≠0\neq 0).
  • Multiply every term by the LCD to clear all fractions.
  • Solve the resulting equation.
  • Check each answer—reject any that is an excluded value (an extraneous solution).

Multiplying both sides by the LCD removes the fractions, but it is not reversible at the excluded values: multiplying by an expression that equals 00 can turn a false statement into a true one. That is how an extraneous "solution" like x=4x = 4 appears, and why every answer must be checked against the excluded values.

Example
Example (real solution): 3x+1=2x−1\dfrac{3}{x+1}=\dfrac{2}{x-1}
xy−3−2−112345678−3−2−1123(5, 1/2)
The two sides as graphs (solid and dashed) meet once, at x=5x = 5.

Cross-multiply (or multiply by the LCD (x+1)(x−1)(x+1)(x-1)):

3(x−1)=2(x+1)3x−3=2x+2x=5.\begin{aligned} 3(x-1) &= 2(x+1) \\ 3x-3 &= 2x+2 \\ x &= 5. \end{aligned}

Since x=5x=5 is not excluded (x≠−1,1x\neq -1,1), the solution is x=5x=5.

Side note
Where it goes wrong

Keeping an answer that makes a denominator zero. In xx−4=4x−4+2\tfrac{x}{x - 4} = \tfrac{4}{x - 4} + 2, the algebra yields x=4x = 4, but both fractions are undefined there. There is no solution.

Example
Example (extraneous!): xx−4=4x−4+2\dfrac{x}{x-4}=\dfrac{4}{x-4}+2

Excluded value: x≠4x\neq 4. Multiply every term by (x−4)(x-4):

x=4+2(x−4)x=4+2x−8x=2x−4−x=−4 ⇒ x=4.\begin{aligned} x &= 4 + 2(x-4) \\ x &= 4 + 2x - 8 \\ x &= 2x - 4 \\ -x &= -4 \ \Rightarrow\ x = 4. \end{aligned}

But x=4x=4 is excluded, so it is extraneous. There is no solution.

Tip

Always check for extraneous solutions! An answer that makes any original denominator 00 must be thrown out—even though it came from correct algebra.

Try it
Try it: solve and check

Solve 1x+1x+2=43x\dfrac{1}{x} + \dfrac{1}{x + 2} = \dfrac{4}{3x}.

Show the answer

Exclude 00 and −2-2. Multiply by 3x(x+2)3x(x + 2): 3(x+2)+3x=4(x+2)3(x + 2) + 3x = 4(x + 2), so 6x+6=4x+86x + 6 = 4x + 8 and x=1x = 1. Check: 1+13=431 + \tfrac13 = \tfrac43.

Applications: Work & Rate Problems

Concept
The Work Formula

If one worker (or pipe) finishes a job in aa units of time and another in bb units, each does a fraction of the job per unit of time: 1a\frac{1}{a} and 1b\frac{1}{b}. Working together for tt units:

1a+1b=1t.\frac{1}{a} + \frac{1}{b} = \frac{1}{t}.

(Their rates add.)

Times do not add, but rates do: in one hour pipe A fills 16\tfrac16 of the tank and pipe B fills 13=26\tfrac13 = \tfrac26, so together they fill 36=12\tfrac36 = \tfrac12 each hour and finish in 22 hours. Adding the times (6+3=96 + 3 = 9) would make two pipes slower than one, which is the clearest sign it is wrong.

Example
Example: Two pipes filling a tank
hour 1ABBhour 2ABBA: 1/6 per hour, B: 2/6, together 3/6: full in 2 hours
Each hour: one sixth from A, two sixths from B.

Pipe A fills a tank in 66 hours; pipe B fills it in 33 hours. Together, how long?

16+13=1t.\frac{1}{6} + \frac{1}{3} = \frac{1}{t}.

The LCD is 66: 16+26=36=12\dfrac{1}{6}+\dfrac{2}{6}=\dfrac{3}{6}=\dfrac{1}{2}. So 1t=12\dfrac{1}{t}=\dfrac{1}{2}, giving t=2t=2 hours.

Side note
Where it goes wrong

Averaging the times: two pipes taking 66 and 33 hours do not take 4.54.5 hours together. Together must be faster than the faster pipe alone; adding rates gives 22 hours.

Example
Example: Solving for an unknown rate

Maria paints a room in 33 hours. With Sam helping, they finish in 22 hours. How long would Sam take alone?

13+1s=12 ⇒ 1s=12−13=3−26=16.\frac{1}{3} + \frac{1}{s} = \frac{1}{2} \ \Rightarrow\ \frac{1}{s} = \frac{1}{2}-\frac{1}{3} = \frac{3-2}{6} = \frac{1}{6}.

So s=6s=6 hours for Sam alone.

Tip

Sanity check: Working together must be faster than either person alone. If your “together” time is bigger than one of the individual times, recheck the setup.

Try it
Try it: an unknown rate

A hose fills a pool in 1010 hours; with a second hose running too, it takes 44 hours. How long would the second hose take alone?

Show the answer

Rates: 110+1s=14\tfrac{1}{10} + \tfrac1s = \tfrac14, so 1s=14−110=5−220=320\tfrac1s = \tfrac14 - \tfrac{1}{10} = \tfrac{5 - 2}{20} = \tfrac{3}{20}, and s=203=623s = \tfrac{20}{3} = 6\tfrac23 hours.

Going Deeper: Advanced Rational Expressions

Concept
Continued & Nested Complex Fractions

A continued fraction stacks fractions inside fractions many layers deep. Simplify it from the bottom up: fully combine the innermost fraction, then work outward one level at a time. The same two tools from before still work—multiply by an LCD, or combine-then-flip—but here you apply them repeatedly.

2+12+12  =  2+1 52   =  2+25  =  125.2 + \dfrac{1}{2 + \dfrac{1}{2}} \;=\; 2 + \dfrac{1}{\ \frac{5}{2}\ } \;=\; 2 + \dfrac{2}{5} \;=\; \dfrac{12}{5}.
Example
Worked Example: Simplify x 1+1 1+1x  \dfrac{x}{\,1+\dfrac{1}{\,1+\dfrac{1}{x}\,}\,}

Start at the deepest level and climb out. First the inner denominator:

1+1x=x+1x.1+\frac{1}{x}=\frac{x+1}{x}.

Substitute and flip that little fraction:

1+1 x+1x =1+xx+1=(x+1)+xx+1=2x+1x+1.1+\dfrac{1}{\ \frac{x+1}{x}\ }=1+\frac{x}{x+1}=\frac{(x+1)+x}{x+1}=\frac{2x+1}{x+1}.

Now the whole expression is x 2x+1x+1 \dfrac{x}{\ \frac{2x+1}{x+1}\ }, so flip once more:

x1⋅x+12x+1=x(x+1)2x+1,x≠0, x≠−1, x≠−12.\frac{x}{1}\cdot\frac{x+1}{2x+1}=\frac{x(x+1)}{2x+1},\qquad x\neq 0,\ x\neq -1,\ x\neq -\tfrac12.

Each inner denominator contributes an excluded value—track them all as you unwind.

Concept
The x+1xx+\dfrac{1}{x} Family

A classic trick: if you know the value of x+1xx+\dfrac{1}{x}, you can find x2+1x2x^2+\dfrac{1}{x^2} and x3+1x3x^3+\dfrac{1}{x^3} without ever solving for xx. The key is that squaring or cubing x+1xx+\frac1x produces a “+2+2” or “+3(⋯ )+3(\cdots)” cross-term that you subtract back off.

(x+1x)2=x2+2+1x2⇒x2+1x2=(x+1x)2−2,(x+1x)3=x3+3(x+1x)+1x3⇒x3+1x3=(x+1x)3−3(x+1x).\begin{aligned} \left(x+\tfrac{1}{x}\right)^2 &= x^2 + 2 + \tfrac{1}{x^2} &&\Rightarrow\quad x^2+\tfrac{1}{x^2}=\left(x+\tfrac{1}{x}\right)^2-2,\\[2pt] \left(x+\tfrac{1}{x}\right)^3 &= x^3 + 3\left(x+\tfrac{1}{x}\right) + \tfrac{1}{x^3} &&\Rightarrow\quad x^3+\tfrac{1}{x^3}=\left(x+\tfrac{1}{x}\right)^3-3\left(x+\tfrac{1}{x}\right). \end{aligned}
Example
Worked Example: If x+1x=5x+\dfrac{1}{x}=5, find x2+1x2x^2+\dfrac{1}{x^2} and x3+1x3x^3+\dfrac{1}{x^3}

Let a=x+1x=5a=x+\dfrac1x=5. Square first:

x2+1x2=a2−2=52−2=25−2=23.x^2+\frac{1}{x^2}=a^2-2=5^2-2=25-2=23.

Now cube, using the formula above:

x3+1x3=a3−3a=53−3(5)=125−15=110.x^3+\frac{1}{x^3}=a^3-3a=5^3-3(5)=125-15=110.

Check the pattern another way: x3+1x3=(x+1x) ⁣(x2−1+1x2)=5 (23−1)=5⋅22=110.x^3+\frac{1}{x^3}=\left(x+\frac1x\right)\!\left(x^2-1+\frac{1}{x^2}\right)=5\,(23-1)=5\cdot 22=110. ✓

Tip

Why it works: multiplying out (x+1x)2\left(x+\frac1x\right)^2 or (x+1x)3\left(x+\frac1x\right)^3, the “x⋅1x=1x\cdot\frac1x=1” cross-terms are constants. Rearranging isolates the “symmetric” power sum you want. This is the same idea as (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2 with b=1ab=\frac1a.

Concept
Telescoping Sums via Partial Fractions

Some long sums of rational terms collapse. Split each term into a difference of simpler fractions; then consecutive pieces cancel and only the first and last survive. The workhorse identity is

1n(n+1)=1n−1n+1.\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}.

Adding from n=1n=1 to n=Nn=N, every −1n+1-\frac{1}{n+1} cancels the next term's +1n+1+\frac{1}{n+1}:

∑n=1N1n(n+1)=(1−12)+(12−13)+⋯+(1N−1N+1)=1−1N+1=NN+1.\sum_{n=1}^{N}\frac{1}{n(n+1)}=\left(1-\tfrac12\right)+\left(\tfrac12-\tfrac13\right)+\cdots+\left(\tfrac1N-\tfrac{1}{N+1}\right)=1-\frac{1}{N+1}=\frac{N}{N+1}.
Example
Worked Example: Evaluate 11⋅2+12⋅3+13⋅4+⋯+19⋅10\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\cdots+\dfrac{1}{9\cdot10}

Each term is 1n(n+1)=1n−1n+1\dfrac{1}{n(n+1)}=\dfrac1n-\dfrac{1}{n+1}. Write them out and watch the middle vanish:

(1−12)+(12−13)+(13−14)+⋯+(19−110).\left(1-\tfrac12\right)+\left(\tfrac12-\tfrac13\right)+\left(\tfrac13-\tfrac14\right)+\cdots+\left(\tfrac19-\tfrac{1}{10}\right).

Only the very first 11 and the very last −110-\tfrac{1}{10} are left:

=1−110=910.=1-\frac{1}{10}=\frac{9}{10}.

(Matches the formula NN+1\frac{N}{N+1} with N=9N=9.)

Tip

Recognizing a telescoping split: to break 1n(n+1)\dfrac{1}{n(n+1)} into An+Bn+1\dfrac{A}{n}+\dfrac{B}{n+1}, clear denominators: 1=A(n+1)+Bn1=A(n+1)+Bn. Setting n=0n=0 gives A=1A=1; setting n=−1n=-1 gives B=−1B=-1. That “cover-up” method is partial-fraction decomposition, and it is the engine behind every telescoping sum.

Concept
Preview: Asymptotes & Holes of Rational Functions

The graph of y=P(x)Q(x)y=\dfrac{P(x)}{Q(x)} is shaped by where the function blows up or flattens. After factoring and canceling:

  • Hole at x=cx=c: a factor (x−c)(x-c) that cancels from top and bottom—the graph is undefined at cc but has no infinite spike (a single missing point).
  • Vertical asymptote at x=cx=c: a factor (x−c)(x-c) left in the denominator after canceling—the graph shoots to ±∞\pm\infty.
  • Horizontal asymptote: compare degrees. Bottom-heavy ⇒y=0\Rightarrow y=0; equal degrees ⇒y=leading coeff topleading coeff bottom\Rightarrow y=\frac{\text{leading coeff top}}{\text{leading coeff bottom}}; top-heavy ⇒\Rightarrow none (a slant asymptote instead).
Example
Example: Locate the features of y=x2−1x2−x−2y=\dfrac{x^2-1}{x^2-x-2}

Factor top and bottom:

y=(x−1)(x+1)(x−2)(x+1).y=\frac{(x-1)(x+1)}{(x-2)(x+1)}.

The factor (x+1)(x+1) cancels, so there is a hole at x=−1x=-1. What remains is x−1x−2\dfrac{x-1}{x-2}, whose leftover denominator factor (x−2)(x-2) gives a vertical asymptote at x=2x=2. Top and bottom have equal degree (11 and 11) with leading coefficients 11 and 11, so the horizontal asymptote is y=11=1y=\frac{1}{1}=1.

Tip

Cancel first, then classify. A factor that cancels makes a hole; a factor that stays in the denominator makes a vertical asymptote. Same value of xx can never be both—decide it after simplifying.

Concept
Average Speed Is a Harmonic Mean

For equal distances traveled at two speeds, the average speed is not the plain average of the speeds. Since time=distancespeed\text{time}=\frac{\text{distance}}{\text{speed}}, cover distance dd out at speed v1v_1 and back at v2v_2:

v‾=total distancetotal time=2d dv1+dv2 =2 1v1+1v2 =2v1v2v1+v2.\overline{v}=\frac{\text{total distance}}{\text{total time}}=\frac{2d}{\ \frac{d}{v_1}+\frac{d}{v_2}\ }=\frac{2}{\ \frac{1}{v_1}+\frac{1}{v_2}\ }=\frac{2v_1v_2}{v_1+v_2}.

This is the harmonic mean—the same “rates add” idea as the work formula.

Example
Worked Example: Round trip at 3030 and 6060 mph

You drive to work at 3030 mph and return along the same road at 6060 mph. The average speed is not 4545. Use the harmonic mean:

v‾=2v1v2v1+v2=2(30)(60)30+60=360090=40 mph.\overline{v}=\frac{2v_1v_2}{v_1+v_2}=\frac{2(30)(60)}{30+60}=\frac{3600}{90}=40\ \text{mph}.

Sanity check with a real distance: let each leg be 6060 miles. Out takes 6030=2\frac{60}{30}=2 h, back takes 6060=1\frac{60}{60}=1 h, total 120120 miles in 33 h =40=40 mph. ✓ The slower leg dominates because you spend more time at the low speed.

Tip

Which mean? Equal times at two speeds ⇒\Rightarrow ordinary average v1+v22\frac{v_1+v_2}{2}. Equal distances ⇒\Rightarrow harmonic mean 2v1v2v1+v2\frac{2v_1v_2}{v_1+v_2}. The harmonic mean is always the smaller of the two—slow stretches cost extra time.

Why It Works

Cancelling works because acbc=ab⋅cc=ab⋅1\tfrac{ac}{bc} = \tfrac ab\cdot\tfrac cc = \tfrac ab\cdot1 — but only when c≠0c \ne 0, since 00\tfrac00 is undefined. That is why x2−9x−3=x+3\tfrac{x^2-9}{x-3} = x + 3 comes with the condition x≠3x \ne 3: at x=3x = 3 the left side is undefined while the right side is 66. The simplified expression is equal to the original everywhere the original exists, and that "everywhere the original exists" is the domain you must state.

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

Formulas, Proofs & Tips

Tip
Excluded values
P(x)Q(x) is undefined where Q(x)=0\frac{P(x)}{Q(x)} \text{ is undefined where } Q(x)=0

What it means. A rational expression exists everywhere except where its denominator vanishes.

Example. 1x−2\dfrac{1}{x-2} is undefined at x=2x=2, so x≠2x\ne 2.

Why it works. Division asks "what times the denominator gives the numerator?". If the denominator is 00, no number works (or every number does, when the numerator is 00 too), so the value cannot be defined.

Tip. Find the excluded values from the original denominator, before cancelling. Cancelling can hide a restriction that still applies.