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IntermediateGrade 825 min15 / 82

Algebraic fractions (rational expressions)

Allowed values of the variable, the basic property of a fraction and cancelling by factoring, the common denominator, the four operations with algebraic fractions, long simplification chains, and fractional equations with the check for excluded values.

Check yourself
In this lesson you will learn
  • Find the values for which an algebraic fraction makes no sense, and tell when a fraction equals zero
  • Simplify fractions by factoring the numerator and the denominator
  • Find the common denominator and carry out the four operations with algebraic fractions, including long chains
  • Solve fractional equations and reject extraneous roots

Leyla rides her bicycle 12 km to her grandmother’s house. If her speed is v km/h, the trip takes 12/v hours. How much time does she save if she rides 2 km/h faster? The answer is 12/v − 12/(v + 2) hours. There is a variable in the denominator: this is an algebraic fraction. By the end of the lesson you will be able to bring it to the form 24/(v(v + 2)): for v = 4, for example, she saves exactly 1 hour.

Working with algebraic fractions is very much like working with ordinary fractions: the same basic property, the same common denominator, the same rules for multiplying and dividing. There are only two new things. First, a denominator can never be zero, so some values of the variable are forbidden. Second, to cancel and to find a common denominator we factor polynomials instead of finding a GCD — as we learned in the lesson “Factoring polynomials”.

Algebraic fractions and allowed values

Definition
Algebraic fraction

A fraction P/Q whose numerator and denominator are polynomials, with a variable in the denominator. For example, (x + 3)/(x − 5), 7/(a² − 9), 2x/(x² + 4). Expressions built with addition, subtraction, multiplication and division (including division by an expression with a variable) are called rational expressions.

Definition
Allowed values of the variable

The values for which the expression makes sense, that is, for which no denominator becomes zero. For the other values the expression is meaningless: (x + 3)/(x − 5), for example, turns into 8/0 at x = 5, and you cannot divide by zero.

P/Q is defined ⇔ Q ≠ 0; P/Q = 0 ⇔ P = 0 and Q ≠ 0P/Q is defined ⇔ Q ≠ 0; P/Q = 0 ⇔ P = 0 and Q ≠ 0
where:
  • Pthe numerator (a polynomial)
  • Qthe denominator (a polynomial)

A fraction equals zero only when its numerator is zero — and at that value the denominator must not be zero.

  1. 1
    List the denominators

    Write down every denominator in the expression, and in a division also the numerator of the divisor: in A ÷ (B/C) neither C nor B may be zero.

  2. 2
    Set them equal to zero

    Factor each one and solve it with the rule a · b = 0.

  3. 3
    Exclude

    Write the values found as x ≠ …; all other numbers are allowed.

Allowed values

For which values of the variable does the expression make sense?
1) (x + 3)/(x − 5)
2) 7/(x² − 9)
3) 2x/(x² + 4)
4) (a − 1)/(a² − 5a)

Show solution
1) x − 5 ≠ 0 ⇒ x ≠ 5.
2) x² − 9 = (x − 3)(x + 3) ≠ 0 ⇒ x ≠ 3, x ≠ −3.
3) Since x² ≥ 0, x² + 4 ≥ 4 > 0 — the denominator is never zero ⇒ x can be any number.
4) a² − 5a = a(a − 5) ⇒ a ≠ 0, a ≠ 5. A zero numerator is not forbidden: at a = 1 the fraction is simply equal to 0.
When is a fraction zero?

For which value of x is the fraction equal to zero?
1) (x − 2)/(x + 3)
2) (x² − 4)/(x − 2)
3) (x² − 3x)/(x − 3)

Show solution
1) x − 2 = 0 ⇒ x = 2; then the denominator is 5 ≠ 0 ⇒ x = 2.
2) x² − 4 = 0 ⇒ x = 2 or x = −2. But at x = 2 the denominator is zero, so this value is forbidden ⇒ x = −2.
3) x(x − 3) = 0 ⇒ x = 0 or x = 3; x = 3 makes the denominator zero ⇒ x = 0.

The basic property and cancelling

As with ordinary fractions, the numerator and the denominator of an algebraic fraction can be multiplied or divided by the same non-zero expression without changing the value of the fraction. Dividing is cancelling (simplifying); multiplying is what we need to bring fractions to a common denominator.

P/Q = (P · M)/(Q · M) (M ≠ 0)P/Q = (P · M)/(Q · M) (M ≠ 0)
where:
  • P, Qthe numerator and the denominator
  • Ma common factor of the numerator and the denominator (a number, a monomial or a polynomial), not zero

The basic property of a fraction. Read from right to left it is cancelling; from left to right, multiplying by an extra factor.

  1. 1
    Factor

    Factor the numerator and the denominator: common factor, special product formulas, grouping, quadratic trinomial.

  2. 2
    Note the allowed values

    Note the values that make the denominator zero before cancelling: after cancelling they can no longer be seen.

  3. 3
    Cancel the common factors

    Remove only whole factors: the same bracket or monomial that stands in both the numerator and the denominator.

  4. 4
    Check

    Substitute an allowed value (say x = 1): the original and the new fraction must have the same value.

Cancelling fractions

Simplify the fraction:
1) 12a³b/(18ab²)
2) (x² − 9)/(x² + 3x)
3) (a² − 4a + 4)/(a² − 4)
4) (x³ − 8)/(x² − 4)

Show solution
1) (6ab · 2a²)/(6ab · 3b) = 2a²/(3b) (a ≠ 0, b ≠ 0).
2) (x − 3)(x + 3)/(x(x + 3)) = (x − 3)/x (x ≠ 0, x ≠ −3).
3) (a − 2)²/((a − 2)(a + 2)) = (a − 2)/(a + 2) (a ≠ ±2).
4) A difference of cubes on top: (x − 2)(x² + 2x + 4)/((x − 2)(x + 2)) = (x² + 2x + 4)/(x + 2) (x ≠ ±2).
Check (2), x = 1: (1 − 9)/(1 + 3) = −2 and (1 − 3)/1 = −2 ✓

Brackets that are opposites of each other often appear in the numerator and the denominator: b − a = −(a − b). Take the minus sign out and then cancel. The sign of a fraction can be written in any of three places: −P/Q = (−P)/Q = P/(−Q).

Opposite brackets

Simplify:
1) (b − a)/(a² − ab)
2) (6 − 2x)/(x² − 6x + 9)
3) (y² − 25)/(10 − 2y)

Show solution
1) b − a = −(a − b), a² − ab = a(a − b) ⇒ −(a − b)/(a(a − b)) = −1/a.
2) 6 − 2x = −2(x − 3), x² − 6x + 9 = (x − 3)² ⇒ −2(x − 3)/(x − 3)² = −2/(x − 3), that is, 2/(3 − x).
3) (y − 5)(y + 5)/(−2(y − 5)) = −(y + 5)/2.

Common denominator: adding and subtracting

Fractions can be added only when their denominators are equal — just like 1/3 + 1/6. If the denominators differ, we “expand” each fraction with the basic property so that the denominators become equal. With numbers we used the LCM; with polynomials its counterpart is the least common denominator built from the factored denominators.

P/Q ± R/Q = (P ± R)/Q P/Q ± R/S = (P · S ± R · Q)/(Q · S)P/Q ± R/Q = (P ± R)/Q P/Q ± R/S = (P · S ± R · Q)/(Q · S)
where:
  • Q, Sthe denominators
  • P, Rthe numerators

With equal denominators, add (subtract) the numerators. With different ones, first bring the fractions to a common denominator. Q · S always works, but the least common denominator is the most economical — the working gets shorter.

  1. 1
    Factor the denominators

    For example, a² − 9 = (a − 3)(a + 3) and 2a + 6 = 2(a + 3).

  2. 2
    Build the common denominator

    Take every different factor, each with its largest exponent — just like the LCM: 2(a − 3)(a + 3).

  3. 3
    Extra factors

    For each fraction: common denominator ÷ its denominator. Multiply its numerator by this extra factor.

  4. 4
    Combine the numerators

    Write the numerators in brackets (especially after a “−” sign!), expand and collect like terms.

  5. 5
    Cancel

    Factor the new numerator: it may share a factor with the denominator.

Like and unlike denominators

Carry out the operations:
1) (3x + 1)/(x − 2) − (x + 5)/(x − 2)
2) 1/(x − 3) + 1/(x + 3)
3) 5/(2a) − 3/a²
4) x/(x² − 4) − 1/(x − 2)

Show solution
1) The denominators are equal: (3x + 1 − (x + 5))/(x − 2) = (2x − 4)/(x − 2) = 2(x − 2)/(x − 2) = 2 (x ≠ 2).
2) Common denominator (x − 3)(x + 3): (x + 3 + x − 3)/((x − 3)(x + 3)) = 2x/(x² − 9).
3) Common denominator 2a², extra factors a and 2: (5 · a − 3 · 2)/(2a²) = (5a − 6)/(2a²).
4) x² − 4 = (x − 2)(x + 2); the extra factor of the second fraction is (x + 2):
(x − (x + 2))/((x − 2)(x + 2)) = −2/(x² − 4).
Opposite brackets, a whole expression and the opening problem

Simplify:
1) 3/(a² − ab) + 3/(b² − ab)
2) a + 1 − a²/(a − 1)
3) 12/v − 12/(v + 2)

Show solution
1) a² − ab = a(a − b), b² − ab = b(b − a) = −b(a − b). Then:
3/(a(a − b)) − 3/(b(a − b)) = (3b − 3a)/(ab(a − b)) = −3(a − b)/(ab(a − b)) = −3/(ab).
2) a + 1 = (a + 1)/1, extra factor (a − 1): ((a + 1)(a − 1) − a²)/(a − 1) = (a² − 1 − a²)/(a − 1) = −1/(a − 1) = 1/(1 − a).
3) (12(v + 2) − 12v)/(v(v + 2)) = 24/(v(v + 2)). For v = 4: 24/(4 · 6) = 1 hour — the time Leyla saves.

Multiplying, dividing and long chains

P/Q · R/S = (P · R)/(Q · S) P/Q ÷ R/S = P/Q · S/R = (P · S)/(Q · R)P/Q · R/S = (P · R)/(Q · S) P/Q ÷ R/S = P/Q · S/R = (P · S)/(Q · R)
where:
  • R/SR/Sthe second fraction; in a division it is turned upside down: S/R
  • Q, S, Rmust not be zero (R only in a division)

Numerator times numerator, denominator times denominator; dividing means multiplying by the reciprocal. But before multiplying, factor everything and cancel across — there is no need to expand the brackets.

Multiplying and dividing

Simplify:
1) 4x²/(15y³) · 5y²/(8x)
2) (a² − 9)/(2a + 4) · (a² + 2a)/(a − 3)
3) (x² − 1)/(x² + 2x + 1) ÷ (x − 1)/(3x + 3)

Show solution
1) (4 · 5 · x² · y²)/(15 · 8 · x · y³) = 20x²y²/(120xy³) = x/(6y).
2) (a − 3)(a + 3) · a(a + 2)/(2(a + 2) · (a − 3)) = a(a + 3)/2.
3) Replace the division by multiplication by the reciprocal:
(x − 1)(x + 1)/(x + 1)² · 3(x + 1)/(x − 1) = 3.
Allowed values: x ≠ −1 (denominators) and x ≠ 1 (numerator of the divisor); the answer 3 holds only for these x.

In exam tasks several operations are combined in one expression. The order of operations is the same as with numbers: brackets first, then multiplication and division, then addition and subtraction. Cancel after every step to keep the expression small. Often the answer turns out surprisingly simple — sometimes it does not even depend on the variable.

Simplification chains

1) Simplify and evaluate for x = 2.7: (1/(x − 2) − 4/(x² − 4)) · (x² + 2x)/3
2) Prove that the value of the expression does not depend on a: ((a + 2)/(a − 2) − (a − 2)/(a + 2)) ÷ 8a/(a² − 4)

Show solution
1) The bracket, common denominator (x − 2)(x + 2):
(x + 2 − 4)/((x − 2)(x + 2)) = (x − 2)/((x − 2)(x + 2)) = 1/(x + 2).
Multiply: 1/(x + 2) · x(x + 2)/3 = x/3.
x = 2.7: 2.7 ÷ 3 = 0.9 — one division instead of a hard calculation.
2) The bracket: ((a + 2)² − (a − 2)²)/((a − 2)(a + 2)). The numerator is a difference of squares: (a + 2 − a + 2)(a + 2 + a − 2) = 4 · 2a = 8a.
So the bracket equals 8a/(a² − 4), and 8a/(a² − 4) · (a² − 4)/(8a) = 1.
The result does not depend on a (for a ≠ 0, a ≠ ±2).

Fractional equations

Definition
Fractional (rational) equation

An equation with the variable in a denominator, for example 3/(x − 2) = 5/(x + 2). We multiply it by the common denominator to get an equation without fractions. This can produce a “root” that makes a denominator zero — an extraneous root, which is not written in the answer.

Where does an extraneous root come from? When you multiply an equation by an expression with the variable, at a point where that expression is zero both sides become 0 and the new equation is “satisfied” — even though the original equation divides by zero there. That is why the check is a compulsory part of the solution.

  1. 1
    Allowed values

    Factor all denominators and write down the forbidden values.

  2. 2
    Multiply by the common denominator

    Multiply every term of the equation by the common denominator — the denominators disappear. (Another way: move everything to the left, write it as one fraction and use P/Q = 0 ⇔ P = 0, Q ≠ 0.)

  3. 3
    Solve the new equation

    Solve the resulting linear equation; if you get a quadratic one, solve it by factoring.

  4. 4
    Check the roots

    Compare every root with the forbidden values from step 1; a root that makes a denominator zero is extraneous — reject it.

  5. 5
    Write the answer

    Write only the roots that remain. If none remain, the equation has no roots.

Equations and extraneous roots

Solve the equation:
1) 3/(x − 2) = 5/(x + 2)
2) x²/(x − 3) = 9/(x − 3)
3) x/(x + 2) + (x + 2)/(x − 2) = 8/(x² − 4)

Show solution
1) x ≠ ±2. Multiply by the common denominator: 3(x + 2) = 5(x − 2) ⇒ 3x + 6 = 5x − 10 ⇒ 2x = 16 ⇒ x = 8. 8 is not forbidden ⇒ x = 8.
2) x ≠ 3. x² = 9 ⇒ x = 3 or x = −3. x = 3 is extraneous ⇒ x = −3.
3) x² − 4 = (x − 2)(x + 2), x ≠ ±2. Multiply by (x − 2)(x + 2):
x(x − 2) + (x + 2)² = 8 ⇒ x² − 2x + x² + 4x + 4 = 8 ⇒ 2x² + 2x − 4 = 0 ⇒ x² + x − 2 = 0.
(x + 2)(x − 1) = 0 ⇒ x = −2 or x = 1. x = −2 is extraneous ⇒ x = 1.
Check: 1/3 + 3/(−1) = −8/3 and 8/(1 − 4) = −8/3 ✓
Interactive
Loading simulation…
The graphs of y = 3/(x − 2) and y = 5/(x + 2). They meet only at x = 8 (both equal 0.5) — the root of equation 1. At x = 2 and x = −2 the graphs break: the fractions make no sense there.

Fractional equations appear naturally in problems about working together, motion and prices: in “time = work ÷ rate” and “time = distance ÷ speed” the unknown ends up in the denominator.

A work problem

Aysel and Leyla make the class newspaper together in 6 hours. Working alone, Aysel does the job 5 hours faster than Leyla. How many hours does each of them need alone?

Show solution
Let Aysel need x hours; then Leyla needs x + 5 hours. In one hour Aysel does 1/x of the job, Leyla 1/(x + 5), and together 1/6:
1/x + 1/(x + 5) = 1/6, x > 0.
Multiply by 6x(x + 5): 6(x + 5) + 6x = x(x + 5) ⇒ 12x + 30 = x² + 5x ⇒ x² − 7x − 30 = 0.
(x − 10)(x + 3) = 0 ⇒ x = 10 or x = −3. Time cannot be negative ⇒ x = 10.
Aysel: 10 hours, Leyla: 15 hours. Check: 1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6 ✓
Check yourself: fill in the gap
  1. 1.7/(x − 4) makes no sense when x =
  2. 2.(x² − 25)/(x + 5) =
  3. 3.(x − 3)/(3 − x) =
  4. 4.1/x + 1/(2x) = /(2x)
  5. 5.(a/b) ÷ (a²/b) = 1/

Key points

  • The denominator of an algebraic fraction can never be zero: find the forbidden values by factoring the denominator. P/Q = 0 ⇔ P = 0 and Q ≠ 0.
  • Cancelling: factor the numerator and the denominator first, then remove only common factors, never terms.
  • For adding and subtracting, build the least common denominator like an LCM; after a “−” sign, put the numerator in brackets.
  • Multiply numerator by numerator and denominator by denominator; dividing means multiplying by the reciprocal. Cancel before you multiply.
  • A fractional equation: allowed values → multiply by the common denominator → solve → reject extraneous roots.

Check yourself

12 questions. Every correct answer earns XP.

1 / 12
For which value of x does 5/(x + 4) make no sense?