Math Vale

⚔️ Algebra Kingdom · Algebra

Rational Expressions & Equations

Treat a fraction of polynomials the way you treat a fraction of numbers: simplify it by factoring, state the values it excludes, multiply, divide, add and subtract, and solve equations that contain one.

In short

  • Factor the top and the bottom first, then cancel — and only whole factors cancel, never terms of a sum.
  • The excluded values are whatever makes a bottom zero, read off the original expression before you simplify; simplifying can hide a restriction but never removes it.
  • Multiply straight across after cancelling; divide by turning the second fraction over; add and subtract only over a common denominator, and rebuild each top as you go.
  • To solve, multiply every term on both sides by the LCD — including the terms with no fraction in them.
  • Every candidate solution has to be checked against the excluded values. One that makes a bottom zero is extraneous and is rejected, however well it fits the cleared equation.

A fraction of polynomials is still a fraction

Echo the Scribe works at the back of the Equation Cave, where every line you say comes back to you half a beat later. She writes fractions with polynomials in them and calls them rational expressions: a top, a bottom, and a division bar meaning exactly what it always meant.

(x2 - 9)/(x + 3) (2x + 6)/(x2 + 3x) 3/(x - 2) + 5/(x + 1)

Everything you already know about 12/18 still holds. You simplify by cancelling a common factor. You multiply straight across. You divide by turning the second fraction over. You add only once both bottoms are the same. Nothing new has been invented; the numbers have simply grown letters.

The one rule that does all the work is this: you cancel factors, never terms. A factor is something the whole line is multiplied by. A term is something the line is added to. In 12/18 you cancel the 6 because 12 = 6 * 2 and 18 = 6 * 3 — both are products. You would never simplify (12 + 5)/(18 + 5) by crossing out the 12 and the 18, and the same is true with letters.

So the first move on any rational expression is always the same: factor the top, factor the bottom, and only then look for something to cancel.

(x2 - 9)/(x + 3) = ((x - 3)(x + 3))/(x + 3) = x - 3

The (x + 3) is a whole bracket on both lines, so it goes. Compare that with (x + 3)/(x + 6), where the x is a term of a sum, not a factor of a product. Nothing cancels; the expression is already as simple as it gets. Echo's version: if you cannot draw a bracket around it, you cannot cross it out.

One trap is worth its own line. A bracket that looks backwards is only a minus sign away from matching:

5 - x = -(x - 5)

So (x2 - 25)/(5 - x) becomes ((x - 5)(x + 5))/(-(x - 5)) = -(x + 5). The bracket cancels and the minus sign stays behind.

The values the cave will not say

A bottom of zero has no meaning, so every rational expression comes with a short list of values the variable is not allowed to take. They are called the excluded values, or the restrictions.

Finding them is one step: set every bottom equal to zero and solve.

(x + 7)/((x - 4)(x + 1)) -> x - 4 = 0 or x + 1 = 0 -> x = 4, x = -1

If the bottom is not factored yet, factor it first — the restrictions are hiding inside the brackets:

3/(x2 + 2x - 15) = 3/((x + 5)(x - 3)) -> x = -5 and x = 3

Two details cost more marks than anything else here.

  • Read the restrictions off the original expression, before you simplify. In (x2 - 9)/(x + 3) the answer is x - 3, which looks harmless at x = -3. But the expression you started with had no value there, so x = -3 stays excluded for good. Simplifying can hide a restriction; it can never remove one.
  • A zero on top is fine. It just makes the whole fraction 0. Only the bottom is fussy.

When a bottom has a number in front of the x, divide it out as well:

2x + 1 = 0 -> 2x = -1 -> x = -1/2

That is a perfectly good excluded value. Restrictions are allowed to be fractions.

The four operations, in the order that keeps them easy

Multiplying. Factor everything, cancel anything that appears on a top and a bottom, and only then multiply what is left. Cancelling first is not a shortcut, it is the whole method — multiply the brackets out and you will have to factor them again.

((x2 + 5x + 6)/(x - 1)) * ((x2 - 3x + 2)/(x + 3)) = (((x + 2)(x + 3))/(x - 1)) * (((x - 1)(x - 2))/(x + 3)) = (x + 2)(x - 2)

Dividing. Turn the *second* fraction over and multiply. Only the second one moves, and it turns over before anything cancels.

((x2 - 4)/(x + 5)) ÷ ((x - 2)/(x + 5)) = ((x2 - 4)/(x + 5)) * ((x + 5)/(x - 2)) = x + 2

Adding and subtracting. These need a common bottom first, and finding the smallest one is the skill. Look at the two bottoms and ask what the smallest expression is that both divide into.

  • Nothing in common: the LCD is their product. For (x + 1) and (x - 4) it is (x + 1)(x - 4).
  • One contains the other: the LCD is the bigger one. For (x - 3) and 2x - 6 = 2(x - 3) it is 2(x - 3), not their product.
  • One needs factoring first: for (x + 2) and x2 - 4 = (x - 2)(x + 2) the LCD is (x - 2)(x + 2).

Then rebuild each fraction over that bottom, add the tops, and leave the bottom in factored form so the restrictions stay visible:

3/(x + 1) + 5/(x - 4) = (3(x - 4))/((x + 1)(x - 4)) + (5(x + 1))/((x + 1)(x - 4)) = (3x - 12 + 5x + 5)/((x + 1)(x - 4)) = (8x - 7)/((x + 1)(x - 4))

Two things never happen. Tops and bottoms are never added straight across — 3/(x + 1) + 5/(x - 4) is not 8/(2x - 3). And when you subtract, the minus sign belongs to the whole top that follows it, so every term inside that bracket changes sign. Say it out loud and let the echo check you.

Solving an equation, and the check that catches a liar

An equation with fractions in it is easier than an expression, because you are allowed to get rid of the bottoms entirely. Multiply both sides by the LCD and every denominator cancels at once.

Echo's rule, the one she repeats to everybody: *multiply every term by the common denominator — every term.* The constant sitting on its own at the end is a term too. It is the one people leave behind.

12/x + 7 = 13 multiply every term by x 12 + 7x = 13x 12 = 6x x = 2

When both sides are a single fraction you can go straight to cross-multiplication, which is the same move wearing a hat:

5/(x - 1) = 3/(x + 3) -> 5(x + 3) = 3(x - 1) -> 5x + 15 = 3x - 3 -> x = -9

Now the part that makes this topic different from every other kind of equation. Multiplying by something that might be zero can invent a solution that was never there. The cleared equation is honest about itself but not about where it came from, so a candidate can satisfy it and still break the original fraction. Such a candidate is called extraneous, and it has to be thrown out.

x/(x - 3) + 2/(x + 3) = 18/(x2 - 9)

The excluded values are x = 3 and x = -3. Clearing gives x2 + 5x - 24 = 0, so the candidates are x = -8 and x = 3. But x = 3 is on the excluded list, so it goes. The only solution is x = -8.

So the last line of every rational equation is the same: write down the excluded values first, then check each candidate against that list. A candidate is rejected only when it makes a bottom zero — not because it is negative, not because it is a fraction, and not because it looks unlikely. Sometimes nothing is rejected and both candidates are genuine solutions; sometimes every candidate goes and the equation has no solution at all.

How to type your answer

A simplified expression is typed the way it is written, using / for the fraction bar and ^ for a power, with a bracket round each polynomial:

  • x - 3
  • (x + 2)/(x + 6)
  • (3x - 11)/((x - 2)(x + 2))
  • 2x/3

Any form equal to the right answer is accepted, so (x + 2)(x - 2) and x2 - 4 both pass — but write yours in simplest form anyway, because that is what the question is asking you to produce, and it is what the next question will need. Never type a times sign as an x: write 3(x + 2) or 2 * 5.

Excluded values are numbers, not expressions. One of them is typed on its own:

4

Two of them are typed as a pair, smaller first, and a fraction is written with a slash:

(-1, 6) (-1/2, 3)

The solution of an equation is a whole number, or a fraction in lowest terms: 7, -9, 9/2. If a question asks which candidate is extraneous, or how many survive, pick the option — do not type a number.

An applied answer carries its own instruction. "Give the number of hours as a fraction in lowest terms" means 12/5, not 2.4; a question that asks for a decimal will say how many places. The unit itself is optional: 12 and 12 gold are both read as 12.

Worked examples

Example 1

Simplify (x2 - 4x - 12)/(x2 - 36), and state the excluded values.

  1. Factor the top: two numbers multiplying to -12 and adding to -4 are -6 and 2, so x2 - 4x - 12 = (x - 6)(x + 2).
  2. Factor the bottom: it is a difference of two squares, so x2 - 36 = (x - 6)(x + 6).
  3. Read the excluded values off this original bottom, before cancelling: x = 6 and x = -6.
  4. Cancel the whole bracket (x - 6), which appears on both lines.
  5. Simplest form is (x + 2)/(x + 6), with x = 6 and x = -6 still excluded. The x = 6 has become invisible, but it has not gone away.

Example 2

Simplify ((x2 - 9)/(x2 + 5x + 6)) ÷ ((x - 3)/(x + 4)).

  1. Turn the second fraction over and multiply: ((x2 - 9)/(x2 + 5x + 6)) * ((x + 4)/(x - 3)). Only the fraction you are dividing by moves.
  2. Factor everything: x2 - 9 = (x - 3)(x + 3) and x2 + 5x + 6 = (x + 2)(x + 3).
  3. The expression is now (((x - 3)(x + 3))/((x + 2)(x + 3))) * ((x + 4)/(x - 3)).
  4. Cancel the bracket (x + 3) and the bracket (x - 3). Both are whole factors, so both may go.
  5. What is left is (x + 4)/(x + 2). The excluded values are x = -3, x = -2 and x = 3 — the last one because dividing by (x - 3)/(x + 4) is impossible when that fraction is 0.

Example 3

Write 3/(x - 2) + 5/(x2 - 4) as a single fraction.

  1. Factor the second bottom before choosing anything: x2 - 4 = (x - 2)(x + 2).
  2. The second bottom already contains the first, so the LCD is (x - 2)(x + 2) — there is no need for a third bracket.
  3. Only the first fraction has to be rebuilt: multiply its top and bottom by (x + 2), giving (3(x + 2))/((x - 2)(x + 2)).
  4. Add the tops over the common bottom: 3(x + 2) + 5 = 3x + 6 + 5 = 3x + 11.
  5. The single fraction is (3x + 11)/((x - 2)(x + 2)), with x = 2 and x = -2 excluded.

Example 4

Solve x/(x - 3) + 2/(x + 3) = 18/(x2 - 9).

  1. Factor the bottom on the right: x2 - 9 = (x - 3)(x + 3), so the LCD is (x - 3)(x + 3).
  2. Write down the excluded values now, while they are in front of you: x = 3 and x = -3.
  3. Multiply every term by the LCD: x(x + 3) + 2(x - 3) = 18.
  4. Expand and collect: x2 + 3x + 2x - 6 = 18, so x2 + 5x - 24 = 0.
  5. Factor and solve: (x + 8)(x - 3) = 0, giving the candidates x = -8 and x = 3.
  6. Check both against the excluded list. x = 3 is excluded, so it is extraneous and is thrown out; x = -8 breaks nothing. The solution is x = -8.

Practice problems, with solutions

Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.

Problem 1

Difficulty 1 of 5

Simplify (8x2 + 28x)/(4x). Give your answer in simplest form.

Answer: 2x + 7

  1. Factor the numerator: 8x2 + 28x = 4x(2x + 7).
  2. (4x(2x + 7))/(4x) = 2x + 7, because the whole factor 4x cancels.
  3. The original bottom is 0 at x = 0, so x = 0 stays excluded even though it has cancelled.

Problem 2

Difficulty 3 of 5

State the two values of x excluded from (7)/(x2 - 11x + 30). Enter the smaller value first, as (a, b).

Answer: (5, 6)

  1. Factor the bottom: x2 - 11x + 30 = (x - 5)(x - 6).
  2. Set each factor to zero: x = 5 and x = 6.
  3. Smaller first, the excluded values are (5, 6).

Problem 3

Difficulty 4 of 5

Simplify (x2 - x - 20)/(x - 1) * (x2 + x - 2)/(x + 4). Give your answer in simplest form.

Answer: (x - 5)(x + 2)

  1. Factor the tops: x2 - x - 20 = (x - 5)(x + 4) and x2 + x - 2 = (x - 1)(x + 2).
  2. Multiplying puts everything over one line, so (x + 4) and (x - 1) each cancel.
  3. The answer is (x - 5)(x + 2), with x = 1 and x = -4 excluded.

Common mistakes

  • Cancelling terms instead of factors: turning (x + 3)/(x + 6) into 1/2, or crossing the x out of (2x + 5)/x. Only a bracket that multiplies the whole line can go.
  • Simplifying first and then reading the restrictions off the simplified version. (x2 - 9)/(x + 3) still excludes x = -3, even though the answer x - 3 does not show it.
  • Adding straight across: writing 3/(x + 1) + 5/(x - 4) as 8/(2x - 3). Tops are added only after both bottoms are identical.
  • Forgetting to distribute the minus when subtracting: in 4/(x - 1) - 3/(x + 2), the top is 4(x + 2) - 3(x - 1), so both terms of 3(x - 1) change sign, not just the first.
  • Leaving a constant term behind when clearing denominators. In 12/x + 7 = 13 the 7 and the 13 are multiplied by x as well; every term means every term.
  • Keeping an extraneous candidate. If a candidate makes a bottom zero it is rejected, even though it satisfies the equation you got after clearing.
  • Turning over the wrong fraction when dividing, or turning both over. The first fraction stays exactly as it is.

What you should be able to do

  • Simplify a rational expression by factoring, and state the values of the variable that are excluded.
  • Multiply and divide rational expressions, cancelling common factors.
  • Add and subtract rational expressions using a least common denominator.
  • Solve a rational equation by clearing denominators, and reject any solution that makes a denominator zero.

Where this fits in the curriculum

Common Core

  • HSA-APR.D.7

    High school — Understand that rational expressions form a system analogous to the rational numbers, closed under addition, subtraction, multiplication and division by a nonzero rational expression; add, subtract, multiply and divide rational expressions.

    HSA-APR.D.7 is a (+) standard — beyond the college- and career-ready threshold, i.e. precalculus rather than Algebra II.

  • HSA-REI.A.2

    High school — Solve simple rational and radical equations in one variable, and give examples showing how extraneous solutions may arise.

  • HSA-APR.D.6

    High school — Rewrite simple rational expressions in different forms: write a(x)/b(x) in the form q(x) + r(x)/b(x), where the degree of r(x) is less than the degree of b(x), using inspection or long division.

  • HSA-SSE.A.2

    High school — Use the structure of an expression to identify ways to rewrite it.

SAT

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