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Algebraic Fractions: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A difference over a single term

Simplify $\dfrac{10x^2 - 35x}{15x}$ completely, and state every value of $x$ that must be excluded.
1. Hint 1
  A fraction simplifies only by canceling a factor of the whole top and the whole bottom, so first write each of them as a product.
2. Hint 2
  Pull the greatest common factor out of $10x^2 - 35x$ , then look for that same factor inside $15x$ .
3. Hint 3
  The excluded values come from the original denominator, not from the simplified one. Ask where $15x$ is zero.
Answer

$\tfrac{2x - 7}{3}$ (or $\tfrac{2x}{3} - \tfrac{7}{3}$ ), for $x \neq 0$ .

Full solution

The numerator is a difference of two terms, so it has to be factored before anything can cancel.
The coefficients $10$ and $35$ share the factor $5$ , and both terms contain $x$ , so the greatest common factor is $5x$ :
$10x^2 - 35x = 5x(2x - 7).$

The denominator contains the same factor, since
$15x = 5x \cdot 3.$
So the fraction is
$\frac{5x(2x - 7)}{5x \cdot 3}.$

The factor $5x$ multiplies the whole top and the whole bottom, and $\tfrac{5x}{5x} = 1$ whenever $x \neq 0$ , so it cancels:
$\frac{2x - 7}{3}.$

Nothing else cancels.
The $7$ is a term of $2x - 7$ , not a factor of it, and $2x - 7$ shares no factor with $3$ .

The original denominator $15x$ is zero only at $x = 0$ , so $x = 0$ is excluded.
The simplified form has a value at $0$ but the original fraction does not, so the condition $x \neq 0$ stays with the answer.

Check at $x = 2$ : the original gives $\tfrac{40 - 70}{30}$ , which is $-1$ , and the simplified form gives $\tfrac{4 - 7}{3}$ , which is also $-1$ .

Answer

$\tfrac{2x - 7}{3}$ (or $\tfrac{2x}{3} - \tfrac{7}{3}$ ), for $x \neq 0$ .

Key idea

Factor the top and the bottom first: a factor of the whole numerator and the whole denominator cancels, and the excluded values come from the original denominator.
Problem 1: Mark complete
Problem 2 A product of two fractions

Multiply $\dfrac{x - 8}{10x} \cdot \dfrac{4x^2}{3x - 24}$ , simplify the result completely, and state every value of $x$ that must be excluded.
1. Hint 1
  A product of fractions is the product of the tops over the product of the bottoms, so a factor on either top can cancel against the same factor on either bottom.
2. Hint 2
  Factor $3x - 24$ first. A binomial then appears on one top and one bottom, and the numbers and powers of $x$ share factors too.
3. Hint 3
  Collect the excluded values from both original denominators before you cancel anything: a factor that cancels leaves no trace in the answer.
Answer

$\tfrac{2x}{15}$ , for $x \neq 0$ and $x \neq 8$ .

Full solution

Record the excluded values before canceling.
The first denominator, $10x$ , is zero at $x = 0$ .
The second, $3x - 24$ , is zero when $3x = 24$ , that is, at $x = 8$ .
So $x \neq 0$ and $x \neq 8$ .

Factor the binomial on the second bottom:
$3x - 24 = 3(x - 8).$
The product is now
$\frac{x - 8}{10x} \cdot \frac{4x^2}{3(x - 8)}.$

Cancel before multiplying.
The binomial $x - 8$ is a factor of the first top and of the second bottom, and it is not zero because $x \neq 8$ , so it cancels.
That leaves
$\frac{1}{10x} \cdot \frac{4x^2}{3}.$

Next, $4x^2 = 2x \cdot 2x$ and $10x = 2x \cdot 5$ , so the shared factor $2x$ cancels as well.
Multiplying what remains, $2x$ on top and $5 \cdot 3$ on the bottom, gives
$\frac{2x}{15}.$

The answer $\tfrac{2x}{15}$ has a value at $0$ and at $8$ , but the original product does not, so both conditions stay with it.

Check at $x = 2$ : the product is $\tfrac{-6}{20} \cdot \tfrac{16}{-18}$ , which is $\tfrac{-96}{-360}$ , or $\tfrac{4}{15}$ , and $\tfrac{2(2)}{15}$ is $\tfrac{4}{15}$ as well.

Answer

$\tfrac{2x}{15}$ , for $x \neq 0$ and $x \neq 8$ .

Key idea

When multiplying fractions, cancel a factor on any top against the same factor on any bottom before multiplying, and keep every value the original denominators excluded.
Problem 2: Mark complete
Problem 3 Two fractions over $x + 7$

Simplify $\dfrac{7x + 20}{x + 7} - \dfrac{3x - 8}{x + 7}$ completely, and state every value of $x$ that must be excluded.
1. Hint 1
  The denominators already match, so the work happens in the numerators. The minus sign between the fractions applies to the whole second numerator.
2. Hint 2
  Write the new numerator as $(7x + 20) - (3x - 8)$ , and distribute the minus to both terms in the second group before collecting like terms.
3. Hint 3
  Once the numerator is collected, factor it and compare it with the denominator: the fraction may simplify further.
Answer

$4$ , for $x \neq -7$ .

Full solution

The denominator $x + 7$ is zero at $x = -7$ , so $x = -7$ is excluded from the start.

Both fractions have the denominator $x + 7$ , so keep it and subtract the numerators.
The minus sign reaches the whole second numerator, so keep that numerator in parentheses:
$\frac{(7x + 20) - (3x - 8)}{x + 7}.$

Distributing the minus turns $-(3x - 8)$ into $-3x + 8$ , so the $-8$ becomes $+8$ .
The numerator is then $7x + 20 - 3x + 8$ , which collects to $4x + 28$ .

The numerator factors as $4(x + 7)$ , which shares the factor $x + 7$ with the denominator.
That factor is not zero because $x \neq -7$ , so it cancels:
$\frac{4(x + 7)}{x + 7} = 4.$

Dropping the parentheses without flipping the $-8$ would give the numerator $4x + 12$ instead, and the fraction would not reduce to a number.

Check at $x = 1$ : the first fraction is $\tfrac{27}{8}$ and the second is $\tfrac{-5}{8}$ .
Their difference is $\tfrac{27 + 5}{8}$ , which is $\tfrac{32}{8}$ , or $4$ .

Answer

$4$ , for $x \neq -7$ .

Key idea

When a fraction is subtracted, the minus sign reaches every term of its numerator, so keep that numerator in parentheses until the sign is distributed.
Problem 3: Mark complete
Problem 4 A quotient of two fractions

Divide $\dfrac{4x - 20}{9x} \div \dfrac{6x - 30}{15x^2}$ , simplify the result completely, and state every value of $x$ that must be excluded.
1. Hint 1
  Dividing by a fraction is multiplying by its reciprocal, and a fraction has a reciprocal only when it is not zero.
2. Hint 2
  Factor $4x - 20$ and $6x - 30$ , flip the second fraction, and cancel shared factors before multiplying what remains.
3. Hint 3
  For the excluded values, look in three places: the two original denominators, and the numerator of the divisor, which becomes a denominator after the flip.
Answer

$\tfrac{10x}{9}$ , for $x \neq 0$ and $x \neq 5$ .

Full solution

Start with the excluded values.
The denominators $9x$ and $15x^2$ are zero only at $x = 0$ .
The divisor $\tfrac{6x - 30}{15x^2}$ must also be nonzero, because nothing can be divided by zero.
Its numerator $6x - 30$ is zero when $6x = 30$ , at $x = 5$ , so $x = 5$ is excluded too, even though $6x - 30$ is not a denominator in the problem as written.

Factor each binomial: $4x - 20 = 4(x - 5)$ and
$6x - 30 = 6(x - 5).$
Then multiply by the reciprocal of the divisor:
$\frac{4(x - 5)}{9x} \cdot \frac{15x^2}{6(x - 5)}.$

Cancel before multiplying.
The binomial $x - 5$ is on a top and a bottom, and it is not zero because $x \neq 5$ , so it cancels.
The numbers $4$ and $6$ share the factor $2$ , leaving $2$ and $3$ .
The numbers $15$ and $9$ share the factor $3$ , leaving $5$ and $3$ .
One $x$ of $x^2$ cancels the $x$ in $9x$ .

What remains is $2 \cdot 5 \cdot x$ on top and $3 \cdot 3$ on the bottom, so the quotient is
$\frac{10x}{9}.$

The simplified form hides the condition $x \neq 5$ , which came only from the divisor's numerator.
At $x = 5$ the divisor is $\tfrac{0}{375}$ , which is $0$ , and dividing by $0$ has no meaning.

Check at $x = 2$ : the first fraction is $-\tfrac{12}{18}$ , or $-\tfrac{2}{3}$ , and the divisor is $-\tfrac{18}{60}$ , or $-\tfrac{3}{10}$ .
Multiplying $-\tfrac{2}{3}$ by the reciprocal $-\tfrac{10}{3}$ gives $\tfrac{20}{9}$ , which matches $\tfrac{10(2)}{9}$ .

Answer

$\tfrac{10x}{9}$ , for $x \neq 0$ and $x \neq 5$ .

Key idea

To divide, multiply by the reciprocal of the divisor; its numerator becomes a denominator, so the values that make that numerator zero are excluded as well.
Problem 4: Mark complete
Problem 5 Denominators $x + 8$ and $2x + 16$

Write $\dfrac{3}{x + 8} - \dfrac{x - 2}{2x + 16}$ as a single fraction in simplest form, and state every value of $x$ that must be excluded.
1. Hint 1
  Subtraction needs a common denominator. Before choosing one, factor each denominator to see what the two already share.
2. Hint 2
  Since $2x + 16 = 2(x + 8)$ , the least common denominator is $2(x + 8)$ . The first fraction needs its top and bottom multiplied by $2$ .
3. Hint 3
  When you subtract, the minus sign applies to both terms of $x - 2$ . Afterwards, check whether the new numerator shares a factor with the denominator.
Answer

$\tfrac{8 - x}{2x + 16}$ , or equivalently $\tfrac{8 - x}{2(x + 8)}$ or $-\tfrac{x - 8}{2x + 16}$ , for $x \neq -8$ .

Full solution

Both denominators are zero at the same value.
The first, $x + 8$ , is zero at $x = -8$ .
The second, $2x + 16$ , is zero when $2x = -16$ , which is also at $x = -8$ .
So $x = -8$ is excluded.

Factor the second denominator:
$2x + 16 = 2(x + 8).$
The first denominator is missing only the factor $2$ , so the least common denominator is $2(x + 8)$ .

Rewrite the first fraction by multiplying its top and bottom by $2$ :
$\frac{3}{x + 8} = \frac{6}{2(x + 8)}.$

Now subtract over the common denominator, keeping the second numerator in parentheses:
$\frac{6 - (x - 2)}{2(x + 8)}.$

The minus sign reaches both terms, so $-(x - 2)$ becomes $-x + 2$ .
The numerator is $6 - x + 2$ , which collects to $8 - x$ .

The numerator $8 - x$ shares no factor with $2(x + 8)$ .
It contains an $8$ and an $x$ , but they are terms of a sum, not factors, so nothing cancels.
The result is
$\frac{8 - x}{2x + 16}.$

Check at $x = 0$ : the original is $\tfrac{3}{8} - \tfrac{-2}{16}$ , which is $\tfrac{3}{8} + \tfrac{1}{8}$ , or $\tfrac{1}{2}$ , and the result gives $\tfrac{8}{16}$ , which is $\tfrac{1}{2}$ as well.

Answer

$\tfrac{8 - x}{2x + 16}$ , or equivalently $\tfrac{8 - x}{2(x + 8)}$ or $-\tfrac{x - 8}{2x + 16}$ , for $x \neq -8$ .

Key idea

Factoring the denominators first shows the least common denominator, and the minus sign in front of a fraction still reaches every term of its numerator.
Problem 5: Mark complete
Problem 6 Two rectangles with equal areas

A rectangle has sides of length $\dfrac{5x}{3x + 27}$ and $\dfrac{x + 9}{10}$ , where $x$ is a positive number and every length is in meters. Find its area as a single fraction in simplest form.

A second rectangle has the same area, and one of its sides has length $\dfrac{2x}{9}$ meters. Find the length of its other side.
1. Hint 1
  The area of a rectangle is the product of its side lengths, and a missing side is the area divided by the side you know.
2. Hint 2
  Factor $3x + 27$ before multiplying: it shares a binomial with the other side, and the numbers $5$ and $10$ share a factor too.
3. Hint 3
  For the second rectangle, divide the area by $\tfrac{2x}{9}$ by multiplying by its reciprocal, and cancel before you multiply.
Answer

The first rectangle's area is $\tfrac{x}{6}$ square meters. The second rectangle's other side is $\tfrac{3}{4}$ meter, for every positive $x$ .

Full solution

Because $x$ is positive, every numerator and denominator here is positive, so every length is a positive number and no value of $x$ in the problem is excluded.

The area is the product of the two sides.
Factor $3x + 27 = 3(x + 9)$ first, so the product is
$\frac{5x}{3(x + 9)} \cdot \frac{x + 9}{10}.$

Cancel before multiplying.
The binomial $x + 9$ is on a top and a bottom, and $5$ and $10$ share the factor $5$ , leaving $1$ and $2$ .
What remains multiplies to
$\frac{x}{3 \cdot 2} = \frac{x}{6}.$
So the area is $\tfrac{x}{6}$ square meters.

The second rectangle's other side is its area divided by the known side.
Dividing by $\tfrac{2x}{9}$ means multiplying by its reciprocal $\tfrac{9}{2x}$ , which exists because $x$ is not zero:
$\frac{x}{6} \cdot \frac{9}{2x}.$

The $x$ cancels, and $9$ and $6$ share the factor $3$ , leaving $3$ on top and $2 \cdot 2$ on the bottom.
So the other side is $\tfrac{3}{4}$ meter.

No $x$ is left in that length, so it is $\tfrac{3}{4}$ meter whatever positive number $x$ is.
The area $\tfrac{x}{6}$ and the known side $\tfrac{2x}{9}$ are each a fixed number times $x$ , so their quotient does not depend on $x$ .

Check with $x = 9$ : the first rectangle's sides are $\tfrac{45}{54}$ , or $\tfrac{5}{6}$ , and $\tfrac{18}{10}$ , or $\tfrac{9}{5}$ , so its area is $\tfrac{45}{30}$ , or $\tfrac{3}{2}$ , which is $\tfrac{9}{6}$ .
The second rectangle's known side is $\tfrac{18}{9}$ , or $2$ , and $2 \cdot \tfrac{3}{4}$ is $\tfrac{3}{2}$ , the same area.

Answer

The first rectangle's area is $\tfrac{x}{6}$ square meters. The second rectangle's other side is $\tfrac{3}{4}$ meter, for every positive $x$ .

Key idea

An area is a product of lengths and a missing side is a quotient, so multiplying and dividing algebraic fractions answers both questions.
Problem 6: Mark complete
Problem 7 The third stretch of a trail

A hiking trail is made of three stretches with a total length of $\dfrac{5x}{6}$ kilometers, where $x$ is a number greater than $6$ . The first stretch is $\dfrac{x}{4}$ kilometers long and the second is $\dfrac{x - 6}{3}$ kilometers long.

Find the length of the third stretch as a single fraction in simplest form, and check your answer by taking $x = 12$ .
1. Hint 1
  The third stretch is what is left of the total once both known stretches are taken away, so write it as one expression with two subtractions.
2. Hint 2
  The least common denominator of $6$ , $4$ and $3$ is $12$ . Rewrite all three fractions over $12$ , keeping $x - 6$ in parentheses.
3. Hint 3
  The minus sign in front of $\tfrac{x - 6}{3}$ reaches both of its terms, so its $-6$ turns positive. After collecting, look for a factor the numerator shares with $12$ .
Answer

$\tfrac{x + 8}{4}$ kilometers; at $x = 12$ both the formula and the stretches give $5$ km.

Full solution

The three stretches add up to the total, so the third is the total minus the other two:
$\frac{5x}{6} - \frac{x}{4} - \frac{x - 6}{3}.$

The least common denominator of $6$ , $4$ and $3$ is $12$ .
Multiply the first fraction's top and bottom by $2$ , the second's by $3$ and the third's by $4$ , keeping $x - 6$ in parentheses:
$\frac{10x}{12} - \frac{3x}{12} - \frac{4(x - 6)}{12}.$

Combine over the common denominator.
The last minus sign reaches both terms of $4(x - 6)$ , which is $4x - 24$ , so it contributes $-4x + 24$ :
$\frac{10x - 3x - 4x + 24}{12}.$

Collecting like terms, $10x - 3x - 4x$ is $3x$ , so the numerator is $3x + 24$ .
It factors as $3(x + 8)$ , and the $3$ cancels with the $3$ in $12 = 3 \cdot 4$ :
$\frac{3(x + 8)}{12} = \frac{x + 8}{4}.$

So the third stretch is $\tfrac{x + 8}{4}$ kilometers.
It is positive for every $x$ greater than $6$ , as a length must be.

Check with $x = 12$ .
The total is $\tfrac{60}{6}$ , or $10$ km, the first stretch is $\tfrac{12}{4}$ , or $3$ km, and the second is $\tfrac{6}{3}$ , or $2$ km, so the third must be $10 - 3 - 2 = 5$ km.
The formula gives $\tfrac{12 + 8}{4} = 5$ km, which agrees.

Forgetting to distribute the minus would give the numerator $3x - 24$ and the answer $\tfrac{x - 8}{4}$ , which at $x = 12$ is $1$ km instead of $5$ .

Answer

$\tfrac{x + 8}{4}$ kilometers; at $x = 12$ both the formula and the stretches give $5$ km.

Key idea

Taking two fractions away from a total is one subtraction over a common denominator, and each minus sign reaches every term of the numerator after it.
Problem 7: Mark complete
Problem 8 Jordan's simplification

Jordan simplifies $\dfrac{6x + 14}{2x}$ . He crosses out the $x$ in $6x$ and the $x$ in $2x$ , which leaves $\dfrac{6 + 14}{2}$ , and he writes his answer as $10$ . To check, he substitutes $x = 1$ into the original fraction, gets $\dfrac{20}{2}$ , which is $10$ , and decides that his answer is right.

Explain why crossing out the two copies of $x$ is not a legal cancellation, and why his check at $x = 1$ could not catch the mistake. Then simplify the fraction correctly, and state the value of $x$ that must be excluded.
1. Hint 1
  Canceling divides the whole numerator and the whole denominator by the same factor. Ask whether $x$ multiplies all of $6x + 14$ .
2. Hint 2
  Try another value, such as $x = 2$ , in the original fraction and compare it with $10$ . Then compare Jordan's fraction $\tfrac{6 + 14}{2}$ with the original at $x = 1$ .
3. Hint 3
  To simplify correctly, take the greatest common factor out of $6x + 14$ and look for the same factor in $2x$ .
Answer

The $x$ multiplies only the term $6x$ , not the whole numerator. His $\tfrac{6 + 14}{2}$ is the original with $x = 1$ put in, so a check at $x = 1$ had to agree. Correct: $\tfrac{3x + 7}{x}$ (or $3 + \tfrac{7}{x}$ ), for $x \neq 0$ .

Full solution

A cancellation divides the whole numerator and the whole denominator by the same factor, and it is legal because that factor over itself is $1$ .
The $x$ that Jordan crossed out on top multiplies only the term $6x$ .
The $14$ has no factor of $x$ , so $x$ is not a factor of the whole numerator $6x + 14$ , and it cannot be canceled.

Crossing it out changes the value.
At $x = 2$ the original fraction is $\tfrac{12 + 14}{4}$ , which is $\tfrac{26}{4}$ , or $\tfrac{13}{2}$ , but Jordan's answer is still $10$ .
A single value at which the two disagree proves they are not equal.

His check could not catch this, because his fraction $\tfrac{6 + 14}{2}$ is exactly the original with $1$ put in place of each $x$ .
So at $x = 1$ his answer and the original are the same number, $10$ , and the check only compared that number with itself.

An agreement at one value is evidence, not proof, and $x = 1$ is a poor test for a lost factor of $x$ , since multiplying by $1$ changes nothing.
A value such as $x = 2$ exposes the slip at once.

To simplify correctly, factor the numerator.
The coefficients $6$ and $14$ share the factor $2$ , so
$6x + 14 = 2(3x + 7).$
The denominator is $2x = 2 \cdot x$ , so the factor the two share is $2$ :
$\frac{2(3x + 7)}{2x} = \frac{3x + 7}{x}.$

The numerator $3x + 7$ and the denominator $x$ share no factor, so this is the simplest form.
Split honestly, it is $\tfrac{3x}{x} + \tfrac{7}{x}$ , which is $3 + \tfrac{7}{x}$ .
The denominator $2x$ is zero at $x = 0$ , so $x = 0$ is excluded.

Check at $x = 2$ : $\tfrac{3(2) + 7}{2}$ is $\tfrac{13}{2}$ , the value the original fraction gave above.

Answer

The $x$ multiplies only the term $6x$ , not the whole numerator. His $\tfrac{6 + 14}{2}$ is the original with $x = 1$ put in, so a check at $x = 1$ had to agree. Correct: $\tfrac{3x + 7}{x}$ (or $3 + \tfrac{7}{x}$ ), for $x \neq 0$ .

Key idea

Canceling needs a factor of the whole numerator and the whole denominator, and since a lost factor of $x$ changes nothing at $x = 1$ , test a simplification at another value.
Problem 8: Mark complete
Problem 9 Two claims to test

Each claim below says that the expression on its left equals the expression on its right.

Claim A: $\dfrac{6x - 10}{9x - 15} = \dfrac{2}{3}$

Claim B: $\dfrac{2}{x} + \dfrac{3}{4} = \dfrac{5}{x + 4}$

Decide whether each claim holds at every value of $x$ at which its left side is defined. For a claim that holds, show why, and state the values of $x$ its left side excludes. For a claim that fails, give a value of $x$ at which both sides are defined but differ, and write what the left side does equal, simplified.
1. Hint 1
  To show that a claim holds, simplify its left side with the rules for fractions. To show that one fails, a single value of $x$ at which both sides are defined but differ is enough.
2. Hint 2
  In claim A, factor the top and the bottom before anything cancels. In claim B, the two fractions on the left need a common denominator before they can be added.
3. Hint 3
  Even a claim that holds excludes values: set each denominator on its left equal to zero. In claim B, the least common denominator of $x$ and $4$ is $4x$ .
Answer

Claim A holds, for $x \neq \tfrac{5}{3}$ . Claim B fails (for example, at $x = 4$ the left side is $\tfrac{5}{4}$ and the right side is $\tfrac{5}{8}$ ); its left side equals $\tfrac{3x + 8}{4x}$ , for $x \neq 0$ .

Full solution

Claim A.
Factor the top and the bottom: $6x - 10 = 2(3x - 5)$ and
$9x - 15 = 3(3x - 5).$
The shared factor $3x - 5$ cancels wherever it is not zero, leaving $\tfrac{2}{3}$ .
The denominator $9x - 15$ is zero when $9x = 15$ , that is, at $x = \tfrac{15}{9}$ , which is $\tfrac{5}{3}$ .
So claim A holds, for
$x \neq \tfrac{5}{3}.$

Claim B.
The right side adds the two numerators and adds the two denominators, which is not how fractions add.
At $x = 4$ the left side is $\tfrac{2}{4} + \tfrac{3}{4}$ , which is $\tfrac{5}{4}$ , while the right side is $\tfrac{5}{8}$ .
Both sides are defined there and they differ, so claim B fails.

To find what the left side does equal, use the least common denominator $4x$ .
Multiply the first fraction's top and bottom by $4$ and the second's by $x$ :
$\frac{8}{4x} + \frac{3x}{4x} = \frac{3x + 8}{4x}.$
The numerator $3x + 8$ shares no factor with $4x$ , so nothing cancels.
This holds for $x \neq 0$ .

Answer

Claim A holds, for $x \neq \tfrac{5}{3}$ . Claim B fails (for example, at $x = 4$ the left side is $\tfrac{5}{4}$ and the right side is $\tfrac{5}{8}$ ); its left side equals $\tfrac{3x + 8}{4x}$ , for $x \neq 0$ .

Key idea

A claimed simplification holds when the fraction rules turn one side into the other, and one value at which both sides are defined but differ is enough to show that it fails.
Problem 9: Mark complete
Problem 10 A fraction with one excluded value

Write an algebraic fraction that is undefined at $x = 6$ and at no other value of $x$ , and that simplifies to $2x - 1$ .

When your fraction is simplified, the result is written as $2x - 1$ together with the condition $x \neq 6$ . Explain why that condition is needed, even though $2x - 1$ itself has a value at $x = 6$ .
1. Hint 1
  Simplifying removes a factor that the top and the bottom share. Here you run that process backwards, putting a shared factor in on purpose.
2. Hint 2
  Choose a denominator that is zero at $x = 6$ and nowhere else.
3. Hint 3
  For the explanation, compare your fraction and $2x - 1$ at $x = 6$ itself, and recall what canceling a factor requires of that factor.
Answer

For example $\tfrac{(2x - 1)(x - 6)}{x - 6}$ ; any denominator that is zero at $x = 6$ and nowhere else works, such as $3(x - 6)$ or $(x - 6)^2$ , with $2x - 1$ times it on top. At $x = 6$ the fraction is undefined while $2x - 1$ is $11$ .

Full solution

Simplifying cancels a factor that the whole numerator and the whole denominator share.
To build a fraction that simplifies to $2x - 1$ , start from $2x - 1$ and put a shared factor in on purpose.

The denominator must be zero at $x = 6$ and at no other value.
The expression $x - 6$ does that, since it is zero exactly when $x = 6$ .

Put $x - 6$ on the bottom, and $2x - 1$ times $x - 6$ on the top:
$\frac{(2x - 1)(x - 6)}{x - 6}.$
For $x \neq 6$ the factor $x - 6$ is shared by the whole top and the whole bottom, so it cancels and leaves $2x - 1$ .

Multiplying out the numerator, every term times every term, gives $2x^2 - 12x - x + 6$ , which collects to $2x^2 - 13x + 6$ .
So $\tfrac{2x^2 - 13x + 6}{x - 6}$ is the same fraction written another way.
Any denominator that is zero at $x = 6$ and nowhere else, such as $3(x - 6)$ or $(x - 6)^2$ , could serve instead, with $2x - 1$ times that denominator on top.

Check at $x = 2$ : the built fraction is $\tfrac{3 \cdot (-4)}{-4}$ , which is $3$ , and $2(2) - 1$ is $3$ as well.

The condition is needed because the two expressions do not agree at $x = 6$ .
There the fraction's denominator is $0$ , so the fraction has no value at all, while $2x - 1$ equals $11$ .
Writing that the fraction equals $2x - 1$ with no condition would claim the fraction is $11$ at $x = 6$ , which is false.

The cancellation shows where the condition comes from.
Canceling $x - 6$ uses the fact that $\tfrac{x - 6}{x - 6} = 1$ , and that is true only when $x - 6$ is not zero.
So the simplified form $2x - 1$ carries $x \neq 6$ as a record of the cancellation.

Answer

For example $\tfrac{(2x - 1)(x - 6)}{x - 6}$ ; any denominator that is zero at $x = 6$ and nowhere else works, such as $3(x - 6)$ or $(x - 6)^2$ , with $2x - 1$ times it on top. At $x = 6$ the fraction is undefined while $2x - 1$ is $11$ .

Key idea

Canceling a factor is valid only where that factor is not zero, so the simplified form carries every restriction of the original fraction.
Problem 10: Mark complete

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