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Adding and Subtracting Rational Expressions: 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 shared denominator

Write $\frac{2x^2+x+8}{x^2+x+5}-\frac{3}{x^2+x+5}$ as one fraction in lowest terms, and state its real domain.
1. Hint 1
  The two terms already use the same denominator.
2. Hint 2
  Subtract the full second numerator, then check for a shared factor.
Answer

$\frac{2x^2+x+5}{x^2+x+5}$ ; all real $x$ .

Full solution

The common denominator is $x^2+x+5=(x+\frac12)^2+\frac{19}4$ , which is positive for every real $x$ .
Combining the numerators gives
$\frac{2x^2+x+8-3}{x^2+x+5}=\frac{2x^2+x+5}{x^2+x+5}$

Neither quadratic has a real zero, since their discriminants $1-40$ and $1-20$ are negative, so neither has a linear factor over the reals.
The numerator is not a constant multiple of the denominator, since
$2(x^2+x+5)=2x^2+2x+10.$
So they share no factor, and the fraction is in lowest terms.

Answer

$\frac{2x^2+x+5}{x^2+x+5}$ ; all real $x$ .

Key idea

Subtracting fractions with a shared denominator changes the numerator, not the denominator.
Problem 1: Mark complete
Problem 2 Preparing a denominator

For $\frac{1}{12(x-1)^2}+\frac{1}{18(x-1)^3}$ , write the least common denominator in factored form.
1. Hint 1
  The numerical coefficient and the polynomial power must both accommodate the two denominators.
2. Hint 2
  Take the least common multiple of $12$ and $18$ , and the higher power of $x-1$ .
Answer

$36(x-1)^3$ .

Full solution

The numerical least common multiple is $36$ .
The highest power of $x-1$ present in a denominator is $3$ .
Thus
$L=36(x-1)^3$

Dividing $L$ by the two denominators gives $3(x-1)$ and $2$ , respectively, confirming that each divides it.

Answer

$36(x-1)^3$ .

Key idea

A least common denominator combines the numerical least common multiple with each factor's highest required power.
Problem 2: Mark complete
Problem 3 A nested fraction

Simplify $\frac{x+8}{2+3/(x-4)}$ for real $x$ , retaining every restriction.
1. Hint 1
  The inner fraction must exist, and the whole denominator must be nonzero.
2. Hint 2
  Combine the denominator into one fraction before dividing by it.
Answer

$\frac{(x+8)(x-4)}{2x-5}$ , with $x\ne4,\frac52$ .

Full solution

The inner denominator requires $x\ne4$ .
On that domain,
$2+\frac3{x-4}=\frac{2x-5}{x-4}$
This entire denominator is zero at $x=\frac52$ , so that input is excluded too.

Dividing by $\frac{2x-5}{x-4}$ means multiplying by $\frac{x-4}{2x-5}$ , which gives $\frac{(x+8)(x-4)}{2x-5}$ , with both restrictions retained.
At $x=1$ , the original is $\frac9{2-1}=9$ , and the new form gives
$\frac{9(-3)}{-3}=9.$

Answer

$\frac{(x+8)(x-4)}{2x-5}$ , with $x\ne4,\frac52$ .

Key idea

A complex fraction must satisfy the restrictions of both its inner fractions and its main denominator.
Problem 3: Mark complete
Problem 4 Cost per printed copy

A print shop charges 90 dollars for a run of $x+4$ copies and 78 dollars for a run of $2x+1$ copies, where $x$ is a positive integer. Write the first cost per copy minus the second as a single fraction in lowest terms, and find that difference at $x=6$ .
1. Hint 1
  Divide each total charge by its own copy count before subtracting.
2. Hint 2
  Use $(x+4)(2x+1)$ as a common denominator.
Answer

$\frac{6(17x-37)}{(x+4)(2x+1)}$ dollars per copy; $3$ dollars per copy at $x=6$ .

Full solution

The required difference is $\frac{90}{x+4}-\frac{78}{2x+1}$ .
Both denominators are positive for positive integer $x$ .
Over the common denominator, the numerator is
$\begin{aligned}&90(2x+1)-78(x+4)\\&=102x-222\end{aligned}$

Thus the difference is $\frac{6(17x-37)}{(x+4)(2x+1)}$ , in lowest terms because $17x-37$ is zero only at $\frac{37}{17}$ , which is neither $-4$ nor $-\frac12$ .
At $x=6$ , the original costs are $9$ and $6$ dollars per copy, whose difference is $3$ dollars per copy, and
$\frac{6\cdot65}{10\cdot13}=3.$

Answer

$\frac{6(17x-37)}{(x+4)(2x+1)}$ dollars per copy; $3$ dollars per copy at $x=6$ .

Key idea

A difference of average costs requires subtracting ratios over a shared denominator.
Problem 4: Mark complete
Problem 5 Three contributions

Write $\frac{1}{x^2+1}+\frac{2}{(x^2+1)^2}-\frac{1}{2}$ as a single fraction in lowest terms for real $x$ .
1. Hint 1
  A common denominator must accommodate every denominator in the expression.
2. Hint 2
  Use a denominator containing two copies of the quadratic and the numerical factor $2$ .
3. Hint 3
  Expand only the combined numerator, keeping brackets around the term being subtracted.
Answer

$\frac{5-x^4}{2(x^2+1)^2}$ ; all real $x$ .

Full solution

The least common denominator is $2(x^2+1)^2$ , positive for all real $x$ .
The rewritten numerators are $2(x^2+1)$ , $4$ , and $(x^2+1)^2$ .
Their combination is
$2(x^2+1)+4-(x^2+1)^2=5-x^4$

The result is $(5-x^4)/(2(x^2+1)^2)$ .
The numerator is not divisible by $x^2+1$ : replacing $x^2$ by $-1$ in its remainder gives $4$ , not $0$ .
No common factor remains.

Answer

$\frac{5-x^4}{2(x^2+1)^2}$ ; all real $x$ .

Key idea

Repeated denominator factors need their highest power once in the common denominator.
Problem 5: Mark complete
Problem 6 A ratio of two combinations

Simplify $\frac{2/(x-14)-1/(x+15)}{1+2/(x+15)}$ for real $x$ , retaining every restriction.
1. Hint 1
  Combine the numerator and denominator separately.
2. Hint 2
  Determine whether the combined denominator can be zero before taking its reciprocal.
Answer

$\frac{x+44}{(x-14)(x+17)}$ , with $x\ne14,-15,-17$ .

Full solution

The inner denominators exclude $14$ and $-15$ .
Combining the top over $(x-14)(x+15)$ gives
$\begin{aligned}N&=\frac{2(x+15)-(x-14)}{(x-14)(x+15)}\\&=\frac{x+44}{(x-14)(x+15)}\end{aligned}$
Combining the bottom gives
$D=\frac{x+15+2}{x+15}=\frac{x+17}{x+15}$

The denominator $D$ is zero at $x=-17$ , so that input is excluded as well.
Dividing $N$ by $D$ gives
$\begin{aligned}&\frac{x+44}{(x-14)(x+15)}\cdot\frac{x+15}{x+17}\\&=\frac{x+44}{(x-14)(x+17)}\end{aligned}$
with
$x\ne14,-15,-17.$
At $x=0$ , the original top is $-\frac17-\frac1{15}=-\frac{22}{105}$ and the bottom is $\frac{17}{15}$ , giving $-\frac{22}{119}$ , as does $\frac{44}{(-14)(17)}$ .

Answer

$\frac{x+44}{(x-14)(x+17)}$ , with $x\ne14,-15,-17$ .

Key idea

The main denominator of a complex fraction must also be nonzero, and that can add restrictions beyond those from the inner fractions.
Problem 6: Mark complete
Problem 7 A polynomial difference

Simplify $\frac{x^2+x}{x^2-1}-\frac1{x-1}$ for real $x$ . State which excluded input or inputs no longer appear in the reduced denominator.
1. Hint 1
  The quadratic denominator contains the other denominator as a factor.
2. Hint 2
  Combine over the factored quadratic and then factor the entire numerator.
Answer

$1$ , with $x\ne-1,1$ ; both exclusions disappear from the reduced denominator.

Full solution

The original denominators exclude $-1$ and $1$ .
Over $(x-1)(x+1)$ , the numerator is
$x^2+x-(x+1)=x^2-1$

Cancel the nonzero factors on the domain to obtain $1$ .
Since the reduced denominator is $1$ , neither exclusion remains visible, but both still apply to the original difference.

Answer

$1$ , with $x\ne-1,1$ ; both exclusions disappear from the reduced denominator.

Key idea

Subtracting rational expressions can cancel a denominator without restoring any excluded input.
Problem 7: Mark complete
Problem 8 An extra common factor

A student combines $1/x$ and $1/(x+1)$ using the denominator $x(x+1)(x-4)$ and then excludes $4$ . Is the additional exclusion valid for the original sum? Explain and give its value at $4$ .
1. Hint 1
  Check whether the proposed extra factor came from an original denominator.
2. Hint 2
  Evaluate the original sum directly at the disputed input.
Answer

No; the original domain is $x\ne0,-1$ , and its value at $4$ is $9/20$ .

Full solution

The original denominators exclude only $0$ and $-1$ .
At $4$ , both fractions exist and
$\frac14+\frac15=\frac9{20}$

Multiplying numerator and denominator by $x-4$ is valid only for $x\ne4$ .
That temporary form has unnecessarily narrowed the domain and does not justify changing the original sum's domain.
The LCD $x(x+1)$ avoids the problem.

Answer

No; the original domain is $x\ne0,-1$ , and its value at $4$ is $9/20$ .

Key idea

Introducing an unnecessary denominator factor can create a restriction that the original expression never had.
Problem 8: Mark complete
Problem 9 A sum that keeps its sign

For real $x\ne6$ , decide whether $\frac1{(x-6)^2}+\frac1{x^2+8}$ is always positive, always negative, or changes sign. Justify your answer twice: once from the two terms, and once from the sum written as a single fraction.
1. Hint 1
  Decide the sign of each term on its own, remembering the excluded input.
2. Hint 2
  For the single fraction, combine over $(x-6)^2(x^2+8)$ and complete the square in the numerator.
Answer

Always positive for $x\ne6$ ; as a single fraction it is $\frac{2((x-3)^2+13)}{(x-6)^2(x^2+8)}$ .

Full solution

From the terms: for $x\ne6$ , $(x-6)^2$ is positive, and $x^2+8$ is positive for every real $x$ .
Both terms are positive, so their sum is positive.

As one fraction, over $(x-6)^2(x^2+8)$ the numerator is
$x^2+8+(x-6)^2=2x^2-12x+44$
Completing the square,
$2x^2-12x+44=2((x-3)^2+13)$
which is at least $26$ .
The denominator is positive for $x\ne6$ , so the fraction is positive throughout its domain.

Answer

Always positive for $x\ne6$ ; as a single fraction it is $\frac{2((x-3)^2+13)}{(x-6)^2(x^2+8)}$ .

Key idea

Combining a sum into one fraction keeps its sign, so the numerator and denominator can confirm what the terms show.
Problem 9: Mark complete
Problem 10 Reduce first, or not

Let $S=\frac{x-16}{(x-16)(x+14)}+\frac{x+14}{(x+14)(x-15)}$ . Give the LCD of the two fractions as written, and the LCD after each fraction is first reduced to lowest terms. Explain why each is a valid choice at its stage, and write $S$ as a single fraction in lowest terms with its domain.
1. Hint 1
  Record the excluded inputs from the fractions as written before reducing anything.
2. Hint 2
  Each fraction can be reduced on its own before any common denominator is chosen.
Answer

As written the LCD is $(x-16)(x+14)(x-15)$ ; after reducing it is $(x+14)(x-15)$ . $S=\frac{2x-1}{(x+14)(x-15)}$ , with $x\ne16,-14,15$ .

Full solution

As written, the denominators are $(x-16)(x+14)$ and $(x+14)(x-15)$ , so the LCD is $(x-16)(x+14)(x-15)$ and the domain is
$x\ne16,-14,15.$

On that domain each fraction reduces: $\frac{x-16}{(x-16)(x+14)}=\frac1{x+14}$ and
$\frac{x+14}{(x+14)(x-15)}=\frac1{x-15}.$
Their LCD is $(x+14)(x-15)$ , and
$\begin{aligned}S&=\frac{(x-15)+(x+14)}{(x+14)(x-15)}\\&=\frac{2x-1}{(x+14)(x-15)}\end{aligned}$

Both choices are valid.
The first is built from the denominators actually present; the second is built from the reduced fractions, which equal the originals on the domain, so it is smaller and saves work.
Reducing first removes the factor $x-16$ from the arithmetic, but not the restriction $x\ne16$ , which must be stated with the answer.

Answer

As written the LCD is $(x-16)(x+14)(x-15)$ ; after reducing it is $(x+14)(x-15)$ . $S=\frac{2x-1}{(x+14)(x-15)}$ , with $x\ne16,-14,15$ .

Key idea

Reducing each fraction first can shrink the LCD, but the restrictions still come from the denominators as written.
Problem 10: Mark complete

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