Rational Expression Practice Questions

Divide the coefficients and subtract the exponents:

\(\dfrac{15x^3}{25x^5} = \dfrac{15}{25} \cdot x^{3-5} = \dfrac{3}{5} \cdot x^{-2} = \dfrac{3}{5x^2}\)

Choice B places \(x^2\) in the numerator instead of the denominator, which would result from subtracting the exponents in the wrong order. Choice D adds the exponents instead of subtracting them.

Factor both the numerator and denominator. The numerator is a difference of squares:

\(x^2 – 16 = (x + 4)(x – 4)\)

For the denominator, find two numbers that multiply to −8 and add to 2:

\(x^2 + 2x – 8 = (x + 4)(x – 2)\)

Cancel the common factor \((x + 4)\):

\(\dfrac{(x + 4)(x – 4)}{(x + 4)(x – 2)} = \dfrac{x – 4}{x – 2}\)

This simplification is valid for all \(x \neq -4\) and \(x \neq 2\).

A rational expression is undefined wherever its denominator equals zero. Set the denominator equal to zero and solve:

\(x^2 – 5x + 6 = 0\)

Factor the trinomial. Find two numbers that multiply to 6 and add to −5:

\((x – 2)(x – 3) = 0\)

So \(x = 2\) and \(x = 3\) must be excluded from the domain. These are the values that would make the denominator, and therefore the entire expression, undefined.

Before multiplying, factor any expressions that can be simplified. The numerator \(x^2 – 9\) is a difference of squares:

\(\dfrac{(x + 3)(x – 3)}{x + 1} \cdot \dfrac{x + 1}{x + 3}\)

Cancel the common factors \((x + 1)\) and \((x + 3)\) that appear in both a numerator and a denominator:

\(\dfrac{(x + 3)(x – 3)}{x + 1} \cdot \dfrac{x + 1}{x + 3} = x – 3\)

Choice C is not fully simplified because \(x^2 – 9\) can be factored, allowing further cancellation.

To divide rational expressions, multiply by the reciprocal of the second fraction:

\(\dfrac{2x}{x + 3} \cdot \dfrac{x^2 + 5x + 6}{6x^2}\)

Factor \(x^2 + 5x + 6 = (x + 2)(x + 3)\) and substitute:

\(= \dfrac{2x}{x + 3} \cdot \dfrac{(x + 2)(x + 3)}{6x^2}\)

Cancel the common factors \((x + 3)\) and \(\frac{2x}{6x^2} = \frac{1}{3x}\):

\(= \dfrac{x + 2}{3x}\)

Choice C incorrectly multiplies straight across instead of flipping the second fraction first.

Since both fractions already share the same denominator, add the numerators directly:

\(\dfrac{3x}{x + 4} + \dfrac{12}{x + 4} = \dfrac{3x + 12}{x + 4}\)

Now factor the numerator:

\(= \dfrac{3(x + 4)}{x + 4} = 3\)

The \((x + 4)\) factors cancel completely, leaving just 3. Always check whether the result can be simplified after combining the numerators.

To add fractions with different denominators, first find the least common denominator (LCD). The LCD of \(x\) and \((x + 1)\) is \(x(x + 1)\). Rewrite each fraction with the LCD:

\(\dfrac{2(x + 1)}{x(x + 1)} + \dfrac{3x}{x(x + 1)}\)

Now add the numerators:

\(= \dfrac{2(x + 1) + 3x}{x(x + 1)} = \dfrac{2x + 2 + 3x}{x(x + 1)}\) \(\:= \dfrac{5x + 2}{x(x + 1)}\)

Choice C incorrectly adds the denominators together (\(x + x + 1 = 2x + 1\)). Remember: you must find a common denominator, not add the existing ones.

The LCD is \((x – 3)(x + 1)\). Rewrite each fraction:

\(\dfrac{5(x + 1)}{(x – 3)(x + 1)} – \dfrac{2(x – 3)}{(x – 3)(x + 1)}\)

Subtract the numerators. Be careful to distribute the negative sign to both terms in \(2(x – 3)\):

\(\dfrac{5(x + 1) – 2(x – 3)}{(x – 3)(x + 1)}\)\(\:= \dfrac{5x + 5 – 2x + 6}{(x – 3)(x + 1)}\)\(\:= \dfrac{3x + 11}{(x – 3)(x + 1)}\)

Choice C results from incorrectly computing \(-2 \times (-3)\) as −6 instead of +6.

To simplify a complex fraction, multiply both the numerator and denominator by the LCD of all the smaller fractions. The LCD of \(\frac{1}{x}\) and \(\frac{1}{y}\) is \(xy\):

\(\dfrac{\left(\dfrac{1}{x} + \dfrac{1}{y}\right) \cdot xy}{\left(\dfrac{1}{x} – \dfrac{1}{y}\right) \cdot xy}\)

Distribute \(xy\) to each term:

\(= \dfrac{y + x}{y – x}\)

This works because \(\frac{1}{x} \cdot xy = y\) and \(\frac{1}{y} \cdot xy = x\).

Multiply every term by the LCD, which is \(3x\), to clear the fractions:

\(3x \cdot \dfrac{5}{x} + 3x \cdot \dfrac{1}{3} = 3x \cdot \dfrac{2}{x}\)

\(15 + x = 6\)

Solve for \(x\):

\(x = 6 – 15 = -9\)

Check by substituting back: \(\frac{5}{-9} + \frac{1}{3} = -\frac{5}{9} + \frac{3}{9} = -\frac{2}{9}\), and \(\frac{2}{-9} = -\frac{2}{9}\) ✓

Multiply both sides by the LCD, \((x – 2)\), to clear the fractions:

\(x + 1 = 3\)

\(x = 2\)

However, substituting \(x = 2\) back into the original equation gives \(\frac{3}{0} = \frac{3}{0}\), which is undefined. This means \(x = 2\) is an extraneous solution. In other words, it emerged from the algebra but does not satisfy the original equation.

Since the only solution is extraneous, the equation has no solution. Always check your answers in the original equation when solving rational equations, because multiplying by the LCD can introduce values that make a denominator zero.

Person A’s rate is \(\frac{1}{6}\) of the room per hour, and Person B’s rate is \(\frac{1}{4}\) of the room per hour. When working together, their rates add up. If \(t\) is the time to finish together:

\(\dfrac{1}{6} + \dfrac{1}{4} = \dfrac{1}{t}\)

Find a common denominator for the left side:

\(\dfrac{2}{12} + \dfrac{3}{12} = \dfrac{5}{12} = \dfrac{1}{t}\)

Solve for \(t\) by taking the reciprocal:

\(t = \dfrac{12}{5} = 2.4 \text{ hours} = 2 \text{ hours 24 minutes}\)

Note that the combined time must always be less than either individual time. If your answer is greater than 4 or 6 hours, something went wrong.

Factor both the numerator and denominator. For the numerator, factor out the GCF of \(3x\):

\(3x^2 + 6x = 3x(x + 2)\)

The denominator is a perfect square trinomial:

\(x^2 + 4x + 4 = (x + 2)^2\)

Write the expression in factored form and cancel one \((x + 2)\) factor:

\(\dfrac{3x(x + 2)}{(x + 2)^2} = \dfrac{3x}{x + 2}\)

Choice C incorrectly cancels the \(x\) from \(3x\) along with the \((x + 2)\). Remember: you can only cancel factors, not individual terms.

To find the LCD, first factor each denominator:

\(x^2 – 4 = (x + 2)(x – 2)\)

\(x^2 – 4x + 4 = (x – 2)^2\)

The LCD must include each factor the greatest number of times it appears in any single denominator:

• \((x + 2)\) appears once (from the first denominator)
• \((x – 2)\) appears at most twice (from the second denominator)

Therefore the LCD is \((x + 2)(x – 2)^2\).

Choice B only uses \((x – 2)\) once, which would not be divisible by \((x – 2)^2\). Choice D uses the maximum power of every factor from both denominators combined, which is larger than necessary.

Factor the first denominator: \(x^2 – 1 = (x + 1)(x – 1)\). The LCD is \((x + 1)(x – 1)\). The first fraction already has this denominator, so only the second fraction needs to be rewritten:

\(\dfrac{x + 3}{(x + 1)(x – 1)} – \dfrac{1 \cdot (x + 1)}{(x – 1)(x + 1)}\)

Subtract the numerators:

\(\dfrac{(x + 3) – (x + 1)}{(x + 1)(x – 1)}\)\(\:= \dfrac{x + 3 – x – 1}{(x + 1)(x – 1)}\)\(\:= \dfrac{2}{(x + 1)(x – 1)}\)

Choice C subtracts only the 1 from the numerator without multiplying by \((x + 1)\) first.