Unit 7 · Checkpoint quiz

Unit 7 checkpoint: Rational expressions

12 questions · about 25-30 minutes · no time limit

Answer every question, then press Grade my answers at the bottom. You'll see what you got right, a worked solution for each question, and links to the lessons worth revisiting. Pencil and paper are encouraged.

This is practice, not a test of you. A low score just shows where to look next.

Simplify \(\dfrac{x^2 - 9}{x^2 + x - 6}\).

Factor the top (difference of squares): \(x^2 - 9 = (x - 3)(x + 3)\).

Factor the bottom: \(x^2 + x - 6 = (x + 3)(x - 2)\).

Cancel the common factor \((x + 3)\): \(\dfrac{(x - 3)(x + 3)}{(x + 3)(x - 2)} = \dfrac{x - 3}{x - 2}\).

Which values of \(x\) must be excluded from \(\dfrac{x + 5}{x^2 - 2x - 15}\)?

\(x = -5\) and \(x = 3\)
\(x = -5\) and \(x = 5\)
\(x = 5\) and \(x = -3\)
\(x = -3\) and \(x = 15\)

Excluded values make the denominator 0. The numerator doesn't matter here.

Factor the denominator: \(x^2 - 2x - 15 = (x - 5)(x + 3)\).

\(x - 5 = 0\) gives \(x = 5\); \(x + 3 = 0\) gives \(x = -3\). Answer: (c).

Simplify \(\dfrac{6 - 2x}{x^2 - 9}\).

Factor the top: \(6 - 2x = 2(3 - x) = -2(x - 3)\). (Taking out \(-2\) instead of \(2\) flips \(3 - x\) into \(x - 3\), which matches the bottom.)

Factor the bottom: \(x^2 - 9 = (x - 3)(x + 3)\).

Cancel \((x - 3)\): \(\dfrac{-2(x - 3)}{(x - 3)(x + 3)} = -\dfrac{2}{x + 3}\).

Multiply and simplify: \(\dfrac{x^2 - 4}{3x} \cdot \dfrac{6x^2}{x + 2}\).

Factor: \(x^2 - 4 = (x - 2)(x + 2)\). So the product is \(\dfrac{(x - 2)(x + 2)}{3x} \cdot \dfrac{6x^2}{x + 2}\).

Cancel \((x + 2)\), and cancel \(3x\) into \(6x^2\): \(\dfrac{6x^2}{3x} = 2x\).

Answer: \(2x(x - 2)\), which is also \(2x^2 - 4x\).

Divide and simplify: \(\dfrac{x^2 + 5x + 6}{x^2 - 1} \div \dfrac{x + 3}{x - 1}\).

Flip the second fraction and multiply: \(\dfrac{x^2 + 5x + 6}{x^2 - 1} \cdot \dfrac{x - 1}{x + 3}\).

Factor: \(\dfrac{(x + 2)(x + 3)}{(x - 1)(x + 1)} \cdot \dfrac{x - 1}{x + 3}\).

Cancel \((x + 3)\) and \((x - 1)\). Answer: \(\dfrac{x + 2}{x + 1}\).

Subtract and simplify: \(\dfrac{3x + 1}{x + 2} - \dfrac{x - 3}{x + 2}\).

Same denominator, so subtract the numerators. Put the second numerator in parentheses: \(\dfrac{3x + 1 - (x - 3)}{x + 2}\).

Distribute the minus: \(3x + 1 - x + 3 = 2x + 4\).

Factor and cancel: \(\dfrac{2x + 4}{x + 2} = \dfrac{2(x + 2)}{x + 2} = 2\).

Subtract and simplify: \(\dfrac{2}{x - 3} - \dfrac{12}{x^2 - 9}\).

Factor: \(x^2 - 9 = (x - 3)(x + 3)\), so the LCD is \((x - 3)(x + 3)\).

Rewrite the first fraction: \(\dfrac{2(x + 3)}{(x - 3)(x + 3)} - \dfrac{12}{(x - 3)(x + 3)} = \dfrac{2x + 6 - 12}{(x - 3)(x + 3)} = \dfrac{2x - 6}{(x - 3)(x + 3)}\).

Factor the top and cancel: \(\dfrac{2(x - 3)}{(x - 3)(x + 3)} = \dfrac{2}{x + 3}\).

Simplify the complex fraction \(\dfrac{1 + \dfrac{2}{x}}{1 - \dfrac{4}{x^2}}\).

The LCD of all the small fractions is \(x^2\). Multiply the top and the bottom by \(x^2\).

Top: \(x^2 \cdot 1 + x^2 \cdot \dfrac{2}{x} = x^2 + 2x\). Bottom: \(x^2 \cdot 1 - x^2 \cdot \dfrac{4}{x^2} = x^2 - 4\).

Factor and cancel: \(\dfrac{x(x + 2)}{(x - 2)(x + 2)} = \dfrac{x}{x - 2}\).

Solve \(\dfrac{2x}{x - 3} = x + \dfrac{6}{x - 3}\). Check for extraneous solutions.

Note \(x \ne 3\). Multiply every term by the LCD \((x - 3)\): \(2x = x(x - 3) + 6\).

Expand and move everything to one side: \(2x = x^2 - 3x + 6\), so \(x^2 - 5x + 6 = 0\), which is \((x - 2)(x - 3) = 0\).

That gives \(x = 2\) or \(x = 3\). But \(x = 3\) makes a denominator 0, so it is extraneous.

Check \(x = 2\): left \(\dfrac{4}{-1} = -4\); right \(2 + \dfrac{6}{-1} = -4\). Answer: \(x = 2\) only.

Ana can paint a fence in 4 hours. Ben can paint the same fence in 6 hours. How many hours will it take them working together?

In one hour Ana paints \(\dfrac{1}{4}\) of the fence and Ben paints \(\dfrac{1}{6}\). Together, in \(t\) hours: \(\dfrac{t}{4} + \dfrac{t}{6} = 1\).

Multiply by the LCD 12: \(3t + 2t = 12\), so \(5t = 12\).

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

Factor \(3x^2 - 10x - 8\).

ac method: \(3 \cdot (-8) = -24\). Two numbers that multiply to \(-24\) and add to \(-10\): \(-12\) and \(2\).

Split and group: \(3x^2 - 12x + 2x - 8 = 3x(x - 4) + 2(x - 4)\).

Answer: \((3x + 2)(x - 4)\).

Simplify \(\dfrac{12x^{-2}y^4}{3x^3y^{-1}}\). Write your answer with positive exponents only.

Numbers: \(12 \div 3 = 4\).

Subtract exponents: \(x^{-2-3} = x^{-5}\) and \(y^{4-(-1)} = y^5\).

Move \(x^{-5}\) to the bottom: answer \(\dfrac{4y^5}{x^5}\).

Want to look something up first? Rational expressions cheat sheet · Glossary

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