Unlock hundreds more features
Save your Quiz to the Dashboard
View and Export Results
Use AI to Create Quizzes and Analyse Results

OR
Sign inSign in with Facebook
Sign inSign in with Google
Quizzes > High School Quizzes > Mathematics

Word Problems with Rational Expressions Quiz

Sharpen problem-solving skills with clear practice steps

Difficulty: Moderate
Grade: Grade 9
Study OutcomesCheat Sheet
Paper art depicting a trivia quiz on Rational Word Riddles for high school students.

What is the domain of the rational expression 1/(x - 5)?
x ≠5
x = 5
x can be any number
x must be 5
The domain of a rational expression excludes values that make the denominator zero. Since x - 5 equals zero when x is 5, x cannot be 5.
A water tank fills at a rate modeled by R(t) = 1/(t - 2) gallons per minute. What is the restriction on t?
t = 2
t > 2
t < 2
t ≠2
The denominator t - 2 cannot be zero, which would occur when t equals 2. Therefore, t must not be 2 to keep the expression defined.
Simplify the expression: (2/x) + (3/x).
3/x
2/x
x/5
5/x
Since the denominators are the same, simply add the numerators: 2 + 3 equals 5. The simplified expression is 5/x.
Simplify the rational expression (4x² - 9)/(2x + 3) and state the restriction on x.
2x + 3, with x ≠-3/2
2x - 3, with x ≠-3/2
4x - 9, with x ≠-3/2
2x - 3, with x ≠3/2
Factor the numerator as (2x - 3)(2x + 3) and notice the common factor (2x + 3) in the denominator. Canceling this factor yields 2x - 3, but x must not be -3/2 because that value would make 2x + 3 equal to zero.
Which of the following best defines a rational expression?
A polynomial that can be factored
A fraction where either the numerator or denominator is a polynomial
A fraction where both the numerator and denominator are polynomials
An irrational number expressed as a fraction
A rational expression is defined as a quotient of two polynomials. Both the numerator and the denominator must be polynomials, which is exactly what the correct answer states.
A worker's time is modeled by T(x) = (x² - 9)/(x - 3) minutes. Simplify T(x) and state the restriction on x.
x - 3, with x ≠-3
x + 3, with x ≠3
x + 3, with x ≠-3
x - 3, with x ≠3
Factor the numerator as (x - 3)(x + 3), which cancels with the denominator x - 3, leaving x + 3. The restriction x ≠3 is necessary to avoid division by zero.
Simplify the rational expression (x² - 16)/(x² - x - 12) and identify the restrictions on x.
(x + 4)/(x + 3) where x ≠4 and x ≠-3
(x - 4)/(x + 3) with x ≠-4
(x + 4)/(x - 3) where x ≠4 and x ≠3
(x + 16)/(x + 12)
Factor the numerator to (x - 4)(x + 4) and the denominator to (x - 4)(x + 3). Canceling the common factor (x - 4) gives (x + 4)/(x + 3), with restrictions x ≠4 and x ≠-3 to prevent division by zero.
A phone plan's per-minute charge is given by the expression (3m + 6)/(m + 2). Simplify this expression.
3, with m ≠-2
3m, with m ≠2
m + 3
3/(m + 2)
Factor the numerator as 3(m + 2); the (m + 2) term cancels with the denominator, leaving 3. Note that m cannot equal -2, as that would make the denominator zero.
Solve the equation: 1/(x + 1) + 1/(x - 1) = 2/(x² - 1). What is the solution set?
No solution
x = 0
x = -1
x = 1
Clearing the denominators leads to an apparent solution x = 1; however, substituting x = 1 into the original equation causes division by zero. Thus, the solution is extraneous and there is no valid solution.
Solve for x: 3/(x + 1) = (x + 2)/(x(x + 1)).
x = 0
x = 2
x = 1
x = -1
Multiplying both sides by x(x + 1) removes the common denominator and simplifies the equation to 3x = x + 2. Solving yields x = 1, with the necessary restrictions that x ≠0 and x ≠-1.
If a garden's length is expressed as L = (2x)/(x - 3) and its width as W = (x + 1)/(x - 3), express the garden's area as a single rational expression.
(2x(x + 1))/(x - 3)²
(2x + x + 1)/(x - 3)
2x(x + 1)/(x - 3)
(2x(x - 3))/(x + 1)
The area is the product of length and width. Multiplying (2x)/(x - 3) by (x + 1)/(x - 3) gives (2x(x + 1))/(x - 3)², noting that x cannot equal 3.
A water outlet is modeled by R(x) = (x² - 4)/(x - 2). Simplify R(x) and specify any restrictions on x.
x + 2, with x ≠2
x + 2, with x ≠-2
x - 2, with x ≠-2
x - 2, with x ≠2
Factor the numerator as (x - 2)(x + 2) and cancel the common factor with the denominator. The simplified expression is x + 2, but x must not be 2 to avoid division by zero.
For the rational expression f(x) = (x² + bx + c)/(x + 3) to have a removable discontinuity at x = -3, what condition must be met?
b + c = 0
9 + 3b + c = 0
9 - 3b + c = 0
b - c = 0
A removable discontinuity occurs when both the numerator and denominator equal zero at a point. Setting x = -3 in the numerator gives 9 - 3b + c = 0, which is the required condition.
Simplify the sum: 1/(x + 2) + 2/(x - 2) into a single rational expression.
(x + 2)/(3x + 2)
(3x + 2)/((x + 2)(x - 2))
(3x + 2)/(x + 2)
(3x - 2)/((x + 2)(x - 2))
Find a common denominator, which is (x + 2)(x - 2), and combine the numerators to obtain 3x + 2. The final expression is (3x + 2)/((x + 2)(x - 2)).
Solve the rational equation: x/(x - 1) = (x + 2)/(x - 1) - 2.
x = -2
x = 4
x = 2
x = 1
With a common denominator of (x - 1), equate the numerators after rewriting the right side as a single fraction. Solving the resulting equation gives x = 2, keeping in mind x ≠1 to avoid division by zero.
Solve the rational equation: (2x/(x - 2)) + (3/(x + 1)) = (5x + 7)/((x - 2)(x + 1)).
x = √(13/2)
x = 13/2
x = √(13/2) and x = -√(13/2)
x = -√(13/2)
Multiplying both sides by (x - 2)(x + 1) clears the denominators and produces the equation 2x² - 13 = 0. Solving for x yields x² = 13/2, so x = ±√(13/2), assuming these do not make any original denominator zero.
The cost function is given by C(x) = (x² - k²)/(x - k). If C(x) simplifies to x + k for x ≠k and the limit as x approaches k equals 10, what is the value of k?
5
2
0
10
The expression factors as (x - k)(x + k) and simplifies to x + k, so the limit as x approaches k is 2k. Setting 2k equal to 10 gives k = 5.
Solve the rational equation: 1/(x + 3) + 2/(x - 1) = 3/(x + 1).
x = 2
x = -7
x = -2
x = 7
Clearing the denominators by multiplying through by (x + 3)(x - 1)(x + 1) results in a linear equation that simplifies to 2x = -14, so x = -7. This solution does not violate any restrictions from the original denominators.
For the function f(x) = (2x² + ax + b)/(x² - 4) to have a removable discontinuity at x = 2, what condition must be met?
8 + 2a + b = 0
8 - 2a + b = 0
4 + a + b = 0
2a + b = 0
A removable discontinuity at x = 2 requires that the numerator also be zero when x = 2. Substituting x = 2 into the numerator gives 8 + 2a + b = 0, which is the necessary condition.
If the expression (x² + 5x + k)/(x + 2) divides evenly with no remainder, what must be the value of k?
k = 5
k = -5
k = -6
k = 6
For the division to have no remainder, the numerator must be zero when x = -2. Substituting x = -2 into x² + 5x + k yields 4 - 10 + k = 0, so k - 6 = 0, and therefore k must be 6.
0
{"name":"What is the domain of the rational expression 1\/(x - 5)?", "url":"https://www.quiz-maker.com/QPREVIEW","txt":"What is the domain of the rational expression 1\/(x - 5)?, A water tank fills at a rate modeled by R(t) = 1\/(t - 2) gallons per minute. What is the restriction on t?, Simplify the expression: (2\/x) + (3\/x).","img":"https://www.quiz-maker.com/3012/images/ogquiz.png"}

Study Outcomes

  1. Analyze word problems by identifying key components of rational expressions.
  2. Apply algebraic techniques to solve rational expression puzzles.
  3. Interpret language clues to form corresponding rational equation models.
  4. Synthesize logical reasoning with mathematical strategies to verify solutions.

Word Problems: Rational Expressions Cheat Sheet

  1. Understand Rational Expressions - Rational expressions are fractions where both the numerator and the denominator are polynomials, like (x² − 1)/(x + 1). They pop up everywhere in algebra and beyond, modeling everything from motion to mixing problems. Embrace them as your new algebraic BFFs! SparkNotes: Rational Expressions
  2. Simplify Rational Expressions - Start by factoring both the numerator and denominator to spot common factors you can cancel, like turning (x² − 9)/(x² − 4) into ((x − 3)(x + 3))/((x − 2)(x + 2)). This not only tidies up your expressions but also reveals hidden structure. Simplifying makes future operations much smoother! MathBits Notebook: Simplify Practice
  3. Perform Operations with Rational Expressions - Want to add or subtract? Find a common denominator, rewrite each fraction, then combine like terms. Multiplication is a breeze - just multiply across top and bottom - and division means flipping the second fraction and multiplying. With these skills in your toolkit, you'll conquer any rational expression challenge! PDESAS: Rational Expression Operations
  4. Solve Rational Equations - Multiply both sides by the least common denominator (LCD) to clear fractions, turning the problem into a simpler polynomial equation. Solve normally, but don't forget to check that your answers don't zero out any denominators. Catching extraneous solutions is the final triumph in your algebra adventure! OpenStax: Solve Rational Equations
  5. Apply to Word Problems - Translate real-life scenarios - like work rates or travel times - into algebraic equations using rational expressions. Assign variables thoughtfully and set up your fraction-based equations to match the story. Solving these enriches your understanding and makes math feel like decoding a mystery! GreeneMath: Word Problems
  6. Understand Domain Restrictions - Identify values that make any denominator zero and exclude them from your solution set, because division by zero is a no-go. For example, in 1/(x − 2), x cannot be 2 - watch out or your equation crashes! Respecting domain rules keeps your answers valid and mathematically sound. SparkNotes: Domain Restrictions
  7. Practice with Real-Life Scenarios - Test your skills by tackling problems that involve mixtures, concentrations, and proportional relationships. Seeing how rational expressions model chemistry labs or financial interest rates makes abstract concepts concrete. Plus, it's way more fun than just flipping symbols on a page! OpenStax: Applications of Rational Expressions
  8. Master Factoring Techniques - Sharpen your ability to factor common patterns like GCFs, trinomials, and differences of squares so you can simplify and solve efficiently. The better you get at factoring, the quicker huge expressions transform into manageable parts. Consider it the secret sauce that powers your rational expression skills! MathBits Notebook: Factoring Techniques
  9. Check for Extraneous Solutions - After solving, plug your answers back into the original equation to ensure they don't turn any denominator into zero. It's like a reality check that separates valid solutions from impostors. This final step is crucial to guarantee your solutions pass with flying colors! OpenStax: Extraneous Solutions Guide
  10. Utilize Online Resources for Practice - Reinforce your learning with interactive quizzes, step-by-step tutorials, and challenging problem sets available online. Regular practice boosts your confidence and helps the concepts stick like glue. Embrace virtual tutors and video guides to make studying feel like a game! Lamar University: Rational Expressions Practice
Make a Quiz (FREE) Copy & Edit This Quiz Create a Quiz with AI

Mathematics Quizzes

Powered by: Quiz Maker