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Quizzes > High School Quizzes > Electives

6 Final Assessment Practice Quiz

Challenge yourself with a dynamic learning experience

Difficulty: Moderate
Grade: Grade 9
Study OutcomesCheat Sheet
Colorful paper art promoting Topic 6 Triumph, a dynamic algebra practice quiz.

Simplify the expression: 3x + 2x.
5x
6x
x
3x
Combine like terms by adding the coefficients. Since 3 + 2 equals 5, the expression simplifies to 5x.
If x = 3, what is the value of 2x + 5?
11
10
9
8
Substitute 3 for x to get 2(3) + 5 which equals 6 + 5 = 11. This problem reinforces the importance of substitution in expressions.
Solve for x: x + 4 = 9.
5
4
9
13
Subtract 4 from both sides to isolate x, giving x = 5. This demonstrates a fundamental step in solving linear equations.
Which of the following equations is in slope-intercept form?
y = 3x + 1
3x + y = 1
x = 3y + 1
1 = 3x + y
Slope-intercept form is written as y = mx + b, where m represents the slope and b represents the y-intercept. Option 1 directly corresponds to this form.
Expand the expression using the distributive property: 2(x + 3).
2x + 6
2x + 3
x + 5
2x + 3x
Apply the distributive property by multiplying 2 with each term inside the parentheses: 2x and 2*3 gives 6, resulting in 2x + 6. This reinforces the basic concept of distribution over addition.
Solve for x: 2x - 3 = 7.
5
7
10
2
Add 3 to both sides to obtain 2x = 10 and then divide both sides by 2 to find x = 5. This question tests the basic method of solving linear equations step by step.
Factor the quadratic expression: x² + 5x + 6.
(x + 2)(x + 3)
(x + 1)(x + 6)
(x - 2)(x - 3)
(x + 3)(x - 2)
The quadratic expression factors into (x + 2)(x + 3) because 2 + 3 equals 5 and 2 multiplied by 3 equals 6. This tests the ability to factor simple quadratic expressions.
Expand and simplify: 4(x - 2) - 3(x + 1).
x - 11
7x - 5
x + 11
7x + 5
First, distribute to get 4x - 8 and -3x - 3, then combine like terms to arrive at x - 11. This reinforces the proper process of distribution and combining like terms.
Solve the equation: 3(2x - 1) = x + 5.
8/5
5/8
2
3
After expanding the left side to obtain 6x - 3 and equating it to x + 5, subtract x from both sides and then add 3 to get 5x = 8, leading to x = 8/5. This challenges students to work carefully through algebraic manipulation.
Solve for x: (5x)/2 = 10.
4
5
2
20
Multiply both sides by 2 to receive 5x = 20, then divide by 5 yielding x = 4. This straightforward problem focuses on inverse operations with fractions.
Simplify the expression: 2x² + 3x².
5x²
6x²
x❴
x²
Add the coefficients for like terms to obtain 5x² because 2 + 3 equals 5. This reinforces the concept that only like terms (same variable and exponent) can be combined.
Solve the system of equations: x + y = 6 and x - y = 2.
(4, 2)
(2, 4)
(3, 3)
(5, 1)
By adding the equations, y is eliminated and you obtain 2x = 8, so x equals 4. Substituting back gives y = 2, resulting in the ordered pair (4, 2).
Solve the inequality: 2x - 5 < 9.
x < 7
x > 7
x < 2
x > 2
Add 5 to both sides to get 2x < 14, then divide by 2 resulting in x < 7. This problem highlights the proper method for solving linear inequalities.
Given f(x) = x² - 4, find f(3).
5
9
3
-5
Substitute x = 3 into the function to get 3² - 4 = 9 - 4 = 5. This reinforces the process of function evaluation.
Factor the expression: 6x² - 9x.
3x(2x - 3)
3(2x - 9x)
x(6x - 9)
3x(2 - 3)
The greatest common factor in both terms is 3x. Factoring 3x out of 6x² - 9x results in 3x(2x - 3), which is the correct factorization.
Solve the quadratic equation: x² - 5x + 6 = 0.
x = 2 or x = 3
x = -2 or x = -3
x = 2 only
x = 3 only
This quadratic factors neatly into (x - 2)(x - 3) = 0, yielding the solutions x = 2 and x = 3. Both roots are valid and must be included in the final answer.
Solve the equation: (x - 1)/(x + 2) = 2/(x + 2).
x = 3
x = -2
x = 1
No solution
Since the denominators are identical (and x cannot be -2 since that would make them undefined), set the numerators equal: x - 1 = 2. Solving this gives x = 3 as the only solution.
Solve for x: |2x - 4| = 6.
x = 5 or x = -1
x = 5 only
x = -1 only
x = 1 or x = -5
The absolute value equation splits into two cases: 2x - 4 = 6 and 2x - 4 = -6. Solving these equations gives x = 5 and x = -1 respectively, both of which are valid solutions.
Solve the equation: (3x - 2)/4 = 2 - (x + 1)/2.
x = 8/5
x = 2
x = 1
x = -8/5
Multiply the entire equation by 4 to eliminate fractions, then simplify and solve the resulting linear equation to find x = 8/5. This problem reinforces careful manipulation of fractions to isolate the variable.
Solve for x: 2(x - 3) + 4 = 3(x + 1) - 2.
x = -3
x = 3
x = -1
x = 1
After expanding both sides, the equation becomes 2x - 6 + 4 = 3x + 3 - 2, which simplifies to 2x - 2 = 3x + 1. Rearranging the terms and solving for x results in x = -3, demonstrating the importance of careful arithmetic manipulation.
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Study Outcomes

  1. Understand key algebraic concepts through practice problems.
  2. Simplify and factor algebraic expressions effectively.
  3. Solve linear equations and inequalities with confidence.
  4. Apply strategies to analyze and graph algebraic functions.
  5. Evaluate problem-solving approaches for test readiness.

Topic 6 Final Assessment Cheat Sheet

  1. Master the Quadratic Formula - Ready to crack any quadratic? Plug values into x = (−b ± √(b² − 4ac)) / (2a) to get lightning-fast solutions and impress your classmates. This trusty formula turns tricky equations into a simple plug-and-play routine. OpenStax: Intermediate Algebra 2e
  2. Understand Factoring Techniques - Factoring is like spotting patterns in a code: recognize a² − b² = (a − b)(a + b) and a² + 2ab + b² = (a + b)² to break down expressions in seconds. Mastering these shortcuts makes solving and simplifying equations feel like a fun puzzle. BYJU'S: Algebra Formulas for Class 9
  3. Grasp the Pythagorean Theorem - In a right triangle, a² + b² = c² tells you how the legs and hypotenuse relate. Whether you're finding a missing side or checking if a triangle is right-angled, this theorem is your ultimate geometry sidekick. Toppers Bulletin: Math Formulas for Grade 9
  4. Learn the Distance Formula - Craving coordinate geometry mastery? Use d = √((x₂ − x₝)² + (y₂ − y₝)²) to measure the straight-line gap between any two points. It's like using a virtual ruler on the xy-plane! Toppers Bulletin: Math Formulas for Grade 9
  5. Apply the Slope Formula - The steepness of a line is just a fraction away: m = (y₂ − y₝) / (x₂ − x₝). This little ratio tells you if your line is rising, falling, or flat - perfect for graphing and algebraic storytelling. Toppers Bulletin: Math Formulas for Grade 9
  6. Explore Arithmetic Sequences - In an arithmetic sequence, add the common difference d each time to get the next term: Tₙ = a + (n − 1)d. Want the sum? Use Sₙ = n/2 [2a + (n − 1)d] to breeze through series problems. Examples.com: Algebraic Formulas
  7. Delve into Geometric Sequences - Multiply by the common ratio r and watch the sequence grow or shrink: Tₙ = ar❿❻¹. To sum the first n terms, apply Sₙ = a(r❿ − 1)/(r − 1) (for r≠1) and marvel at exponential patterns. Examples.com: Algebraic Formulas
  8. Practice Completing the Square - Turn x² + bx into a perfect square by adding and subtracting (b/2)², giving (x + b/2)² − (b/2)². This trick not only solves quadratics but also reveals the vertex of a parabola. Examples.com: Algebraic Formulas
  9. Understand Sum and Difference of Cubes - Factor cubes like a pro: a³ + b³ = (a + b)(a² − ab + b²) and a³ − b³ = (a − b)(a² + ab + b²). These identities simplify cubic expressions into neat, solvable pieces. Examples.com: Algebraic Formulas
  10. Familiarize with the Binomial Theorem - Expand any power of a binomial using (a + b)❿ = Σₖ₌₀❿ (n choose k) a❿❻ᵝ bᵝ. From Pascal's Triangle to probability, this theorem is the algebraic equivalent of a superpower. OpenStax: College Algebra 2e
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