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Quizzes > High School Quizzes > Mathematics

Chapter 9 Algebra 1 Practice Test

Review Chapter 8 Concepts and Strengthen Skills

Difficulty: Moderate
Grade: Grade 8
Study OutcomesCheat Sheet
Colorful paper art promoting the Mid-Chapter Algebra Challenge trivia quiz for high school students.

Solve for x: 2x + 3 = 11.
x = 4
x = 3
x = 5
x = 6
Subtracting 3 from both sides of the equation gives 2x = 8, and dividing both sides by 2 yields x = 4. This is a straightforward linear equation solved by isolation of the variable.
Simplify the expression: 3 + 2(x - 1).
2x + 1
2x - 1
2x + 3
x + 3
Distribute the 2 into (x - 1) to get 2x - 2, then add 3, which results in 2x + 1. This combines like terms in a simple algebraic expression.
Identify the coefficient of x in the expression: 7x - 4.
7
4
-4
1
The coefficient is the number multiplying the variable x. In this expression, 7 is the multiplier for x, making it the coefficient.
Apply the distributive property: What is the expanded form of 3(x + 4)?
3x + 12
3x + 4
x + 12
3x - 12
Multiplying 3 by x gives 3x and multiplying 3 by 4 gives 12, so the expanded expression is 3x + 12. This is a direct application of the distributive property.
Evaluate the expression: 5 - 2(3).
-1
1
11
8
First, calculate 2 multiplied by 3 to get 6, then subtract 6 from 5 resulting in -1. This question tests basic order of operations.
Solve for x: 3x - 5 = 10.
x = 5
x = 15
x = -5
x = 10/3
Adding 5 to both sides gives 3x = 15 and dividing by 3 then yields x = 5. This is a basic linear equation that requires two steps to solve.
Solve for x: 2x + 4 = x + 9.
x = 5
x = 3
x = 9
x = 13
Subtract x from both sides to get x + 4 = 9, then subtract 4 to find x = 5. This question tests the ability to isolate the variable.
Simplify the expression: 4(2x - 3) - 2(x - 5).
6x - 2
6x + 2
8x - 7
2x - 2
Distribute to get 8x - 12 from the first term and -2x + 10 from the second, then combine like terms to obtain 6x - 2. This process demonstrates proper distribution and combination of like terms.
Solve for x: (x/3) + 2 = 5.
x = 9
x = 15
x = 6
x = 7
Subtracting 2 from both sides gives x/3 = 3; multiplying both sides by 3 results in x = 9. This problem reinforces solving equations that include fractions.
Solve for x: -3x = 12.
x = -4
x = 4
x = -36
x = 36
Dividing both sides of the equation by -3 gives x = -4. This illustrates the handling of negative coefficients when isolating variables.
Solve for x: 5x + 7 = 2x + 16.
x = 3
x = 9
x = 7
x = 8
Subtract 2x from both sides to obtain 3x + 7 = 16, then subtract 7 and divide by 3 to get x = 3. This question tests the ability to consolidate like terms and solve a two-step equation.
Simplify the expression: (x^2) * (x^3).
x^5
x^6
x^2
x^3
When multiplying expressions with the same base, you add the exponents. Therefore, (x^2) * (x^3) simplifies to x^(2+3) which is x^5.
Simplify the expression: (2^3) * (2^2).
32
16
64
8
Adding the exponents, (2^3)*(2^2) gives 2^(3+2) which equals 2^5. Since 2^5 is 32, that is the correct answer.
What is the simplified value of √(16)?
4
2
8
16
The square root of 16 is 4 because 4 multiplied by 4 equals 16. This question verifies the understanding of basic radical evaluation.
Solve for x: 2(x + 3) = 14.
x = 4
x = 7
x = 14
x = -4
Dividing both sides by 2 gives x + 3 = 7, and subtracting 3 from both sides yields x = 4. The problem demonstrates solving a simple two-step equation.
Solve for x: (x + 2)/3 = (2x - 1)/5.
x = 13
x = 7
x = 8
x = 11
Cross-multiply to get 5(x + 2) = 3(2x - 1), which simplifies to 5x + 10 = 6x - 3. Subtracting 5x from both sides and then adding 3 yields x = 13.
Factor the quadratic expression: x^2 + 5x + 6.
(x + 2)(x + 3)
(x + 1)(x + 6)
(x - 2)(x - 3)
(x + 3)(x + 3)
The numbers 2 and 3 multiply to give 6 and add to give 5, so the quadratic factors as (x + 2)(x + 3). This factoring process is fundamental to solving quadratic equations.
Solve the quadratic equation: x^2 - 5x + 6 = 0.
x = 2 or x = 3
x = -2 or x = -3
x = 1 or x = 6
x = -1 or x = -6
The given quadratic factors as (x - 2)(x - 3) = 0, which means that x = 2 or x = 3. Factoring is an efficient method to find the solutions of a quadratic equation.
A rectangle has a perimeter of 30 and its length is twice its width. What are its dimensions?
Width = 5, Length = 10
Width = 10, Length = 20
Width = 3, Length = 6
Width = 15, Length = 30
Let the width be w; then the length is 2w and the perimeter is 2(w + 2w) = 6w. Setting 6w equal to 30 gives w = 5 and thus the length is 10.
Solve for y: 4(y - 2) - 3(y + 1) = 2y - 8.
y = -3
y = 3
y = -5
y = 5
Expanding the left side gives 4y - 8 - 3y - 3, which simplifies to y - 11. Setting y - 11 equal to 2y - 8 and solving for y yields y = -3. This question combines distribution and combining like terms.
0
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Study Outcomes

  1. Analyze and simplify algebraic expressions using fundamental algebra principles.
  2. Solve linear equations and inequalities by applying appropriate methods.
  3. Evaluate and interpret algebraic functions to determine key characteristics.
  4. Apply problem-solving strategies to approach and solve algebra challenges.
  5. Assess personal understanding to identify strengths and areas for improvement in algebra concepts.

Algebra 1 Chapter 9 Mid Test Answers Cheat Sheet

  1. Standard form of a quadratic function - Quadratics rock the mathematical world with their signature shape: y = ax2 + bx + c, where a ≠ 0. This formula tells you if the parabola opens up (a > 0) or down (a < 0) and sets the stage for all your graphing adventures. Dive into standard form
    2. Quizlet Flashcards
  2. Axis of symmetry - Think of the axis of symmetry as your math mirror: it's the vertical line x = –b/(2a) that splits the parabola in two perfect halves. Every point on one side reflects exactly to a matching point on the other. Mastering this concept helps you find the vertex in a snap! Explore the axis
    3. Quizlet Flashcards
  3. Graphing to find roots - Roots (or zeros) are where the parabola crashes through the x-axis. By sketching the curve and spotting the intercepts, you visually solve the quadratic without a formula - perfect for pattern lovers! See graphing tips
    4. Quizlet Flashcards
  4. Completing the square - Transform x2 + bx into a perfect square by adding (b/2)2 to both sides. This trick not only solves equations but also reveals the vertex form for maximum clarity. Master the square
    5. Quizlet Flashcards
  5. Quadratic formula - With x = [ - b ± √(b2 - 4ac)] / (2a), you have a universal key to unlock any quadratic equation. Memorize it once and solve forever - no more guessing! Unlock the formula
    6. Quizlet Flashcards
  6. Graph transformations - Shifts, stretches, and reflections make each quadratic uniquely yours. Adding c moves it up/down, changing a stretches or squeezes, and flipping the sign flips it - play with parameters for instant visual feedback! Transform your graph
    7. Quizlet Flashcards
  7. Special functions - The absolute value f(x)=|x| creates a neat V-shape, while the greatest integer f(x)=[x] hops stepwise along. Both expand your understanding of function behavior beyond smooth curves. Check out special cases
    8. Quizlet Flashcards
  8. Discriminant magic - b2 - 4ac reveals root secrets: positive means two real hits, zero means a single special touch, and negative signals a complex party underground. It's your quadratic mood detector! Decode the discriminant
    9. Quizlet Flashcards
  9. Double root phenomenon - When b2 - 4ac = 0, both solutions collapse into one point, and the parabola merely kisses the x-axis. Recognizing this makes solving feel like discovering a hidden gem. Spot double roots
    10. Quizlet Flashcards
  10. Vertex: max or min - The vertex is your turning point: the lowest dip if a > 0 or the highest peak if a < 0. Use x = - b/(2a) and plug back in to snag its exact coordinates. Find the vertex
    11. Quizlet Flashcards
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