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Algebra 2 Flashcards: Creating Solving One Variable Equations Inequalities

Study Creating Solving One Variable Equations Inequalities in Algebra 2 with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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What this deck covers

This deck focuses on Creating Solving One Variable Equations Inequalities, giving you a quick way to review the definitions, rules, and examples that matter most for Algebra 2.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Algebra 2 Flashcards: Creating Solving One Variable Equations Inequalities

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QUESTION

What value must be excluded from solutions of x+5x2−9=1\frac{x+5}{x^2-9}=1x2−9x+5​=1?

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ANSWER

x≠3x\ne 3x=3 and x≠−3x\ne -3x=−3. Denominator x2−9=(x−3)(x+3)x^2-9=(x-3)(x+3)x2−9=(x−3)(x+3) cannot equal zero.

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Flashcard 1: What value must be excluded from solutions of x+5x2−9=1\frac{x+5}{x^2-9}=1x2−9x+5​=1?

Answer: x≠3x\ne 3x=3 and x≠−3x\ne -3x=−3. Denominator x2−9=(x−3)(x+3)x^2-9=(x-3)(x+3)x2−9=(x−3)(x+3) cannot equal zero.

Flashcard 2: What is the vertex xxx-coordinate of y=x2−8x+1y=x^2-8x+1y=x2−8x+1?

Answer: x=4x=4x=4. Use x=−b2a=−(−8)2(1)=4x=-\frac{b}{2a}=-\frac{(-8)}{2(1)}=4x=−2ab​=−2(1)(−8)​=4.

Flashcard 3: What is the slope-intercept form of a linear equation in one variable context (function form)?

Answer: y=mx+by=mx+by=mx+b. Standard form for linear functions with slope mmm and yyy-intercept bbb.

Flashcard 4: What condition on bbb in y=abxy=ab^xy=abx represents exponential decay?

Answer: 0<b<10<b<10<b<1. Base between 000 and 111 creates decreasing exponential function.

Flashcard 5: What does b2−4ac=0b^2-4ac=0b2−4ac=0 tell you about the real solutions of ax2+bx+c=0ax^2+bx+c=0ax2+bx+c=0?

Answer: One real solution (double root). Zero discriminant means the parabola touches the xxx-axis once.

Flashcard 6: What is the axis of symmetry for y=a(x−h)2+ky=a(x-h)^2+ky=a(x−h)2+k?

Answer: x=hx=hx=h. Vertical line through the vertex of any parabola.

Flashcard 7: What is a simple rational equation form that requires excluding values making the denominator 000?

Answer: p(x)q(x)=k\frac{p(x)}{q(x)}=kq(x)p(x)​=k with q(x)≠0q(x)\ne 0q(x)=0. Fraction equals constant with domain restrictions.

Flashcard 8: What is the key restriction to state when solving 1x−3=5\frac{1}{x-3}=5x−31​=5?

Answer: x≠3x\ne 3x=3. Denominator cannot equal zero in rational equations.

Flashcard 9: What is the general exponential function form used to model growth or decay in one variable?

Answer: y=abxy=ab^xy=abx. Base bbb and initial value aaa for exponential modeling.

Flashcard 10: What is the interval notation for the solution set of x≥3x\ge 3x≥3?

Answer: [3,∞)[3,\infty)[3,∞). Bracket includes 333, parenthesis extends to infinity.

Flashcard 11: What is the vertex xxx-coordinate of y=x2−8x+1y=x^2-8x+1y=x2−8x+1?

Answer: x=4x=4x=4. Use x=−b2a=−(−8)2(1)=4x=-\frac{b}{2a}=-\frac{(-8)}{2(1)}=4x=−2ab​=−2(1)(−8)​=4.

Flashcard 12: What is the solution set of ∣x−1∣≤4|x-1|\le 4∣x−1∣≤4 in interval notation?

Answer: [−3,5][-3,5][−3,5]. Distance from 111 is at most 444 units.

Flashcard 13: What is the solution to the absolute value equation ∣2x+1∣=7|2x+1|=7∣2x+1∣=7?

Answer: x=3x=3x=3 or x=−4x=-4x=−4. Solve 2x+1=72x+1=72x+1=7 and 2x+1=−72x+1=-72x+1=−7 separately.

Flashcard 14: What is the interval notation for the solution set of x<−2x< -2x<−2?

Answer: (−∞,−2)(-\infty,-2)(−∞,−2). Parenthesis excludes −2-2−2, extends from negative infinity.

Flashcard 15: What does b2−4ac<0b^2-4ac<0b2−4ac<0 tell you about the real solutions of ax2+bx+c=0ax^2+bx+c=0ax2+bx+c=0?

Answer: No real solutions (two complex). Negative discriminant means the parabola doesn't cross the xxx-axis.

Flashcard 16: What is the solution to the inequality 2x≥82^{x}\ge 82x≥8?

Answer: x≥3x\ge 3x≥3. Rewrite as 2x≥232^x\ge 2^32x≥23, so x≥3x\ge 3x≥3.

Flashcard 17: What is the solution to the exponential equation 5⋅2x=405\cdot 2^{x}=405⋅2x=40?

Answer: x=3x=3x=3. Divide by 555 to get 2x=8=232^x=8=2^32x=8=23.

Flashcard 18: What is the solution to the exponential equation 3x=193^{x}=\frac{1}{9}3x=91​?

Answer: x=−2x=-2x=−2. Rewrite as 3x=3−23^x=3^{-2}3x=3−2, so x=−2x=-2x=−2.

Flashcard 19: What is the solution to the exponential equation 2x=162^x=162x=16?

Answer: x=4x=4x=4. Rewrite as 2x=242^x=2^42x=24, so x=4x=4x=4.

Flashcard 20: What is the solution to xx−1=2\frac{x}{x-1}=2x−1x​=2, with restrictions stated?

Answer: x=2x=2x=2, with x≠1x\ne 1x=1. Multiply by (x−1)(x-1)(x−1) to get x=2(x−1)x=2(x-1)x=2(x−1).

Flashcard 21: What is the compound interest formula for amount after ttt years with nnn compounds per year?

Answer: A=P(1+rn)ntA=P\left(1+\frac{r}{n}\right)^{nt}A=P(1+nr​)nt. Principal PPP, rate rrr, compounds nnn times, over ttt years.

Flashcard 22: What is the continuous growth formula for amount after time ttt at rate rrr?

Answer: A=PertA=Pe^{rt}A=Pert. Principal PPP grows continuously at rate rrr over time ttt.

Flashcard 23: What inequality symbol results after multiplying both sides of an inequality by a negative number?

Answer: The inequality sign reverses. Multiplying by negative flips the inequality direction.

Flashcard 24: What is the interval notation for the solution set of x≥3x\ge 3x≥3?

Answer: [3,∞)[3,\infty)[3,∞). Bracket includes 333, parenthesis extends to infinity.

Flashcard 25: What is the interval notation for the solution set of x<−2x< -2x<−2?

Answer: (−∞,−2)(-\infty,-2)(−∞,−2). Parenthesis excludes −2-2−2, extends from negative infinity.

Flashcard 26: What is the interval notation for the compound inequality −1<x≤4-1<x\le 4−1<x≤4?

Answer: (−1,4](-1,4](−1,4]. Parenthesis excludes −1-1−1, bracket includes 444.

Flashcard 27: Identify the first step to create an equation from a word problem involving a changing quantity.

Answer: Define a variable for the unknown. Establishes what quantity you're solving for.

Flashcard 28: What operation clears denominators in 2x+1=5\frac{2}{x}+1=5x2​+1=5 to create an equivalent equation?

Answer: Multiply both sides by the LCD. Eliminates fractions by multiplying by common denominator.

Flashcard 29: What is the standard approach to solve ∣x−4∣=9|x-4|=9∣x−4∣=9?

Answer: Solve x−4=9x-4=9x−4=9 and x−4=−9x-4=-9x−4=−9. Absolute value equals positive creates two linear equations.

Flashcard 30: What is the standard approach to solve ∣x−4∣<9|x-4|<9∣x−4∣<9?

Answer: Write −9<x−4<9-9<x-4<9−9<x−4<9. Absolute value less than positive creates compound inequality.