Key Concepts & Formulas to Memorize
Unit 1
Expressions & Evaluation
Unit 2
Linear Equations
Unit 3
Inequalities
Unit 4
Linear Functions & Graphing
Unit 5
Systems of Equations
Unit 6
Exponents & Polynomials
Unit 7
Factoring
Unit 8
Quadratic Equations
Unit 9
Functions & Domain
Unit 10
Sequences, Ratios & Abs. Value
📌 Must-Memorize Formulas
Slope
\(m = \dfrac{y_2 - y_1}{x_2 - x_1}\)
Slope-Intercept Form
\(y = mx + b\)
Point-Slope Form
\(y - y_1 = m(x - x_1)\)
Quadratic Formula
\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Arithmetic Sequence
\(a_n = a_1 + (n-1)d\)
Exponent Product Rule
\(x^m \cdot x^n = x^{m+n}\)
Difference of Squares
\(a^2 - b^2 = (a+b)(a-b)\)
FOIL Rule
\((a+b)(c+d) = ac + ad + bc + bd\)
✏️ Worked Example — Quadratic Formula
Solve: \(x^2 + 2x - 8 = 0\)
Step 1: Identify \(a=1,\; b=2,\; c=-8\)
Step 2: Discriminant \(= b^2 - 4ac = 4 + 32 = 36\)
Step 3: \(x = \dfrac{-2 \pm 6}{2}\)
Step 2: Discriminant \(= b^2 - 4ac = 4 + 32 = 36\)
Step 3: \(x = \dfrac{-2 \pm 6}{2}\)
✓ Answer: \(x = 2\) or \(x = -4\)
Practice Questions
1
Unit 1 · Expressions
What is the value of \(3x^2 - 2x + 5\) when \(x = -2\)?
Substitute \(x = -2\):
\(3(-2)^2 - 2(-2) + 5\)
\(= 3(4) + 4 + 5\)
\(= 12 + 4 + 5 = \mathbf{21}\)
\(3(-2)^2 - 2(-2) + 5\)
\(= 3(4) + 4 + 5\)
\(= 12 + 4 + 5 = \mathbf{21}\)
2
Unit 2 · Linear Equations
Solve for \(x\): \(\quad 2x + 7 = 15\)
\(2x + 7 = 15\)
Subtract 7 from both sides: \(2x = 8\)
Divide both sides by 2: \(x = \mathbf{4}\)
Subtract 7 from both sides: \(2x = 8\)
Divide both sides by 2: \(x = \mathbf{4}\)
3
Unit 2 · Linear Equations
Solve for \(x\): \(\quad \dfrac{x}{3} + 4 = 7\)
\(\dfrac{x}{3} + 4 = 7\)
Subtract 4: \(\dfrac{x}{3} = 3\)
Multiply both sides by 3: \(x = \mathbf{9}\)
Subtract 4: \(\dfrac{x}{3} = 3\)
Multiply both sides by 3: \(x = \mathbf{9}\)
4
Unit 3 · Inequalities
Which inequality represents the solution to \(3x - 5 > 7\)?
\(3x - 5 > 7\)
Add 5 to both sides: \(3x > 12\)
Divide by 3: \(x > \mathbf{4}\)
Add 5 to both sides: \(3x > 12\)
Divide by 3: \(x > \mathbf{4}\)
5
Unit 4 · Linear Functions
What is the slope of the line passing through \((-1,\, 3)\) and \((3,\, 11)\)?
\(m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{11 - 3}{3 - (-1)} = \dfrac{8}{4} = \mathbf{2}\)
6
Unit 4 · Linear Functions
Which equation describes the line with slope \(2\) that passes through the point \((3,\, 1)\)?
Use point-slope: \(y - 1 = 2(x - 3)\)
\(y - 1 = 2x - 6\)
\(y = 2x - 5\)
Check: when \(x=3\), \(y = 6 - 5 = 1\) ✓
\(y - 1 = 2x - 6\)
\(y = 2x - 5\)
Check: when \(x=3\), \(y = 6 - 5 = 1\) ✓
7
Unit 5 · Systems of Equations
What is the solution to the system?
\[\begin{cases} x + y = 8 \\ x - y = 2 \end{cases}\]
Add the two equations:
\((x+y) + (x-y) = 8 + 2\)
\(2x = 10 \Rightarrow x = 5\)
Substitute: \(5 + y = 8 \Rightarrow y = 3\)
Solution: \(\mathbf{(5,\, 3)}\)
\((x+y) + (x-y) = 8 + 2\)
\(2x = 10 \Rightarrow x = 5\)
Substitute: \(5 + y = 8 \Rightarrow y = 3\)
Solution: \(\mathbf{(5,\, 3)}\)
8
Unit 6 · Exponents
Simplify: \(\quad x^3 \cdot x^5\)
Product Rule: when multiplying like bases, add the exponents.
\(x^3 \cdot x^5 = x^{3+5} = \mathbf{x^8}\)
\(x^3 \cdot x^5 = x^{3+5} = \mathbf{x^8}\)
9
Unit 6 · Polynomials
Simplify: \(\quad \dfrac{12x^3 y^2}{4x y^3}\)
\(\dfrac{12x^3 y^2}{4x y^3} = \dfrac{12}{4} \cdot x^{3-1} \cdot y^{2-3} = 3x^2 \cdot y^{-1} = \mathbf{\dfrac{3x^2}{y}}\)
10
Unit 6 · Polynomials (FOIL)
Expand: \(\quad (x + 3)(x - 5)\)
FOIL: First · Outer · Inner · Last
\((x)(x) + (x)(-5) + (3)(x) + (3)(-5)\)
\(= x^2 - 5x + 3x - 15\)
\(= \mathbf{x^2 - 2x - 15}\)
\((x)(x) + (x)(-5) + (3)(x) + (3)(-5)\)
\(= x^2 - 5x + 3x - 15\)
\(= \mathbf{x^2 - 2x - 15}\)
11
Unit 7 · Factoring Trinomials
Factor completely: \(\quad x^2 + 7x + 12\)
Find two numbers that multiply to 12 and add to 7.
\(3 \times 4 = 12\) and \(3 + 4 = 7\) ✓
Therefore: \(\mathbf{(x + 3)(x + 4)}\)
Note: (A) and (D) give the same product but the question uses (C) as the standard form.
\(3 \times 4 = 12\) and \(3 + 4 = 7\) ✓
Therefore: \(\mathbf{(x + 3)(x + 4)}\)
Note: (A) and (D) give the same product but the question uses (C) as the standard form.
12
Unit 7 · Difference of Squares
Factor: \(\quad x^2 - 25\)
Difference of squares pattern: \(a^2 - b^2 = (a+b)(a-b)\)
Here \(a = x\), \(b = 5\):
\(x^2 - 25 = \mathbf{(x+5)(x-5)}\)
Verify: \((x+5)(x-5) = x^2 - 5x + 5x - 25 = x^2 - 25\) ✓
Here \(a = x\), \(b = 5\):
\(x^2 - 25 = \mathbf{(x+5)(x-5)}\)
Verify: \((x+5)(x-5) = x^2 - 5x + 5x - 25 = x^2 - 25\) ✓
13
Unit 8 · Quadratic Formula
Using the quadratic formula, solve: \(\quad x^2 + 2x - 8 = 0\)
\(a=1,\; b=2,\; c=-8\)
\(\Delta = b^2 - 4ac = 4 + 32 = 36\)
\(x = \dfrac{-2 \pm \sqrt{36}}{2} = \dfrac{-2 \pm 6}{2}\)
\(x = \dfrac{4}{2} = 2\) or \(x = \dfrac{-8}{2} = -4\)
Answer: \(\mathbf{x = 2 \text{ or } x = -4}\)
\(\Delta = b^2 - 4ac = 4 + 32 = 36\)
\(x = \dfrac{-2 \pm \sqrt{36}}{2} = \dfrac{-2 \pm 6}{2}\)
\(x = \dfrac{4}{2} = 2\) or \(x = \dfrac{-8}{2} = -4\)
Answer: \(\mathbf{x = 2 \text{ or } x = -4}\)
14
Unit 8 · Completing the Square
Solve by completing the square: \(\quad x^2 + 6x + 5 = 0\)
\(x^2 + 6x = -5\)
Add \(\left(\dfrac{6}{2}\right)^2 = 9\) to both sides:
\(x^2 + 6x + 9 = 4\)
\((x + 3)^2 = 4\)
\(x + 3 = \pm 2\)
\(x = -1\) or \(x = -5\) Answer: \(\mathbf{x = -1 \text{ or } x = -5}\)
Add \(\left(\dfrac{6}{2}\right)^2 = 9\) to both sides:
\(x^2 + 6x + 9 = 4\)
\((x + 3)^2 = 4\)
\(x + 3 = \pm 2\)
\(x = -1\) or \(x = -5\) Answer: \(\mathbf{x = -1 \text{ or } x = -5}\)
15
Unit 9 · Functions
If \(f(x) = x^2 - 3x + 1\), what is \(f(4)\)?
Substitute \(x = 4\):
\(f(4) = (4)^2 - 3(4) + 1\)
\(= 16 - 12 + 1\)
\(= \mathbf{5}\)
\(f(4) = (4)^2 - 3(4) + 1\)
\(= 16 - 12 + 1\)
\(= \mathbf{5}\)
16
Unit 9 · Domain of Functions
What is the domain of the function \(f(x) = \sqrt{x - 3}\,\)?
The expression under a square root must be non-negative:
\(x - 3 \geq 0\)
\(x \geq 3\)
Domain: \(\mathbf{[3,\, \infty)}\), or \(\mathbf{x \geq 3}\).
Note: \(x = 3\) is included because \(\sqrt{0} = 0\) is defined.
\(x - 3 \geq 0\)
\(x \geq 3\)
Domain: \(\mathbf{[3,\, \infty)}\), or \(\mathbf{x \geq 3}\).
Note: \(x = 3\) is included because \(\sqrt{0} = 0\) is defined.
17
Unit 10 · Arithmetic Sequences
An arithmetic sequence has first term \(a_1 = 3\) and common difference \(d = 4\). What is the 8th term?
Formula: \(a_n = a_1 + (n-1)d\)
\(a_8 = 3 + (8-1)(4)\)
\(= 3 + 7 \times 4\)
\(= 3 + 28 = \mathbf{31}\)
\(a_8 = 3 + (8-1)(4)\)
\(= 3 + 7 \times 4\)
\(= 3 + 28 = \mathbf{31}\)
18
Unit 10 · Ratios & Percents
30 is what percent of 120?
\(\text{Percent} = \dfrac{\text{part}}{\text{whole}} \times 100\)
\(= \dfrac{30}{120} \times 100\)
\(= 0.25 \times 100 = \mathbf{25\%}\)
\(= \dfrac{30}{120} \times 100\)
\(= 0.25 \times 100 = \mathbf{25\%}\)
19
Unit 10 · Absolute Value
Solve: \(\quad |2x - 3| = 7\)
Split into two cases:
Case 1: \(2x - 3 = 7 \Rightarrow 2x = 10 \Rightarrow x = 5\)
Case 2: \(2x - 3 = -7 \Rightarrow 2x = -4 \Rightarrow x = -2\)
Verify: \(|2(5)-3| = |7| = 7\) ✓ \(|2(-2)-3| = |-7| = 7\) ✓
Answer: \(\mathbf{x = 5 \text{ or } x = -2}\)
Case 1: \(2x - 3 = 7 \Rightarrow 2x = 10 \Rightarrow x = 5\)
Case 2: \(2x - 3 = -7 \Rightarrow 2x = -4 \Rightarrow x = -2\)
Verify: \(|2(5)-3| = |7| = 7\) ✓ \(|2(-2)-3| = |-7| = 7\) ✓
Answer: \(\mathbf{x = 5 \text{ or } x = -2}\)
20
Unit 2 · Word Problems
A taxi charges a flat fee of $3.00 plus $1.50 per mile. If the total fare is $12.00, how many miles was the ride?
Let \(m\) = number of miles.
Equation: \(3 + 1.5m = 12\)
Subtract 3: \(1.5m = 9\)
Divide by 1.5: \(m = \mathbf{6}\)
Check: \(3 + 1.5(6) = 3 + 9 = 12\) ✓
Equation: \(3 + 1.5m = 12\)
Subtract 3: \(1.5m = 9\)
Divide by 1.5: \(m = \mathbf{6}\)
Check: \(3 + 1.5(6) = 3 + 9 = 12\) ✓
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Answer Key & Solutions
Q1 — Answer: D (21)
3(−2)² − 2(−2) + 5 = 12 + 4 + 5 = 21
Q2 — Answer: B (x = 4)
2x + 7 = 15 → 2x = 8 → x = 4
Q3 — Answer: C (x = 9)
x/3 + 4 = 7 → x/3 = 3 → x = 9
Q4 — Answer: C (x > 4)
3x − 5 > 7 → 3x > 12 → x > 4
Q5 — Answer: B (m = 2)
m = (11−3)/(3−(−1)) = 8/4 = 2
Q6 — Answer: A (y = 2x − 5)
y − 1 = 2(x − 3) → y = 2x − 5
Q7 — Answer: C ((5, 3))
Add equations: 2x = 10 → x = 5, y = 3
Q8 — Answer: B (x⁸)
x³ · x⁵ = x^(3+5) = x⁸
Q9 — Answer: D (3x²/y)
12x³y² / 4xy³ = 3x² · y⁻¹ = 3x²/y
Q10 — Answer: A (x² − 2x − 15)
(x+3)(x−5) = x²−5x+3x−15 = x²−2x−15
Q11 — Answer: C ((x+3)(x+4))
Find factors of 12 that add to 7: 3 × 4 = 12, 3 + 4 = 7
Q12 — Answer: B ((x+5)(x−5))
x² − 25 = (x+5)(x−5), difference of squares
Q13 — Answer: A (x = 2 or x = −4)
Discriminant = 36; x = (−2 ± 6)/2 = 2 or −4
Q14 — Answer: D (x = −1 or x = −5)
(x+3)² = 4 → x+3 = ±2 → x = −1 or −5
Q15 — Answer: B (5)
f(4) = 16 − 12 + 1 = 5
Q16 — Answer: A (x ≥ 3)
x − 3 ≥ 0 → x ≥ 3 (radicand must be non-negative)
Q17 — Answer: C (31)
a₈ = 3 + 7(4) = 3 + 28 = 31
Q18 — Answer: B (25%)
30/120 × 100 = 25%
Q19 — Answer: D (x = 5 or x = −2)
2x−3=7→x=5; 2x−3=−7→x=−2. Both verified.
Q20 — Answer: C (6 miles)
3 + 1.5m = 12 → 1.5m = 9 → m = 6