Algebra Essentials

Algebra is the largest topic area on the OAR Math Skills Test. You will encounter everything from simple one-step equations to more complex multi-step problems involving quadratics and systems of equations.

Solving Single-Variable Equations

One-Step Equations

Two-Step Equations

Strategy: Undo addition/subtraction first, then multiplication/division.

Example: 4x + 7 = 31

1. Subtract 7: 4x = 24
2. Divide by 4: x = 6

Example: (x - 3)/5 = 4

1. Multiply by 5: x - 3 = 20
2. Add 3: x = 23

Multi-Step Equations

Example: 3(2x - 4) = 5x + 2

1. Distribute: 6x - 12 = 5x + 2
2. Subtract 5x: x - 12 = 2
3. Add 12: x = 14

Example: 2(x + 3) - 4(x - 1) = 10

1. Distribute: 2x + 6 - 4x + 4 = 10
2. Combine: -2x + 10 = 10
3. Subtract 10: -2x = 0
4. Divide: x = 0

Exponents and Roots

Exponent Rules

Common Powers to Memorize

Square Roots

Simplifying Radicals

Find the largest perfect square factor:

• √72 = √(36 × 2) = 6√2
• √50 = √(25 × 2) = 5√2
• √48 = √(16 × 3) = 4√3

FOIL and Polynomial Multiplication

FOIL Method

(a + b)(c + d) = ac + ad + bc + bd

Example: (x + 3)(x + 5)

• First: x × x = x²
• Outer: x × 5 = 5x
• Inner: 3 × x = 3x
• Last: 3 × 5 = 15
• Result: x² + 5x + 3x + 15 = x² + 8x + 15

Special Products

Example: (x + 4)² = x² + 8x + 16

Example: (x + 7)(x - 7) = x² - 49

Factoring

Factoring Trinomials

To factor x² + bx + c, find two numbers that multiply to c and add to b.

Example: x² + 7x + 12

• Need: two numbers that multiply to 12 and add to 7
• Answer: 3 and 4
• Factored: (x + 3)(x + 4)

Example: x² - 5x + 6

• Need: two numbers that multiply to 6 and add to -5
• Answer: -2 and -3
• Factored: (x - 2)(x - 3)

Factoring with Leading Coefficient

For ax² + bx + c where a ≠ 1, use the AC method:

Example: 2x² + 7x + 3

1. Multiply a × c = 2 × 3 = 6
2. Find factors of 6 that add to 7: 6 and 1
3. Rewrite: 2x² + 6x + x + 3
4. Factor by grouping: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)

Quadratic Equations

The Quadratic Formula

For ax² + bx + c = 0:

x = (-b ± √(b² - 4ac)) / 2a

Example: Solve 2x² - 5x - 3 = 0

• a = 2, b = -5, c = -3
• Discriminant: (-5)² - 4(2)(-3) = 25 + 24 = 49
• x = (5 ± 7) / 4
• x = 12/4 = 3 or x = -2/4 = -1/2

The Discriminant

b² - 4ac tells you the nature of the roots:

Systems of Equations

Substitution Method

1. Solve one equation for one variable
2. Substitute into the other equation
3. Solve and back-substitute

Example:

• y = 2x + 1
• 3x + y = 16

Substitute: 3x + (2x + 1) = 16 → 5x + 1 = 16 → 5x = 15 → x = 3 Back-substitute: y = 2(3) + 1 = 7 Solution: (3, 7)

Elimination Method

1. Align equations
2. Multiply to make one variable's coefficients opposites
3. Add equations to eliminate that variable

Example:

• 2x + 3y = 12
• 4x - 3y = 6

Add directly (y cancels): 6x = 18 → x = 3 Substitute: 2(3) + 3y = 12 → 3y = 6 → y = 2 Solution: (3, 2)

Inequalities

Solving Linear Inequalities

The same as equations, with one critical exception: flip the inequality sign when multiplying or dividing by a negative number.

Example: -3x + 6 > 15

1. Subtract 6: -3x > 9
2. Divide by -3 (flip!): x < -3

Compound Inequalities

• "And" (intersection): Both conditions must be true: -2 < x < 5
• "Or" (union): At least one condition is true: x < -1 or x > 4