Algebra — Expressions, Equations, and Inequalities

Algebra is a compact way to describe patterns and solve unknowns. The main job is to keep operations balanced: simplify like terms, undo operations in reverse order, and check that a result satisfies the original statement. This article covers the core Grade 11 tools used in later mathematics, science, and technical work.

Expressions: Simplify Before You Solve

An expression has no equals sign, so it cannot be “solved.” It can be simplified. Combine only like terms: terms with the same variable part and the same exponents.

3x + 5x - 2      = 8x - 2
4a² - 7a + 3a²  = 7a² - 7a

Use the distributive property to remove parentheses:

a(b + c) = ab + ac

Worked example: simplifying an expression

Simplify 3(2x - 5) - 4x + 7.

3(2x - 5) - 4x + 7
= 6x - 15 - 4x + 7     distribute 3
= 2x - 15 + 7          combine x terms
= 2x - 8               combine constants

The simplified expression is 2x - 8.

Exponent Rules

Exponent rules apply when the base is the same. They are not permission to add exponents across addition: x² + x³ cannot be simplified to x⁵.

Rule	Example
`a^m × aⁿ = a^m+n`	`x³ × x⁴ = x⁷`
`a^m / aⁿ = a^m-n`, `a ≠ 0`	`y⁶ / y² = y⁴`
`(a^m)ⁿ = a^mn`	`(p³)² = p⁶`
`(ab)ⁿ = aⁿbⁿ`	`(2x)³ = 8x³`
`a⁰ = 1`, `a ≠ 0`	`5x⁰ = 5`
`a^-n = 1/aⁿ`, `a ≠ 0`	`x^-2 = 1/x²`

Worked example: negative exponents

Simplify (6x⁵y^-2) / (3x²y).

(6x⁵y⁻²) / (3x²y)
= (6 / 3)x⁽⁵⁻²⁾y⁽⁻²⁻¹⁾
= 2x³y⁻³
= 2x³ / y³

The final form uses positive exponents. Here, x ≠ 0 and y ≠ 0.

Factoring: Reverse the Expansion

Factoring writes an expression as a product. It is useful because a product is zero only when at least one factor is zero.

Greatest common factor

First look for a factor shared by every term.

6x² - 9x
= 3x(2x - 3)

Trinomials

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

Worked example: factoring a trinomial

Factor x² - 5x + 6.

x² - 5x + 6
= (x - 2)(x - 3)

Check the middle term:

(x - 2)(x - 3)
= x² - 3x - 2x + 6
= x² - 5x + 6

When the coefficient of x² is not 1, grouping is reliable.

Worked example: factoring by grouping

Factor 2x² + 7x + 3.

2x² + 7x + 3
= 2x² + 6x + x + 3     split 7x into 6x + x
= 2x(x + 3) + 1(x + 3)
= (2x + 1)(x + 3)

Linear Equations

An equation says two expressions have the same value. Do the same operation to both sides to preserve that equality. A linear equation has its variable only to the first power.

Worked example: solve a linear equation

Solve 4(2x - 1) + 3 = 19.

4(2x - 1) + 3 = 19
8x - 4 + 3 = 19        distribute 4
8x - 1 = 19            combine constants
8x = 20                add 1 to both sides
x = 20 / 8
x = 5 / 2

Check: 4(2 × 5/2 - 1) + 3 = 4(4) + 3 = 19. Therefore x = 5/2.

Quadratic Equations

A quadratic equation can be written as ax² + bx + c = 0, where a ≠ 0. Put one side equal to zero before using either method below.

Solving by factoring

Use the zero-product property: if AB = 0, then A = 0 or B = 0.

Worked example: quadratic equation by factoring

Solve x² - 5x + 6 = 0.

x² - 5x + 6 = 0
(x - 2)(x - 3) = 0
x - 2 = 0  or  x - 3 = 0
x = 2      or  x = 3

Both values make the original equation zero, so the solutions are x = 2 and x = 3.

The quadratic formula

When a quadratic does not factor neatly, use:

x = (-b ± √(b² - 4ac)) / (2a)

The quantity b² - 4ac is the discriminant. A positive discriminant gives two real solutions, zero gives one repeated real solution, and a negative discriminant gives no real solutions.

Worked example: quadratic formula

Solve 2x² + x - 4 = 0.

a = 2, b = 1, c = -4

x = (-b ± √(b² - 4ac)) / (2a)
x = (-1 ± √(1² - 4(2)(-4))) / (2(2))
x = (-1 ± √(1 + 32)) / 4
x = (-1 ± √33) / 4

The two exact solutions are (-1 + √33)/4 and (-1 - √33)/4.

Linear Inequalities and Intervals

An inequality compares values instead of declaring them equal. Solve it much like an equation, except that multiplying or dividing by a negative number reverses the inequality sign.

Symbol	Meaning	Interval notation
`x < a`	less than `a`	`(-∞, a)`
`x ≤ a`	at most `a`	`(-∞, a]`
`x > a`	greater than `a`	`(a, ∞)`
`x ≥ a`	at least `a`	`[a, ∞)`

Round parentheses exclude an endpoint; square brackets include it. Infinity is never included, so it always uses a parenthesis.

Worked example: linear inequality

Solve -3x + 5 ≥ 14.

-3x + 5 ≥ 14
-3x ≥ 9             subtract 5 from both sides
x ≤ -3              divide by -3 and reverse the sign

The solution is x ≤ -3, or (-∞, -3]. On a number line, place a closed dot at -3 and shade left.

Quadratic Inequalities

For a quadratic inequality, move everything to one side, factor if possible, find the zeros, then test the intervals separated by those zeros. A sign chart prevents guessing.

Worked example: solve a quadratic inequality

Solve x² - x - 6 < 0.

x² - x - 6 < 0
(x - 3)(x + 2) < 0

Critical values: x = -2 and x = 3

Test x = -3:  (-3 - 3)(-3 + 2) = (+)   not included
Test x = 0:   (0 - 3)(0 + 2) = (-)     included
Test x = 4:   (4 - 3)(4 + 2) = (+)     not included

The product is negative between the zeros. Because the inequality is strict, the endpoints are excluded: -2 < x < 3, or (-2, 3).

Worked example: endpoints included

Solve x² - 4x ≥ 0.

x² - 4x ≥ 0
x(x - 4) ≥ 0

Critical values: x = 0 and x = 4

Test x = -1:  (-1)(-5) = (+)   included
Test x = 2:   (2)(-2) = (-)    not included
Test x = 5:   (5)(1) = (+)     included

The zeros themselves are included because the inequality is ≥. The solution is x ≤ 0 or x ≥ 4, written (-∞, 0] ∪ [4, ∞).

Common Mistakes

Combining unlike terms: 3x + 2x² is not 5x². The powers differ.
Dropping a negative while distributing: -2(x - 4) = -2x + 8, not -2x - 8.
Using the zero-product property too early: it applies only when the product equals zero. First rewrite a quadratic equation as ... = 0.
Forgetting the second quadratic-formula solution: ± means calculate both the plus and minus cases.
Not reversing an inequality: dividing -2x < 8 by -2 gives x > -4.
Using brackets for a strict inequality: x < 3 is (-∞, 3), not (-∞, 3].

Practice Set

Simplify 5(3x - 2) - 4x + 1.
Simplify (12a⁴b^-1) / (3a²b²).
Factor 3x² - 12x.
Solve 5x - 7 = 3x + 9.
Solve x² + 2x - 15 = 0.
Solve x² + x - 1 = 0 using the quadratic formula.
Solve and give interval notation: 2 - 4x < 10.
Solve and give interval notation: x² - 9 ≤ 0.

Answers

5(3x - 2) - 4x + 1 = 15x - 10 - 4x + 1 = 11x - 9.
(12a⁴b^-1) / (3a²b²) = 4a²b^-3 = 4a²/b³, where a ≠ 0 and b ≠ 0.
3x² - 12x = 3x(x - 4).
5x - 7 = 3x + 9, so 2x = 16 and x = 8.
x² + 2x - 15 = (x + 5)(x - 3) = 0, so x = -5 or x = 3.
x = (-1 ± √(1 - 4(1)(-1))) / 2 = (-1 ± √5) / 2.
2 - 4x < 10, so -4x < 8, then x > -2: (-2, ∞).
x² - 9 = (x - 3)(x + 3) ≤ 0. The product is non-positive from -3 through 3: [-3, 3].