Variables, expressions and algebraic operations

Learning outcome

Identify terms and coefficients, simplify expressions, substitute signed values correctly, and preserve restrictions when dividing algebraic quantities.

Concepts and assumptions

Unless a question narrows the domain, variables represent real numbers. A variable can take different allowed values; a constant has a fixed value. An expression describes a quantity without asserting equality. An equation asserts that two expressions are equal.

Terms are separated by addition or subtraction at the outermost level. In 5x² − 3x + 7, the coefficients of x² and x are 5 and −3; 7 is the constant term. Multiplication joins factors: 5x² means 5 × x × x.

Like terms have exactly the same variable factors and powers. They combine because the distributive rule gives 3x + 2x = (3 + 2)x. However, x and x² generally represent different quantities and cannot be combined into one like term.

Distribution also explains bracket removal: multiply every term inside by the outside factor. Subtracting a bracket means multiplying its entire contents by −1.

Evaluate brackets, then powers, then multiplication/division from left to right, then addition/subtraction from left to right. Use brackets around negative substituted values: (−2)² = 4, whereas −2² means −(2²) = −4.

Division requires a nonzero denominator. Only common multiplicative factors can be cancelled; separate terms cannot simply be crossed out. A simplified expression retains the original domain even when its new appearance hides an excluded value.

Worked examples

Example 1 — Combining like terms. Simplify 5x − 3y + 7 − 2x + 4y − 9, then evaluate at x = 2, y = −1.

Collect matching terms: (5 − 2)x + (−3 + 4)y + (7 − 9) = 3x + y − 2. Substitution gives 3(2) + (−1) − 2 = 6 − 1 − 2 = 3.

Example 2 — Removing brackets. Simplify 3(2x − 5) − 2(x + 4) + 7.

Distribute both outside factors: 6x − 15 − 2x − 8 + 7. Combine: (6 − 2)x + (−15 − 8 + 7) = 4x − 16. The second bracket contributes −2x − 8, not −2x + 8.

Example 3 — Cancelling factors. Simplify 12a²b/(3ab), stating its domain.

The denominator is nonzero only when a ≠ 0 and b ≠ 0. Write the numerator as (3ab)(4a). Cancelling the nonzero factor 3ab gives 4a, with both restrictions retained. At a = −2, b = 5, its value is 4(−2) = −8.

Common mistakes

Combining unlike powers; losing negative signs; omitting substitution brackets; cancelling across addition; allowing previously excluded values after simplification.

Practice questions

1. Simplify 7p − 4q + 3 − 2p + q − 8.
2. Simplify 4(2x − 3) − 3(x + 1).
3. Evaluate 2x² − 3xy + y² at x = −2, y = 3.
4. For real m and n, state the domain and simplify (18m²n + 6mn²)/(6mn).

Worked solutions

1. Collect coefficients and constants: (7 − 2)p + (−4 + 1)q + (3 − 8) = 5p − 3q − 5.
2. Expanding gives 8x − 12 − 3x − 3 = 5x − 15.
3. Substitute with brackets: 2(−2)² − 3(−2)(3) + 3² = 8 + 18 + 9 = 35.
4. Require m ≠ 0 and n ≠ 0. Divide termwise: 18m²n/(6mn) + 6mn²/(6mn) = 3m + n. Both exclusions remain.