Addition and subtraction of algebraic expressions: Worked examples

This is the practice arena for the parent topic. There we learned why only like terms combine and why the minus sign in front of a bracket flips everything via the Distributive Property. Here we hunt down every kind of problem that can appear, so no exam question can surprise you.

The scenario matrix

Every cell below is a distinct situation this topic can produce. The worked examples that follow are each tagged with the cell they cover.

#	Cell (case class)	What makes it different	Example
A	All-positive addition	No sign traps at all	Ex 1
B	Subtraction bracket	The $- 1$ must flip every term	Ex 2
C	Coefficient goes to zero	A whole term vanishes	Ex 3
D	Coefficient goes negative	Answer term has a minus in front	Ex 3
E	Multiple variables & powers	Sorting many different "boxes"	Ex 4
F	Degenerate: nothing to combine	Answer is unchanged (no like terms)	Ex 5
G	Fractions / decimals as coefficients	Same rule, uglier numbers	Ex 6
H	Real-world word problem	Translate → combine → interpret with units	Ex 7
I	Exam twist (nested / triple brackets)	Two minus signs, order of operations	Ex 8

We meet cells C and D together in one example (they naturally co-occur), so 8 examples cover all 9 cells.

Worked examples

Worked example Example 1 — Cell A: all-positive addition

Statement: Add $(4 p + 9 q) + (6 p + 2 q)$ .

Forecast: Before reading on — guess the coefficient of $p$ and of $q$ in the answer.

Step 1. Drop the brackets: $4 p + 9 q + 6 p + 2 q$ . Why this step? A $+$ in front of a bracket means "add everything inside exactly as it is" — no sign changes. See Order of Operations.

Step 2. Group like terms: $(4 p + 6 p) + (9 q + 2 q)$ . Why this step? Terms can be reordered freely (commutative + associative), so we park the $p$ 's together and the $q$ 's together — like sorting apples from oranges.

Step 3. Add coefficients: $10 p + 11 q$ . Why this step? $4 + 6 = 10$ and $9 + 2 = 11$ . The variable part is unchanged — we only counted "how many."

Verify: Let $p = 1, q = 1$ . Original: $(4 + 9) + (6 + 2) = 13 + 8 = 21$ . Answer: $10 + 11 = 21$ . ✓ (Testing at $p = q = 1$ turns each term into just its coefficient, a quick sanity check.)

Worked example Example 3 — Cells C & D: a term hits zero, another goes negative

Statement: Simplify $(3 x^{2} + 7 x) - (3 x^{2} + 10 x)$ .

Forecast: One power of $x$ will completely disappear. Which — $x^{2}$ or $x$ ?

Step 1. Distribute the minus: $3 x^{2} + 7 x - 3 x^{2} - 10 x$ . Why this step? The whole second bracket is multiplied by $- 1$ : $+ 3 x^{2} \to - 3 x^{2}$ , $+ 10 x \to - 10 x$ .

Step 2. Combine $x^{2}$ terms: $(3 - 3) x^{2} = 0 x^{2} = 0$ . Why this step? When the coefficient is $0$ , the whole term is $0$ and we drop it — this is the "degenerate / vanishing" case (Cell C).

Step 3. Combine $x$ terms: $(7 - 10) x = - 3 x$ . Why this step? $7 - 10 = - 3$ : a negative coefficient survives (Cell D). Negatives are allowed — they just mean "the opposite direction."

Step 4. Final answer: $- 3 x$ .

Verify: Let $x = 4$ . Original: $(48 + 28) - (48 + 40) = 76 - 88 = - 12$ . Answer: $- 3 (4) = - 12$ . ✓

Worked example Example 7 — Cell H: real-world word problem (with a figure)

Statement: A rectangular garden is $(3 x + 2)$ metres long and $(x + 5)$ metres wide. A path removes a strip so the usable length becomes $(3 x + 2) - (x - 1)$ metres. Write the usable length, then find the total fencing (perimeter) needed around the usable rectangle.

Forecast: Will the usable length be longer or shorter than the original $3 x + 2$ ?

Step 1. Usable length $= (3 x + 2) - (x - 1) = 3 x + 2 - x + 1$ . Why this step? The strip's length is subtracted, so its bracket is multiplied by $- 1$ : $- (x - 1) = - x + 1$ .

Step 2. Combine: $(3 - 1) x + (2 + 1) = 2 x + 3$ metres. Why this step? $3 x - x = 2 x$ ; $2 + 1 = 3$ . So the strip made the length shorter — good sanity match with intuition.

Step 3. Perimeter $= 2 (length) + 2 (width) = 2 (2 x + 3) + 2 (x + 5)$ . Why this step? A rectangle has two lengths and two widths; the $2 (\dots)$ uses the Distributive Property.

Step 4. Expand and combine: $4 x + 6 + 2 x + 10 = 6 x + 16$ metres. Why this step? $4 x + 2 x = 6 x$ ; $6 + 10 = 16$ . Units stay metres throughout — we never mix metres with square metres.

Verify: Let $x = 3$ m. Usable length $= 2 (3) + 3 = 9$ ; width $= 3 + 5 = 8$ . Perimeter $= 2 (9) + 2 (8) = 34$ . Formula: $6 (3) + 16 = 34$ m. ✓

Worked example Example 8 — Cell I: exam twist (nested / triple brackets)

Statement: Simplify $2 x - [3 x - (x - 4)]$ .

Forecast: Two minus signs are hiding here. Guess the final coefficient of $x$ .

Step 1. Work the innermost bracket first: $- (x - 4) = - x + 4$ . Why this step? Order of Operations says resolve the deepest bracket before the outer one. This is a $- 1$ distribution.

Step 2. Inside the square bracket now: $3 x - x + 4 = 2 x + 4$ . Why this step? Combine like terms inside before dealing with the outer minus — keeps the expression small.

Step 3. Apply the outer minus: $2 x - [2 x + 4] = 2 x - 2 x - 4$ . Why this step? The square bracket is multiplied by $- 1$ : $2 x + 4 \to - 2 x - 4$ .

Step 4. Combine: $(2 - 2) x - 4 = 0 x - 4 = - 4$ . Why this step? The $x$ 's cancel to zero (Cell C reappears!), leaving a pure constant $- 4$ .

Verify: Let $x = 5$ . Original: $2 (5) - [3 (5) - (5 - 4)] = 10 - [15 - 1] = 10 - 14 = - 4$ . Answer: $- 4$ . ✓

Recall Self-test: name the cell

Which case is $(4 a + 3 b) - (4 a + 9 b)$ ? ::: Cell C+D — the $a$ terms vanish to $0$ , and the $b$ terms give a negative $- 6 b$ . Which case is $(x + 2) + (y + 3)$ ? ::: Cell F — no like terms, answer is $x + y + 5$ (only the constants merge). What must you do the instant you see $- (\dots)$ ? ::: Multiply every term inside by $- 1$ (flip all signs) before combining.

Next: these skills feed straight into Solving Linear Equations, Polynomials, and Factoring Algebraic Expressions.