Addition and subtraction of algebraic expressions: Exercises

This page is your training ground for the parent topic. Work each problem on paper first, then open the collapsible solution. The exercises climb five levels — from just spotting like terms up to building your own expressions. Every skill here leans on Combining Like Terms, the Distributive Property, and Order of Operations.

Definition Like-term family

A like-term family is a group of terms that all share the exact same variable part — the same letters raised to the same powers (for example, all the $x^{2}$ terms, or all the $ab$ terms, or the plain numbers with no variable at all). Terms in one family can be combined into a single term; terms in different families cannot. Throughout this page we sort every expression into families first, then combine within each family.

Why combining works this way: a term like $6 x$ literally means " $6$ copies of the unknown number $x$ ." Adding $2 x$ means " $2$ more copies," giving $8$ copies — i.e. $8 x$ . You are counting how many of the same object you have, so the coefficient (the count) changes while the variable part (the object) is untouched. That is why the exponents never move during addition or subtraction: you are not multiplying the objects together, only tallying them.

Level 1 — Recognition

Goal: just identify what is "like" and what is not. No heavy arithmetic yet.

Exercise L1.1

Which of these pairs are like terms? Answer yes/no for each. (a) $7 x$ and $- 2 x$ (b) $4 x^{2}$ and $4 x$ (c) $5 ab$ and $- 9 ab$ (d) $3 x^{2} y$ and $3 x y^{2}$

Recall Solution L1.1

Like terms need the same variables raised to the same powers — the number in front (the coefficient) does not matter.

(a) Yes — both are $x^{1}$ . Only the coefficients ( $7$ and $- 2$ ) differ, which is allowed.
(b) No — $x^{2}$ vs $x^{1}$ . Same letter, different power.
(c) Yes — both are $ab$ (i.e. $a^{1} b^{1}$ ).
(d) No — $x^{2} y$ means $x \cdot x \cdot y$ , while $x y^{2}$ means $x \cdot y \cdot y$ . Different power on each letter.

Exercise L1.2

In the expression $6 m - 4 n + 2 m + 7 - n$ , group the terms into like-term families.

Recall Solution L1.2

Scan the expression once and sort each term by its variable part (this is exactly what "sort into like-term families" means):

$m$ family: $6 m$ and $2 m$
$n$ family: $- 4 n$ and $- n$
constant family (no variable): $7$

Notice $- n$ carries a coefficient of $- 1$ (an invisible $1$ ). We keep the sign glued to the term.

Level 2 — Application

Goal: actually combine like terms and handle the parentheses.

Exercise L2.1

Simplify $(8 x + 5) + (3 x - 2)$ .

Recall Solution L2.1

Step 1 — drop the parentheses. A $+$ in front of a bracket changes nothing inside: $8 x + 5 + 3 x - 2$ Step 2 — group into like-term families. $x$ family: $8 x, 3 x$ . Constant family: $5, - 2$ . (Why group? Only terms in the same family are copies of the same object, so only they can be tallied together.) Step 3 — combine coefficients. We add the counts and leave the object ( $x$ ) alone, because we are only counting how many $x$ 's we have, not multiplying $x$ 's together: $(8 + 3) x + (5 - 2) = 11 x + 3$ Answer: $11 x + 3$ .

Exercise L2.2

Simplify $(9 a - 4 b) - (3 a - 7 b)$ .

Recall Solution L2.2

Step 1 — distribute the minus. The $-$ in front of the second bracket is a multiply-by- $- 1$ on every term inside. Inside the bracket the terms are $+ 3 a$ and $- 7 b$ , so: $- (3 a - 7 b) = (- 1) (3 a) + (- 1) (- 7 b) = - 3 a + 7 b .$ The whole expression becomes: $9 a - 4 b - 3 a + 7 b$ So $+ 3 a$ turned into $- 3 a$ , and $- 7 b$ turned into $+ 7 b$ — both signs flipped. Step 2 — group into like-term families: $a$ family $9 a, - 3 a$ ; $b$ family $- 4 b, + 7 b$ . Step 3 — combine coefficients (the letters $a$ and $b$ stay put; we only tally the counts): $(9 - 3) a + (- 4 + 7) b = 6 a + 3 b$ Answer: $6 a + 3 b$ .

Exercise L2.3

Simplify $4 t^{2} + 6 t - t^{2} + t - 5$ .

Recall Solution L2.3

Sort into families — remember $t^{2}$ and $t$ are different objects, so they live in different families:

$t^{2}$ family: $4 t^{2} - t^{2} = (4 - 1) t^{2} = 3 t^{2}$
$t$ family: $6 t + t = (6 + 1) t = 7 t$
Constant family: $- 5$

In each family we add the coefficients and keep the exponent unchanged, because addition just counts more of the same power — it never creates a new power. Answer: $3 t^{2} + 7 t - 5$ .

Level 3 — Analysis

Goal: many variables, powers, and nested brackets — you must sort carefully.

Exercise L3.1

Simplify $(5 x^{2} y - 2 x y^{2} + 3 x) - (2 x^{2} y + x y^{2} - 5 x)$ .

Recall Solution L3.1

Step 1 — distribute the minus over the second bracket (every sign inside flips): $5 x^{2} y - 2 x y^{2} + 3 x - 2 x^{2} y - x y^{2} + 5 x$ Step 2 — sort into like-term families. Remember $x^{2} y$ and $x y^{2}$ are different families (different powers on each letter):

$x^{2} y$ family: $5 x^{2} y - 2 x^{2} y = 3 x^{2} y$
$x y^{2}$ family: $- 2 x y^{2} - x y^{2} = - 3 x y^{2}$
$x$ family: $3 x + 5 x = 8 x$

Within each family only the coefficients combine; the variable part is the shared object and stays fixed. Answer: $3 x^{2} y - 3 x y^{2} + 8 x$ .

Exercise L3.2

Simplify $2 (3 p - 4) - 3 (p - 5)$ . (Distribute first, then combine.)

Recall Solution L3.2

Step 1 — Distributive Property on each bracket: $2 \cdot 3 p - 2 \cdot 4 - 3 \cdot p + (- 3) (- 5) = 6 p - 8 - 3 p + 15$ Note the last term: $- 3 \times (- 5) = + 15$ . Step 2 — combine within families ( $p$ family and constant family). We add the counts and keep $p$ unchanged, since these are copies of the same object: $p family: 6 p - 3 p = 3 p; constants: - 8 + 15 = 7$ Answer: $3 p + 7$ .

Exercise L3.3

Simplify $[(4 a + 3 b) - (a - b)] + (2 a - 5 b)$ .

Recall Solution L3.3

Work the inner bracket first (Order of Operations): $(4 a + 3 b) - (a - b) = 4 a + 3 b - a + b = 3 a + 4 b$ Now add the last bracket, combining each family (coefficients add, letters stay): $(3 a + 4 b) + (2 a - 5 b) = (3 + 2) a + (4 - 5) b = 5 a - b$ Answer: $5 a - b$ .

Level 4 — Synthesis

Goal: combine addition/subtraction with other algebra ideas — perimeter models, missing terms, verification.

Figure — Addition and subtraction of algebraic expressions

Figure s01: a blueprint-style rectangle. The two horizontal sides (top and bottom) are each labelled in amber as the length $3 x + 2$ ; the two vertical sides (left and right) are each labelled in cyan as the width $x - 1$ . The centre of the rectangle shows the assembled perimeter formula $P = 2 (3 x + 2) + 2 (x - 1) = 8 x + 2$ , so you can see how the four side-expressions add up.

Exercise L4.1

A rectangle has length $(3 x + 2)$ and width $(x - 1)$ . Write a simplified expression for its perimeter. (Perimeter $= 2 \times length + 2 \times width$ .) The figure above shows the four sides labelled — two lengths and two widths.

Recall Solution L4.1

Step 1 — build the expression. The perimeter walks around all four sides: two lengths and two widths. $P = 2 (3 x + 2) + 2 (x - 1)$ Step 2 — distribute (the $2$ multiplies every term inside each bracket): $P = 6 x + 4 + 2 x - 2$ Step 3 — combine within families. The $x$ terms are copies of the same object, so we add their counts and keep $x$ ; the constants are plain numbers: $x family: 6 x + 2 x = 8 x; constants: 4 - 2 = 2$ Answer: $P = 8 x + 2$ .

Exercise L4.2

Find the missing expression $M$ such that $(4 x^{2} - 3 x + 7) + M = 6 x^{2} + x - 2.$

Recall Solution L4.2

To undo an addition, subtract. Isolate $M$ (remove $(4 x^{2} - 3 x + 7)$ from both sides): $M = (6 x^{2} + x - 2) - (4 x^{2} - 3 x + 7)$ Distribute the minus (every sign in the second bracket flips): $M = 6 x^{2} + x - 2 - 4 x^{2} + 3 x - 7$ Combine within families (coefficients add, powers stay fixed):

$x^{2}$ family: $6 x^{2} - 4 x^{2} = 2 x^{2}$
$x$ family: $x + 3 x = 4 x$
constants: $- 2 - 7 = - 9$

Answer: $M = 2 x^{2} + 4 x - 9$ . Check: $(4 x^{2} - 3 x + 7) + (2 x^{2} + 4 x - 9) = 6 x^{2} + x - 2.$ ✓

Exercise L4.3

A triangle has sides $(2 y + 1)$ , $(3 y - 4)$ , and $(y + 5)$ (see figure). Find its perimeter, then find the perimeter when $y = 3$ .

Figure s02: a blueprint-style triangle with its three sides labelled in amber as $2 y + 1$ , $3 y - 4$ , and $y + 5$ . Inside the triangle the assembled perimeter is shown as $P = (2 y + 1) + (3 y - 4) + (y + 5) = 6 y + 2$ , illustrating that the perimeter is just the sum of the three side-expressions.

Recall Solution L4.3

Perimeter = sum of the three sides: $(2 y + 1) + (3 y - 4) + (y + 5)$ Combine within families (add the $y$ -counts, keeping $y$ ; add the plain numbers):

$y$ family: $2 y + 3 y + y = 6 y$
constants: $1 - 4 + 5 = 2$

Symbolic answer: $6 y + 2$ . At $y = 3$ : $6 (3) + 2 = 18 + 2 = 20$ . Answer: perimeter $= 6 y + 2$ , and $= 20$ when $y = 3$ .

Level 5 — Mastery

Goal: full-length simplification and reasoning, no scaffolding.

Exercise L5.1

Simplify completely: $(3 a^{2} b - 5 a b^{2} + 2 ab) - (a^{2} b - 3 a b^{2} - 4 ab) + (2 a^{2} b + a b^{2}) .$

Recall Solution L5.1

Step 1 — open all three brackets (middle one has a $-$ in front, so flip its signs): $3 a^{2} b - 5 a b^{2} + 2 ab - a^{2} b + 3 a b^{2} + 4 ab + 2 a^{2} b + a b^{2}$ Step 2 — sort into three families ( $a^{2} b$ , $a b^{2}$ , $ab$ are all different objects), then add the coefficients while the variable part stays fixed:

$a^{2} b$ family: $3 a^{2} b - a^{2} b + 2 a^{2} b = (3 - 1 + 2) a^{2} b = 4 a^{2} b$
$a b^{2}$ family: $- 5 a b^{2} + 3 a b^{2} + a b^{2} = (- 5 + 3 + 1) a b^{2} = - a b^{2}$
$ab$ family: $2 ab + 4 ab = 6 ab$

Answer: $4 a^{2} b - a b^{2} + 6 ab$ .

Exercise L5.2

Let $P = 5 x^{2} - 3 x + 1$ , $Q = - 2 x^{2} + 4 x - 6$ , and $R = x^{2} - x + 2$ . Compute $P - Q + R$ and evaluate at $x = 2$ .

Recall Solution L5.2

Step 1 — write it out with sign care. $- Q$ flips every sign of $Q$ : $P - Q + R = (5 x^{2} - 3 x + 1) + (2 x^{2} - 4 x + 6) + (x^{2} - x + 2)$ Step 2 — combine each family (add coefficients, keep each power):

$x^{2}$ family: $5 + 2 + 1 = 8 \Rightarrow 8 x^{2}$
$x$ family: $- 3 - 4 - 1 = - 8 \Rightarrow - 8 x$
constant family: $1 + 6 + 2 = 9$

Symbolic answer: $8 x^{2} - 8 x + 9$ . At $x = 2$ : $8 (4) - 8 (2) + 9 = 32 - 16 + 9 = 25$ . Answer: $8 x^{2} - 8 x + 9$ , equal to $25$ when $x = 2$ .

Exercise L5.3

Simplify and state the final number of distinct terms: $2 (x^{2} - x y) - 3 (x y - y^{2}) + (x^{2} + 5 x y - y^{2}) .$

Recall Solution L5.3

Step 1 — distribute each coefficient: $2 x^{2} - 2 x y - 3 x y + 3 y^{2} + x^{2} + 5 x y - y^{2}$ (careful: $- 3 (- y^{2}) = + 3 y^{2}$ .) Step 2 — combine families (coefficients add within each object):

$x^{2}$ family: $2 x^{2} + x^{2} = 3 x^{2}$
$x y$ family: $- 2 x y - 3 x y + 5 x y = 0 x y = 0$ — this family cancels completely!
$y^{2}$ family: $3 y^{2} - y^{2} = 2 y^{2}$

Answer: $3 x^{2} + 2 y^{2}$ . The $x y$ terms vanish, so there are 2 distinct terms.

Recall Quick self-test (cloze)

Two terms are like terms only if they share the same variables raised to the same powers. Subtracting a bracket means multiplying every term inside by == $- 1$ . When combining like terms, the exponents stay the same== while the coefficients add. To find $M$ in $A + M = B$ , compute $M === B - A ==$ .

Ready for more theory? Head back to the parent note, or push forward into Polynomials and Factoring Algebraic Expressions.