Like and unlike terms — simplification: Exercises

This page is your self-testing gym for the parent topic. Every problem hides its full solution inside a collapsible box — try first, then peek. Problems climb five levels: Recognition → Application → Analysis → Synthesis → Mastery.

Before you start, one promise: every word below builds only on ideas the parent note already earned. If you meet a symbol you're unsure of, here is the one-line refresher.

The figure below is the mental picture for the entire page: sorting terms into bins by variable part, then summing coefficients inside each bin.

Figure — Like and unlike terms — simplification

Level 1 — Recognition

Goal: can you SPOT which terms are allowed to combine, before any arithmetic?

Recall Solution L1.1

Match the variable part exactly — same letters, same powers.

(a) Like. Both variable parts are $x^{2}$ . Coefficients $3$ and $- 9$ may differ; that's fine.
(b) Unlike. $x y$ (both letters) is not the same as $x$ (one letter). Missing the $y$ makes them different bins.
(c) Like. Both are pure numbers (constants). Their variable part is "nothing", and "nothing = nothing", so they share a bin.
(d) Unlike. $a^{2} b$ has $a$ squared; $a b^{2}$ has $b$ squared. Powers are attached to different letters, so the variable parts do not match.

Recall Solution L1.2

Look only at the power of $m$ . Three terms carry $m^{3}$ ; one carries $m^{2}$ .

$6 m^{3}, - m^{3}, \frac{1}{2} m^{3}$ all live in the $m^{3}$ bin.
$2 m^{2}$ is the outsider — different power, different bin. Answer: $2 m^{2}$ .

Level 2 — Application

Goal: run the combine move cleanly, watching every sign.

Recall Solution L2.1

All three are in the $x$ bin. Note $x$ means $1 x$ (an invisible coefficient of $1$ ). $9 x - 4 x + 1 x = (9 - 4 + 1) x = 6 x$ Answer: $6 x$ . The variable part $x$ stays — only the count changed.

Recall Solution L2.2

Sort into bins, keeping each sign glued to its term.

$a$ bin: $8 a - 5 a = (8 - 5) a = 3 a$
$b$ bin: $3 b + 7 b = (3 + 7) b = 10 b$ $a$ and $b$ are unlike, so we cannot merge the two results. $= 3 a + 10 b$ Answer: $3 a + 10 b$ .

Recall Solution L2.3

Three bins: $x^{2}$ , $x$ , constants.

$x^{2}$ : $5 x^{2} - 3 x^{2} = 2 x^{2}$
$x$ : $6 x - 8 x = - 2 x$
constants: $- 2 + 9 = 7$ Write in descending powers (standard form — you'll want this in 4.1.01-Polynomials-introduction): $= 2 x^{2} - 2 x + 7$ Answer: $2 x^{2} - 2 x + 7$ .

Level 3 — Analysis

Goal: handle mixed products, hidden $1$ s, and terms that look alike but aren't.

Recall Solution L3.1

Three distinct bins because $pq$ , $p$ , and $q$ are all different variable parts.

$pq$ bin: $4 pq - 1 pq = 3 pq$ (the lone $pq$ is $1 pq$ )
$p$ bin: $2 p - 3 p = - 1 p = - p$
$q$ bin: $5 q$ (alone) $= 3 pq - p + 5 q$ Answer: $3 pq - p + 5 q$ .

Recall Solution L3.2

Read each variable part as a stacked recipe: $x^{2} y$ vs $x y^{2}$ — different, so two bins.

$x^{2} y$ bin: $7 x^{2} y - 2 x^{2} y = 5 x^{2} y$
$x y^{2}$ bin: $3 x y^{2} + 1 x y^{2} = 4 x y^{2}$ These two are unlike, so they stay apart. $= 5 x^{2} y + 4 x y^{2}$ Answer: $5 x^{2} y + 4 x y^{2}$ .

Recall Solution L3.3

$t$ bin: $6 t + 2 t - 8 t = (6 + 2 - 8) t = 0 t = 0$
$s$ bin: $- 4 s + 4 s = 0 s = 0$ Every bin collapses to zero. A coefficient of $0$ erases the term entirely ( $0 \cdot t = 0$ ). $= 0$ Answer: $0$ . Not " $0 t$ ", not "empty" — just the number zero.

Level 4 — Synthesis

Goal: combine the like-term skill with the distributive property and negatives — the machinery of 2.1.03-Addition-and-subtraction-of-algebraic-expressions.

Recall Solution L4.1

First open the brackets (each outside number multiplies everything inside — see 2.2.01-Distributive-property). Watch the $- 2$ hitting both inner terms. $3 (2 x + 4) = 6 x + 12$ $- 2 (x - 5) = - 2 x + 10$ Now combine like terms:

$x$ bin: $6 x - 2 x = 4 x$
constants: $12 + 10 = 22$ $= 4 x + 22$ Answer: $4 x + 22$ .

Recall Solution L4.2

Subtracting a bracket flips every sign inside it (it's $- 1$ distributed across the group). $= 5 a^{2} + 3 a - 1 - 2 a^{2} + 4 a - 6$ Bins:

$a^{2}$ : $5 a^{2} - 2 a^{2} = 3 a^{2}$
$a$ : $3 a + 4 a = 7 a$
constants: $- 1 - 6 = - 7$ $= 3 a^{2} + 7 a - 7$ Answer: $3 a^{2} + 7 a - 7$ .

Recall Solution L4.3

Distribute first: $2 (3 x y + x) = 6 x y + 2 x, - (x y - 4 x) = - x y + 4 x$ Rebuild and sort: $6 x y + 2 x - x y + 4 x + 5 x y$

$x y$ bin: $6 x y - x y + 5 x y = 10 x y$
$x$ bin: $2 x + 4 x = 6 x$ $= 10 x y + 6 x$ Answer: $10 x y + 6 x$ .

Level 5 — Mastery

Goal: reason about like terms in reverse, and connect to equations.

Recall Solution L5.1

The left side is all one bin ( $x$ ). Combine coefficients keeping $k$ symbolic: $(7 + k - 3) x = (k + 4) x$ For this to equal $9 x$ , the coefficients must match: $k + 4 = 9 ⟹ k = 5$ Answer: $k = 5$ . (This is exactly the reasoning behind 3.1.01-Solving-linear-equations-one-variable.)

Recall Solution L5.2

All three are $x^{2}$ terms. Combine: $(4 + c - 9) x^{2} = (c - 5) x^{2}$ For the whole thing to be $0$ no matter what $x$ is, the coefficient must be $0$ : $c - 5 = 0 ⟹ c = 5$ Answer: $c = 5$ .

Recall Solution L5.3

Distribute each bracket carefully: $2 (x^{2} - 3 x + 1) = 2 x^{2} - 6 x + 2$ $- 3 (x^{2} - x) = - 3 x^{2} + 3 x$ Assemble everything: $2 x^{2} - 6 x + 2 - 3 x^{2} + 3 x + 4 x + 5$ Sort into bins:

$x^{2}$ : $2 x^{2} - 3 x^{2} = - x^{2}$
$x$ : $- 6 x + 3 x + 4 x = 1 x = x$
constants: $2 + 5 = 7$ $= - x^{2} + x + 7$ Answer: $- x^{2} + x + 7$ .

Wrap-up

Recall One-line recap of the method

Sort every term into a bin by its exact variable part ::: then add coefficients inside each bin, keep the variable part unchanged, and leave different bins separate.

Connections

2.1.01-Variables-constants-and-coefficients — spotting the coefficient (with its sign) is step zero here.
2.2.01-Distributive-property — the reverse-distribute move powering every combine.
2.1.03-Addition-and-subtraction-of-algebraic-expressions — L4 problems are exactly this skill.
3.1.01-Solving-linear-equations-one-variable — L5 matching-coefficients is equation solving in disguise.
4.1.01-Polynomials-introduction — writing answers in descending powers is polynomial standard form.