1. From numbers to algebra
You already know that a variable is a letter standing in for a number. The leap algebra asks you to make is small but transformative: treat the letter exactly like a number, even when you don't know what number it is. Every rule you learned for arithmetic still holds.
Concretely, addition is still commutative ($a + b = b + a$) and associative ($(a + b) + c = a + (b + c)$). Multiplication is still commutative ($ab = ba$) and associative ($(ab)c = a(bc)$). Multiplication still distributes over addition ($a(b + c) = ab + ac$). These three families — commutativity, associativity, distributivity — are the structural backbone of every manipulation in this topic.
In pre-algebra you mostly read expressions left to right and computed. Now expressions are objects you transform. Two expressions that look completely different on the page can describe the same thing — and the entire game is recognizing when that's true.
2. Equivalent expressions
Two expressions are equivalent when they evaluate to the same number for every legal value of their variables. We write $A \equiv B$ to mean "$A$ and $B$ are equivalent expressions."
That definition is doing more work than it looks. "For every legal value" matters — not just "for the value I tried." A single value can falsify equivalence (find one input where the two sides disagree and you're done), but a single value can never prove it.
Two ways to establish equivalence
Test by substitution. If you suspect $A$ and $B$ are equivalent, plug in a few values. If any disagree, they aren't equivalent. If they agree on every value you try, you have evidence — not proof.
Prove by algebraic manipulation. Apply the laws of arithmetic to transform one side into the other. Every step must itself preserve equivalence. The classic example:
$$ 2(x + 3) \;\equiv\; 2x + 6 $$Here distributing the $2$ is a single equivalence-preserving move. The two expressions are the same function of $x$ — they just look different on the page.
Domain matters
The qualifier "for every legal value" is where most subtle equivalence errors live. Consider:
$$ \frac{x^2 - 1}{x - 1} \;\stackrel{?}{\equiv}\; x + 1 $$Factor the numerator: $x^2 - 1 = (x - 1)(x + 1)$. Cancel the common factor of $x - 1$:
$$ \frac{(x - 1)(x + 1)}{x - 1} = x + 1 \quad \text{provided } x \neq 1 $$The left-hand side is undefined at $x = 1$ (division by zero); the right-hand side equals $2$ there. So they agree on almost every value — but not all. The two are equivalent only on the domain $x \neq 1$.
Every time you cancel a factor, ask: could that factor be zero? If yes, the equivalence carries a restriction. It doesn't go away — you carry it with you for the rest of the problem.
3. Multi-variable expressions
An expression can hold more than one variable. Each variable is independent — you can vary one while holding the others fixed, or all of them at once. The arithmetic rules don't change; you just have more letters in play.
Take $3xy + 2x - y$. To evaluate at $(x, y) = (2, -1)$, substitute and compute:
$$ 3(2)(-1) + 2(2) - (-1) \;=\; -6 + 4 + 1 \;=\; -1 $$The order of substitution doesn't matter — what matters is that each variable gets replaced with its assigned value, parentheses preserve signs, and the rest is arithmetic.
What counts as a like term
Two terms are like when their variable parts match exactly — same letters, same exponents on each letter. Only the numerical coefficients can differ. Coefficients you combine; variable parts you copy.
| Pair | Like? | Why |
|---|---|---|
| $3xy$ and $5xy$ | Yes | Same variables, same exponents ($x^1 y^1$ both). |
| $3xy$ and $3x^2 y$ | No | Different exponent on $x$. |
| $-7a^2 b$ and $\tfrac{1}{2} a^2 b$ | Yes | Same $a^2 b$ structure; coefficients can be anything. |
| $2xy$ and $2yx$ | Yes | Multiplication is commutative — $xy$ and $yx$ are the same variable part. |
| $4x$ and $4y$ | No | Different variables entirely. |
Once you see them, combining is straightforward: $3xy + 5xy = 8xy$, $-7a^2 b + \tfrac{1}{2} a^2 b = -\tfrac{13}{2} a^2 b$. The variable part rides along unchanged.
4. Factoring out a common factor
Distribution takes a product and turns it into a sum: $a(b + c) = ab + ac$. Factoring is exactly that move played in reverse — looking at a sum and recognizing it as a hidden product.
$$ 6x^2 + 9x \;=\; 3x(2x + 3) $$The factor we pulled out, $3x$, is the greatest common factor (GCF) of the two terms. The rest goes inside the parentheses, and you can always check by re-distributing.
The GCF of a few monomials
The GCF of a collection of monomial terms is built piece by piece:
- Take the GCF of the numerical coefficients.
- For each variable that appears in every term, take it to the smallest exponent it shows.
So for $12x^3 y^2 + 18x^2 y^4$:
- $\gcd(12, 18) = 6$.
- $x$ appears in both, smallest exponent is $2$: take $x^2$.
- $y$ appears in both, smallest exponent is $2$: take $y^2$.
The GCF is $6x^2 y^2$, and the factored form is:
$$ 12x^3 y^2 + 18x^2 y^4 \;=\; 6x^2 y^2 \,(2x + 3y^2) $$Pulling out a GCF is the first and easiest kind of factoring. The harder kinds — factoring trinomials, difference of squares, grouping — all come later in Polynomials. They are the same move scaled up: rewrite a sum as a product. If you can do GCF cleanly, the rest is just pattern recognition on top.
5. Substitution at depth
In pre-algebra, substitution meant "plug in a number for a variable." In algebra it generalizes: you can substitute any expression for a variable, as long as you do it consistently and use parentheses to preserve the structure.
Suppose $u = x^2 + 1$. Then anywhere we see $u$, we can put $(x^2 + 1)$ — parentheses included, every time. Watch what happens with $u^2$:
$$ u^2 \;=\; (x^2 + 1)^2 \;=\; (x^2 + 1)(x^2 + 1) \;=\; x^4 + 2x^2 + 1 $$The parentheses aren't decoration; they're the entire point. Drop them and you'd write $x^2 + 1^2 = x^2 + 1$, which is wrong by miles. Parentheses say "treat this whole thing as one quantity, the same way $u$ was one quantity."
Why this move is everywhere
Substituting an expression for a variable is one of the most general moves in mathematics. You will meet it again and again:
- Change of variables. A complicated expression often simplifies dramatically when you give one piece a new name and substitute it back at the end.
- u-substitution in calculus. Choosing a clever $u$ turns an unfriendly integral into a routine one — the same algebraic move you're learning here.
- Function composition. $f(g(x))$ is substitution: take whatever $f$ does to its input, then feed it $g(x)$ as that input.
- Problem-solving in general. Naming a complicated piece, doing algebra on the name, then substituting back is how messy problems get tame.
6. Why care about equivalence
Here is the secret that ties this topic to everything that follows in algebra: equations don't get solved by guessing — they get rewritten. Each step in a solution replaces one expression (or one equation) with an equivalent one that's slightly simpler, until the answer is sitting in front of you.
That's the deal you sign every time you do algebra. You may add the same thing to both sides, multiply both sides by the same nonzero number, distribute, factor, combine like terms, substitute — but every move has to preserve equivalence, or the chain breaks and your "solution" is meaningless.
To solve an equation is to find a chain of equivalent equations whose last link makes the answer obvious.
For example, working with the equation $3(x + 2) = 18$:
$$ \begin{aligned} 3(x + 2) &= 18 \\ 3x + 6 &= 18 &&\text{(distribute — equivalence-preserving)}\\ 3x &= 12 &&\text{(subtract 6 from both sides)} \\ x &= 4 &&\text{(divide both sides by 3)} \end{aligned} $$Every line is the same equation, just dressed differently. The last line is the version where the answer is unmissable. That's the whole game — and it's exactly what the next topic, Linear Equations, will codify into a recipe.
7. Common pitfalls
$\tfrac{x^2 - 1}{x - 1}$ simplifies to $x + 1$ only on the domain $x \neq 1$. Cancellation is real and useful, but it can create or destroy values where the expression is defined. Always note any restriction the cancellation introduces.
$\dfrac{x + y}{x}$ is not $1 + y$. The correct split is $\dfrac{x + y}{x} = 1 + \dfrac{y}{x}$. Division distributes over the numerator's sum but not the way most beginners hope — every term has to keep its denominator. Forgetting this is one of the most common ways students convince themselves a wrong simplification is right.
$(a + b)^2$ is not $a^2 + b^2$. The middle term doesn't vanish: $(a + b)^2 = a^2 + 2ab + b^2$. This bites especially hard during substitution, when "$u$" is a sum and you write $u^2$ blindly.
The exponents on every variable must match for terms to be combinable. Same letters isn't enough — same letters at the same powers. Mismatched exponents stay separate.
8. Worked examples
Try each one before opening the solution. The goal is to check whether your sequence of moves matches a clean chain of equivalences, not just whether you got the right last line.
Example 1 · Factor out the greatest common factor of $8x^3 + 12x^2 - 20x$
Step 1. GCF of coefficients: $\gcd(8, 12, 20) = 4$.
Step 2. The variable $x$ appears in every term, smallest exponent is $1$: take $x$.
Step 3. GCF is $4x$. Pull it out and divide each term:
$$ 8x^3 + 12x^2 - 20x \;=\; 4x(2x^2 + 3x - 5) $$Check. Re-distribute: $4x \cdot 2x^2 = 8x^3$, $4x \cdot 3x = 12x^2$, $4x \cdot (-5) = -20x$. ✓
Example 2 · Are $\;(x + 2)(x + 3)\;$ and $\;x^2 + 5x + 6\;$ equivalent?
Strategy. First test by substitution. Then prove by manipulation.
Substitution test. Try $x = 1$: left $= (3)(4) = 12$. Right $= 1 + 5 + 6 = 12$. ✓ Try $x = -2$: left $= (0)(1) = 0$. Right $= 4 - 10 + 6 = 0$. ✓ Try $x = 5$: left $= (7)(8) = 56$. Right $= 25 + 25 + 6 = 56$. ✓
Three matches is strong evidence — but not proof. To prove, expand the left side by distributing:
$$ (x + 2)(x + 3) \;=\; x \cdot x + x \cdot 3 + 2 \cdot x + 2 \cdot 3 \;=\; x^2 + 3x + 2x + 6 \;=\; x^2 + 5x + 6 $$Every step is equivalence-preserving, so the two expressions are equivalent for every real $x$. ✓
Example 3 · Evaluate $\;2a^2 b - 3ab + b^2\;$ at $(a, b) = (-1, 4)$
Step 1. Substitute, keeping every replacement in parentheses:
$$ 2(-1)^2(4) - 3(-1)(4) + (4)^2 $$Step 2. Handle exponents first: $(-1)^2 = 1$, $(4)^2 = 16$.
$$ 2(1)(4) - 3(-1)(4) + 16 $$Step 3. Multiply, then add:
$$ 8 - (-12) + 16 \;=\; 8 + 12 + 16 \;=\; 36 $$So the expression equals $36$ at $(a, b) = (-1, 4)$.
Example 4 · If $\;u = 2x - 1$, write $\;u^2 + 3u\;$ in terms of $x$
Step 1. Substitute the expression for $u$ — parentheses around the whole thing, every time:
$$ u^2 + 3u \;=\; (2x - 1)^2 + 3(2x - 1) $$Step 2. Expand $(2x - 1)^2 = (2x)^2 - 2(2x)(1) + 1^2 = 4x^2 - 4x + 1$.
Step 3. Distribute the $3$: $3(2x - 1) = 6x - 3$.
Step 4. Combine:
$$ 4x^2 - 4x + 1 + 6x - 3 \;=\; 4x^2 + 2x - 2 $$So $u^2 + 3u \equiv 4x^2 + 2x - 2$ once $u$ is unpacked.
Example 5 · Simplify $\;4x^2 y - 3xy^2 + 7x^2 y + 2xy^2 - x^2 y\;$
Step 1. Group like terms — match the variable parts exactly.
- $x^2 y$ terms: $4x^2 y$, $7x^2 y$, $-x^2 y$.
- $xy^2$ terms: $-3xy^2$, $2xy^2$.
Step 2. Combine each group by adding the coefficients:
$$ (4 + 7 - 1)\, x^2 y \;+\; (-3 + 2)\, xy^2 \;=\; 10 x^2 y - xy^2 $$The variable parts $x^2 y$ and $xy^2$ stay separate — different exponent patterns means they aren't like terms with each other.