Equivalent Expressions Practice

Simplifying is only half of each question. The other half is stating what the variable cannot be — and those restrictions come from the expression you were given, not the one you end up with. Work each on paper before opening an answer.

1. Expand and simplify

$(2x-3)(x+5)-(x-4{)}^2$

Answer 1

$(2x^2+7x-15)-(x^2-8x+16)=x^2+15x-31$ The bracket around the second expansion matters: every term of $x^2-8x+16$ changes sign. Dropping that bracket is the most common error in this question and it is invisible afterwards.

2. Simplify, and state the restrictions

$\frac{x^2-9}{x^2+7x+12}$

Answer 2

Factor both: $\frac{(x-3)(x+3)}{(x+3)(x+4)}=\frac{x-3}{x+4}$

Restrictions: $x\ne -3$ and $x\ne -4$.

The $-3$ is the one people lose. It came from the original denominator, and cancelling it does not make it legal — the original expression is undefined there, so the simplified one must carry the same hole.

3. Add these

$\frac{3}{x-2}+\frac{5}{x+1}$

Answer 3

Common denominator $(x-2)(x+1)$: $\frac{3(x+1)+5(x-2)}{(x-2)(x+1)}=\frac{8x-7}{(x-2)(x+1)}$ Restrictions: $x\ne 2$, $x\ne -1$. Leave the denominator factored — it is more useful factored, and it shows the restrictions at a glance.

4. Multiply and simplify

$\frac{x^2-4}{x^2-x-6}\times \frac{x+3}{x+2}$

Answer 4

$\frac{(x-2)(x+2)}{(x-3)(x+2)}\times \frac{x+3}{x+2}=\frac{(x-2)(x+3)}{(x-3)(x+2)}$ Restrictions: $x\ne 3$, $x\ne -2$. Factor everything before cancelling anything — cancelling across an unfactored sum is the error this question exists to catch.

5. Divide

$\frac{2x^2+6x}{x^2-1}÷\frac{4x}{x-1}$

Answer 5

Multiply by the reciprocal: $\frac{2x(x+3)}{(x-1)(x+1)}\times \frac{x-1}{4x}=\frac{x+3}{2(x+1)}$ Restrictions: $x\ne 1$, $x\ne -1$, and $x\ne 0$ — that last one because $4x$ was a divisor. A restriction from the expression you divided BY is the one students forget every time.

6. Prove they are equivalent, or show they are not

$\frac{x^2-1}{x-1}\text{and}x+1$

Answer 6

The first simplifies to $x+1$, so they agree for every value except $x=1$, where the first is undefined and the second equals 2. They are not the same function: their graphs differ by one point, and the first has a hole at $(1,2)$.

This is exactly why restrictions are marked. “Equivalent” means the same value everywhere, and a single missing point breaks it.

7. Simplify

$\frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{xy}}$

Answer 7

Multiply the top and bottom by $xy$: $\frac{y+x}{1}=x+y$ Restrictions: $x\ne 0$, $y\ne 0$. Multiplying through by the common denominator turns a complex fraction into an ordinary one in a single step, which beats simplifying top and bottom separately.

Curriculum connection

A3.2

verify, through investigation with and without technology, that $\sqrt{ab}=\sqrt{a}\times \sqrt{b}$, $a\ge 0$, $b\ge 0$, and use this relationship to simplify radicals (e.g., $\sqrt{24}$) and radical expressions obtained by adding, subtracting, and multiplying [e.g., $(2+\sqrt{6})(3-\sqrt{12})$]

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A3.3

simplify rational expressions by adding, subtracting, multiplying, and dividing, and state the restrictions on the variable values Sample problem: Simplify $\frac{2x}{4x^2+6x}-\frac{3}{2x+3}$, and state the restrictions on the variable.

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A3.1

simplify polynomial expressions by adding, subtracting, and multiplying Sample problem: Write and simplify an expression for the volume of a cube with edge length $2x+1$.

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