Rewrite rational expressions and preserve restrictions

Your feedback matters:

Lesson progressPractice problems 0/3

Difficulty: Intermediate
Estimated time: 30 minutes
Domains: Advanced Math
Techniques: Denominator-restrictionsFactor-and-cancelLeast-common-denominatorSplitting-fractionsGraph-overlap

What you’ll learn

Tell when a question wants an equivalent form or the excluded values, and pick the quicker method, algebra or a graph.
Find every value of $x$ that makes an original denominator zero.
Factor, then cancel only whole common factors.
Add and subtract fractions using a least common denominator.
Split the top of a fraction over its denominator, without ever splitting the bottom.

Prerequisites

Why this matters on the SAT

Rewrite the fraction and keep its restrictions visible

Some SAT questions never ask for a value of $x$ . They ask which expression is equivalent to a fraction, "for all values of $x$ for which the expression is defined." That phrase means the new form has to match the original at every $x$ the original allows.

Here's a small case. The fraction $\frac{x^2}{x}$ isn't defined at $x=0$ , but at every other $x$ it equals $x$ . So $x$ counts as an equivalent form, even though $x$ on its own has no trouble at $0$ . When the $x$ 's canceled, the excluded value $x=0$ disappeared from view. On a graph, $y=\frac{x^2}{x}$ is the line $y=x$ with one point missing at $x=0$ . A missing point like that is called a hole.

So the new form doesn't have to show every value the original rules out. Still, write those values down, because you'll need them whenever a question asks about the domain, about where the expression is undefined, or about plugging in a value.

Solution to the example

First, the values $x$ can't take. The denominators are zero at $x=2$ and $x=-2$ , so both are out.

To add the fractions, you need one denominator for both. The smallest one that works is the product of the two:

(x-2)(x+2)=x^2-4.

Multiply each fraction's top and bottom by the factor it's missing, then add the tops:

\begin{aligned} \frac{1}{x-2}+\frac{3}{x+2} &=\frac{x+2}{(x-2)(x+2)} +\frac{3(x-2)}{(x+2)(x-2)}\\[1.4em] &=\frac{x+2+3x-6}{x^2-4}\\[1.4em] &=\frac{4x-4}{x^2-4}\\[1.4em] &=\frac{4(x-1)}{x^2-4}. \end{aligned}

The answer is A, and $x=2$ and $x=-2$ are still excluded.

SAT example

Which expression is equivalent to

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

for all values of $x$ for which the expression is defined?

A
$\dfrac{4(x-1)}{x^2-4}$
B
$\dfrac{4(x+1)}{x^2-4}$
C
$\dfrac{4x-4}{2x}$
D
$\dfrac{3x-4}{x^2-4}$

Choose algebra, Desmos, or both

A rational expression is a fraction with a polynomial on top and a polynomial on the bottom, like $\frac{x^2-9}{x^2-3x}$ . SAT questions about them ask you to spot an equivalent form, build a form they describe, combine fractions, or find the values $x$ can't take. If a question asks you to solve for $x$ instead, it's a rational equation, and Solve rational equations covers that.

Some of these take a line or two by hand. Others come with four long answer choices and messy algebra, and that's where Desmos can save you time. Here's one.

Example: Let the graph test long choices

Desmos example

Which expression is equivalent to

\frac{2x+3}{x-4}-\frac{x-1}{x+2}

for all values of $x$ for which the original expression is defined?

A
$\dfrac{x^2+2x+10}{x^2-2x-8}$
B
$\dfrac{x^2+12x+2}{x^2-2x-8}$
C
$\dfrac{3x^2+2x+10}{x^2-2x-8}$
D
$\dfrac{x^2+12x-2}{x^2-2x-8}$

Fast Desmos solution

Combining these by hand means two products and a subtraction, with plenty of room for a sign slip. The graph is quicker.

Put the original on line 1 and leave it alone. On line 2, type choice A, then edit that same line into B, then C, and so on. You're looking for graph overlap: a curve that lands exactly on top of the original everywhere, not one that only crosses it at a few points. Choice B covers it completely, so the answer is B.

The graph found the formula, but it didn't hand you the excluded values. Those come from the original denominators, $x-4$ and $x+2$ , so $x=4$ and $x=-2$ are out. If you'd like to see the algebra behind choice B, its numerator works out like this:

(2x+3)(x+2)-(x-1)(x-4)=x^2+12x+2.

For more calculator practice, see Equivalent expressions by graph overlap and Restrictions, piecewise functions, and rational expressions.

Calculator loads as you approach

Leave line 1 alone. Edit line 2 through the choices until the two curves match completely.

Ask which way is shorter from start to finish. Count the typing, the algebra, and finding the excluded values.

Start with graph overlap when

The choices are long, fixed formulas in $x$ , like the four in the example above.
Typing the original once and editing one line is quicker than multiplying out and combining everything by hand, as it was there.
You only have to pick one of the written choices, not build a new form yourself.

Start with algebra when

Factoring, canceling, or finding the common denominator takes a line or two, as it does for $\frac{6x^2-24}{3x+6}$ .
The question wants you to build something exact yourself, like a new form, the $k$ in $2+\frac{k}{x-3}$ , or the full list of excluded values.
The expressions have letters besides $x$ , like $a$ or $c$ , or two forms might differ only at a hole. A picture can’t settle either one.

On many questions you’ll use both: the graph picks the formula, and the original denominators give you the excluded values.

Check your understanding:

A question gives four long formulas in $x$ as choices, and it also asks for every excluded value. How would you get each part?

Write down the original exclusions first

A denominator can't be $0$ , because dividing by zero isn't defined. So before you simplify anything, find where the original denominator is zero. Take

\frac{x^2-9}{x^2-3x}.

Factor the denominator:

x^2-3x=x(x-3).

It's zero when $x=0$ or $x=3$ , so

x\ne0 \quad\text{and}\quad x\ne3.

Now factor the top and simplify:

\begin{aligned} \frac{x^2-9}{x^2-3x} &=\frac{(x-3)(x+3)}{x(x-3)}\\[1.4em] &=\frac{x+3}{x}, \qquad x\ne0,3. \end{aligned}

Look at what's left. You can still see that $x=0$ is a problem, since $x$ sits in the denominator. But the factor $x-3$ , the one that explained $x=3$ , is gone. Put $x=3$ into the original and you get $\frac{0}{0}$ , which is undefined. Put it into $\frac{x+3}{x}$ and you get $2$ . The short formula works at $x=3$ , but the original doesn't, so you have to carry $x\ne3$ forward yourself. On the graph of the original, that's a hole at the point $(3,2)$ .

This is an easy value to lose, and not because the math is hard. Once $x-3$ is gone, nothing on the page reminds you that $x=3$ was ever a problem. The fix takes a few seconds, and it's the habit for this whole topic: exclusions first, then simplify.

Check your understanding:

The expression $\frac{(t+5)(t-1)}{(t-1)(t+2)}$ simplifies to $\frac{t+5}{t+2}$ . Which values are excluded, and why?

Cancel factors, not terms

Canceling means dividing the top and the bottom by the same nonzero thing. That only works when that thing multiplies the whole top and the whole bottom:

\frac{AB}{AC}=\frac{B}{C}, \qquad A\ne0.

That's why you factor first. In

\frac{x^2+5x+6}{x^2+4x+3},

it's tempting to cross out the two $x^2$ terms. But each $x^2$ is only one piece of a sum, so it isn't multiplying the whole top or the whole bottom. Factor both polynomials instead:

\begin{aligned} \frac{x^2+5x+6}{x^2+4x+3} &=\frac{(x+2)(x+3)}{(x+1)(x+3)}\\[1.4em] &=\frac{x+2}{x+1}, \qquad x\ne-3,-1. \end{aligned}

Now the matching piece is the whole binomial $x+3$ , a factor of both, so it cancels.

Common mistake:

Crossing out matching pieces inside sums changes the fraction. In $\frac{x+4}{x}$ , the two $x$ ’s are terms, not factors of the whole top and bottom. Test it at $x=2$ : the fraction is $\frac{6}{2}=3$ , but crossing out the $x$ ’s leaves $4$ . Factor first, and if no whole factor matches, nothing cancels.

Try it yourself:

Simplify $\frac{2x^2-8}{x^2+x-6}$ and list every excluded value. Factor the top and the bottom before you cancel anything.

Check your understanding:

What’s the simplified form of the Try it yourself expression, and which values are excluded?

Combine fractions with the least common denominator

You can add or subtract fractions only when their denominators match. The least common denominator, or LCD, is the smallest denominator they all fit into. To build it, take every factor that shows up, and use each one the most times it appears in any single denominator.

Take

\frac{2}{x(x-1)}-\frac{1}{x-1}.

The second denominator, $x-1$ , is already inside the first. So the LCD is $x(x-1)$ itself. Multiplying the two denominators, $x(x-1)^2$ , would still work, but you'd have an extra $x-1$ to cancel at the end.

First, the exclusions from the original denominators:

x\ne0,1.

Only the second fraction needs to change. It's missing a factor of $x$ , so multiply its top and bottom by $x$ . That's the same as multiplying by $1$ , so the value stays the same and only the form changes:

\frac{1}{x-1}\cdot\frac{x}{x} =\frac{x}{x(x-1)}.

Now the denominators match, so subtract the tops:

\begin{aligned} \frac{2}{x(x-1)}-\frac{1}{x-1} &=\frac{2}{x(x-1)}-\frac{x}{x(x-1)}\\[1.4em] &=\frac{2-x}{x(x-1)}, \qquad x\ne0,1. \end{aligned}

Whenever you multiply a bottom, multiply its top too. Changing only the denominator changes the fraction's value.

Check your understanding:

What’s the LCD of $\frac{1}{(x+4)(x-2)}$ and $\frac{3}{x+4}$ , and which fraction needs another factor?

Split a numerator, never a denominator

Adding fractions with the same denominator also works in reverse:

\frac{A+B}{C}=\frac{A}{C}+\frac{B}{C}, \qquad C\ne0.

Each piece of the top is divided by the same bottom, so you can split the pieces apart. That's handy for writing a fraction as a number plus a leftover fraction. Take $\frac{3x+7}{x-2}$ . Write the top using the bottom: $3(x-2)=3x-6$ , and you need $13$ more to reach $3x+7$ . So

\frac{3x+7}{x-2} =\frac{3(x-2)+13}{x-2} =3+\frac{13}{x-2}, \qquad x\ne2.

This number-plus-remainder form puts useful numbers in plain sight. If a question asks you to write $\frac{3x+7}{x-2}$ as $a+\frac{b}{x-2}$ , you can read off $a=3$ and $b=13$ . It also tells you about the graph: when $x$ is far from $2$ , the fraction $\frac{13}{x-2}$ is close to $0$ , so the curve levels off near $y=3$ .

The bottom is different. Splitting a sum in the denominator is false:

\frac{5}{x+2} \ne \frac{5}{x}+\frac{5}{2}.

Try $x=2$ . The left side is $\frac{5}{4}$ , but the right side is $\frac{5}{2}+\frac{5}{2}=5$ . In general, combining the right side gives

\frac{5}{x}+\frac{5}{2} =\frac{10+5x}{2x},

a different expression altogether. A sum in the denominator is one whole divisor, so it stays together. Split the top, never the bottom.

Check your understanding:

Rewrite $\frac{4x-1}{x+3}$ in the form $a+\frac{b}{x+3}$ . What are $a$ , $b$ , and the excluded value?

Example: Simplify first, then combine

Here the moves come in a row: exclusions, then canceling, then combining.

Worked example

Which choice gives an equivalent form together with all values excluded by the original expression?

\frac{x^2-1}{x^2+x-2}-\frac{2}{x+2}

A
$\dfrac{x-1}{x+2}$ , with $x\ne-2,1$
B
$\dfrac{x-1}{x+2}$ , with $x\ne-2$
C
$\dfrac{x+3}{x+2}$ , with $x\ne-2,1$
D
$\dfrac{x-1}{x-1}$ , with $x\ne-2,1$

Step 1

Write down the exclusions

Factor every original denominator. The first one is

x^2+x-2=(x+2)(x-1),

and the second, $x+2$ , is already factored. So across the whole expression, the denominators are zero at

x=-2 \quad\text{and}\quad x=1.

Both values stay excluded, whatever happens next.

Step 2

Factor, then cancel

Factor the first numerator:

x^2-1=(x-1)(x+1).

Now the first fraction simplifies:

\frac{(x-1)(x+1)}{(x+2)(x-1)} =\frac{x+1}{x+2}, \qquad x\ne-2,1.

Only the whole factor $x-1$ cancels.

Step 3

Subtract the numerators

The expression is now

\frac{x+1}{x+2}-\frac{2}{x+2}.

The denominators already match, so subtract the tops:

\frac{x+1-2}{x+2} =\frac{x-1}{x+2}.

Step 4

Put the full answer together

The equivalent form is

\frac{x-1}{x+2}, \qquad x\ne-2,1,

so the answer is A.

Common mistake:

Choice B is the tempting one, because its formula is right: it matches the original at every allowed input. What’s missing is $x=1$ , which vanished when $x-1$ canceled, and the question asks for every original exclusion. Writing the exclusions down in step 1 is what kept $x=1$ from getting lost.

Practice problems

Try these on your own. Before you start each one, look at the choices and decide: is the algebra short, or would the graph check them faster?

Factor and cancel all the way

Practice problem

Which choice gives an equivalent form together with all values excluded by the original expression?

\frac{6x^2-24}{3x+6}

Calculator loads as you approach

The factoring is quick, so algebra is faster here. Graph your final formula if you’d like a check.

Match the formula and keep the exclusions

Practice problem

Which choice gives an equivalent form together with all values excluded by the original expression?

\frac{3}{x-4}+\frac{2}{x+1}

Calculator loads as you approach

Graph the original once, then edit one line through the choices.

Cancel, then split off the remainder

Practice problem

The expression

\frac{2x^2+x-6}{x^2-x-6}

can be written in the form

2+\frac{k}{x-3}.

Which choice gives the value of $k$ and all values excluded by the original expression?

Calculator loads as you approach

Algebra is quicker here: you need an exact k, and a graph can hide the excluded input x=-2.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

These questions ask for an equivalent form or the excluded values, not a value of $x$ .
Exclusions first: find every zero of every original denominator before you simplify.
Cancel factors, not terms, so factor the whole top and bottom first.
A value excluded by a canceled factor stays excluded, even if the shorter formula works there. When it does, the graph shows only a hole.
To add or subtract, build the LCD and multiply each top by whatever its bottom was missing.
Split the top, never the bottom.
Start with graph overlap for long fixed choices in $x$ , and with algebra when it's short or you have to build an exact answer.
Often you'll use both: the graph finds the formula, and the original denominators give the exclusions.

Related lesson

Solve rational equations picks up when the question asks for the values of $x$ that make an equation true.

Next lesson

Use polynomial identities, factors, and unknown coefficients

Use an identity, a given factor, or matching coefficients to find unknown constants.

Start next lesson

Practice

Practice this lesson

516 SAT questions use what this lesson teaches. Practice a few in a study session at the difficulty you choose.

Start practice