Algebra 2 · Rational expressions · Polynomials worksheets · Quadratics worksheets
Practice solving 25 rational equations, identifying excluded values, and checking for extraneous solutions. Includes four worked examples, an interactive graph, a randomized drill on excluded values, and a printable answer key with every excluded value written out.
Part of a nine guide unit. See all the rational expressions and rational functions worksheets, listed in a recommended order.
Two files. The worksheet holds all 25 practice problems plus three challenge problems, with room to work and a two page answer key that shows the excluded values, the cleared equation, and the verdict on every candidate. The cheat sheet is one page for a binder or a classroom wall.
Free to copy for your own classroom or your own kitchen table. If you want to share it, please link to this page rather than rehosting the files, so students always land on the current version.
A rational equation is an equation with a variable in at least one denominator. That single feature changes the whole procedure, because a denominator is allowed to be any number except zero. Before you solve anything, the equation has already told you which values of the variable are off limits.
Take this equation:
5x − 2 = 3x + 4
If x were 2, the left side would read 5 divided by 0, which has no value. If x were −4, the right side would do the same thing. So 2 and −4 are not candidates for anything. They are excluded values.
The method for solving is short. Multiply every term by the least common denominator, which clears the fractions and leaves a polynomial equation. Solve that. The polynomial equation is easier to solve, but its roots still need to be checked against the original restrictions. That check is where extraneous solutions get caught.
If factoring the denominators is the part that slows you down, work through the factoring polynomials flowchart guide first, and simplifying rational expressions after that. Every equation on this page needs one of those two skills before the algebra starts.
I teach this topic as two lists. Write the excluded values before solving, so you can compare them against your answers at the end.
Factor every denominator, set each factor equal to zero, and write the results down. This list depends only on the original equation. Nothing you do later can change it.
Multiply through by the LCD, solve the polynomial equation, and write the results down. These are candidate solutions. Check each one against the original equation.
Now compare the lists. Any candidate that also appears on the excluded list is thrown out. Any candidate that does not appear on the excluded list gets substituted into the original equation to confirm it. What remains is the solution set, and it can be empty.
Step 6 has to use the original equation. Substituting into the cleared polynomial equation proves nothing, because every candidate satisfies that one by construction. That is the check that catches extraneous solutions, and doing it in the wrong equation is the single most common way students lose the point.
List 1. The denominators are x + 1 and x − 3, so the excluded values are −1 and 3.
Clear the fractions. With a single fraction on each side you can cross multiply, which is the same thing as multiplying both sides by (x + 1)(x − 3).
4(x − 3) = 6(x + 1)
4x − 12 = 6x + 6
−18 = 2x
x = −9
List 2. One candidate, −9. It is not on List 1, so substitute it into the original equation.
4−9 + 1 = 4−8 = −12 and 6−9 − 3 = 6−12 = −12
Both sides equal −12. Solution x = −9
Cross multiplication is safe here only because each side is a single fraction. As soon as a side holds a sum of two fractions, cross multiplying is wrong, and you need the LCD instead.
Factor first. x2 − 3x − 4 = (x − 4)(x + 1), so all three denominators are built from the same two factors and the LCD is (x − 4)(x + 1).
List 1. x = 4 and x = −1. Write those down now, in the margin, before touching the algebra.
Multiply every term by the LCD. The first fraction keeps x + 1, the second keeps x − 4, and the third loses its denominator entirely.
x(x + 1) + (x − 4) = 20
x2 + x + x − 4 = 20
x2 + 2x − 24 = 0
(x + 6)(x − 4) = 0
List 2. x = −6 and x = 4.
| Candidate | On List 1? | Substitute into the original | Verdict |
|---|---|---|---|
| x = −6 | No | −6−10 + 1−5 = 35 − 15 = 25, and 2050 = 25 | Keep |
| x = 4 | Yes | Both x − 4 and x2 − 3x − 4 become 0, so two of the three fractions have no value | Reject |
Solution x = −6. The value 4 is an extraneous solution. It is a perfectly good root of x2 + 2x − 24 = 0, and it is not a root of the equation the question actually asked about.
Factor and list. x2 − 4 = (x − 2)(x + 2), so List 1 is 2 and −2, and the LCD is (x − 2)(x + 2).
Clear and solve. Watch the subtraction. The second numerator needs brackets before it gets distributed.
5(x + 2) − 3(x − 2) = 20
5x + 10 − 3x + 6 = 20
2x + 16 = 20
x = 2
Compare the lists. The only candidate is 2, and 2 is on List 1.
No solution Write the answer as no solution, or as the empty set. Do not write x = 2 with a note beside it, and do not leave the answer blank. An equation with no solution is a complete answer, and on most rubrics it carries the same credit as a numerical one.
There is a second thing worth noticing here. The candidate came out of correct algebra. Nothing went wrong in the solving. The equation simply had no value of x that worked, and the clearing step could not tell you that on its own.
Set it up as rates. In one hour pump A fills 1t of the tank and pump B fills 1t + 3. Together they fill 12 of it.
1t + 1t + 3 = 12
List 1. t = 0 and t = −3. The LCD is 2t(t + 3).
2(t + 3) + 2t = t(t + 3)
4t + 6 = t2 + 3t
t2 − t − 6 = 0
(t − 3)(t + 2) = 0
List 2. t = 3 and t = −2. Neither one is on List 1, so neither one is extraneous. Substitute both and they both check:
| Candidate | Check in the original equation | Status as a number | Status as an answer |
|---|---|---|---|
| t = 3 | 13 + 16 = 12 | True solution | Use it |
| t = −2 | −12 + 1 = 12 | True solution | Discard |
Answer Pump A takes 3 hours and pump B takes 6 hours.
Read that second row again, because it is the distinction this page exists for. t = −2 satisfies the equation. It is not extraneous. It is discarded for a different reason: a pump cannot take a negative number of hours to fill a tank. That is a modeling decision about the word problem, not an algebra check about the domain.
Two different reasons to throw a candidate out. Extraneous means the value breaks the equation, because it makes a denominator zero. Non viable means the value satisfies the equation but contradicts the situation, like a negative time, a negative length, or a speed faster than the boat. Word problems can produce either one, so on an applied question I ask students to write both reasons explicitly. Graders give credit for the reason, not just the crossing out.
Most algebra steps can be undone. If you add 7 to both sides, you can subtract 7 and get back exactly where you were, so the new equation and the old one have the same solutions. Multiplying both sides by 5 works the same way, because you can divide by 5 afterward.
Multiplying both sides by x − 4 is different. Clearing denominators gives an equation that is equivalent to the original one only for the values the original equation allows. At x = 4 the original equation has no value at all, so there is nothing there for the new equation to be equivalent to.
The polynomial equation you are left with is defined at every real number, including the ones the original equation excludes. It can therefore have roots sitting at those excluded values. Those roots belong to the polynomial equation and not to the equation you were asked about, which is why they get rejected rather than reported.
| Step | Reversible? | Can it create a solution that was not there? |
|---|---|---|
| Add or subtract the same quantity | Always | No |
| Multiply or divide by a nonzero constant | Always | No |
| Multiply by an expression containing the variable | Only where that expression is not zero | Yes |
| Square both sides | Only where both sides share a sign | Yes |
The last two rows are the whole reason rational equations and radical equations are grouped together in the same standard, though the two steps fail for different reasons. Clearing denominators produces an equation that is equivalent only where the denominators are nonzero. Squaring both sides does not change what is defined, it loses the sign, so a = b and −a = b both turn into a2 = b2. Either way the final equation can carry roots the first one did not have, so both need a check at the end. When a student asks me why radical equations also require checking, this table is what I draw.
Seen on a graph it is plainer. Treat the left side as one function and the right side as another. Solutions occur where both sides are defined and have the same value. An excluded candidate cannot be a solution, because at least one side is undefined there.
What the graph looks like at an excluded value depends on one thing: whether the factor in the denominator cancels against the numerator.
| Left side | At the excluded value | What the graph does |
|---|---|---|
| xx − 2 | Numerator is 2, denominator is 0 | Vertical asymptote at x = 2. The curve runs off to infinity on both sides and never reaches that line. |
| x2 − 4x − 2 | Numerator is 0 and denominator is 0 | Hole at x = 2. The expression equals x + 2 everywhere else, so the graph is a straight line with one point missing. |
The second graph looks like a straight line, so the missing point can be easy to overlook. Cancelling x − 2 changes what the expression looks like and does not change what it is allowed to do. The original expression still has x − 2 in a denominator, so x = 2 is excluded, and it stays on List 1 for the rest of the problem.
This matters because the two graphs can appear to meet at exactly that missing point. Substituting still settles it: if a denominator in the original equation is zero there, the value is rejected, whatever the picture suggests. Problem 19 includes a removable hole. The last visualizer preset shows a case where that hole occurs at an apparent intersection.
Find the excluded values before you simplify. Write List 1 from the denominators as they were handed to you. A student who cancels first and then reads the excluded values off the simplified expression loses every hole in the problem, because the factor that caused them is gone from the page.
The next section lets you look at both cases on a graph.
Pick an equation. The graph draws the left side in blue and the right side in orange, with a dashed vertical line at every excluded value. Then click a candidate to see what happens when you put it back into the original equation. A solution sits where both sides are defined and equal. An excluded candidate sits on a dashed line, where at least one side has no value. The last preset is there because that missing value is not always an asymptote. Sometimes the factor cancels and the graph has a hole instead, in the exact spot where the two curves look like they meet.
Choose a candidate above to run the substitution check.
Two habits come out of looking at these graphs. First, compare any answer against List 1 before you write it down, since you already recorded those values. Second, read the factored form of the cleared equation. If one of its factors is also a factor of the LCD, that root will be excluded.
Read the denominators, factor them in your head, and type every value the equation forbids. Separate them with commas, and use a slash for fractions such as 5/2.
Type the values that would make any denominator zero, then press Check.
Order does not matter. If a value is forbidden by two different denominators, you only need to list it once. Fractions can be written as 3/2 or 3 / 2.
A note on the fractional answers this drill produces. A denominator such as 3x + 4 forbids x = −43, not −4. Excluded values are allowed to be fractions, negatives, or zero, and they carry no information about what the answer will look like.
These six come up in my own sessions, in roughly this order of frequency. Each one is written the way I say it at the table, with the fix beside it.
A student finishes the algebra, gets x = 4, and then looks back at the original equation to see whether 4 was allowed. By that point the page is covered in work and the original denominators are three lines up. The excluded values get missed, not because the student does not know the rule, but because nothing on the page is holding them.
The fractions get multiplied correctly and a lone whole number sits there untouched. In xx + 3 + 1 = 7x + 3, the 1 has to become x + 3. Write the multiplication above every term, including the ones that are already whole.
When the second fraction is subtracted, its whole numerator changes sign, not only the first term of it. I have students write the brackets before they distribute anything, every time, even when the numerator is a single number.
A denominator of x2 − 9 gets read as one excluded value instead of two. So does x2 + 5x, which excludes both 0 and −5. Factoring the denominator is not optional bookkeeping here. It is how you find the list.
The drill above practices this step on its own, with no solving attached.
A student substitutes x = 4 into x2 + 2x − 24 = 0, finds that it works, and keeps it. It does work. Every candidate works in that equation, which is where it came from. The check has to happen in the equation with the denominators still in it.
Cross multiplication is a shortcut for one specific shape, a single fraction equal to a single fraction. Applied to 1x + 1x + 2 = 34 it produces nonsense, because the left side is not one fraction yet. My rule for this is short: if you can count more than two fractions in the equation, you are using the LCD.
The whole method, sized for one side of paper. This is the same content as the printable PDF, written out here so it is searchable and readable on a phone.
A cancelling factor still excludes. x2 − 4x − 2 equals x + 2 everywhere except at x = 2, where the graph has a hole. Take the list from the denominator before you cancel.
Cross multiply only when the equation is one fraction equal to one fraction.
Multiply by the LCD in every other case, including when a whole number or a variable term sits outside a fraction.
Clearing denominators gives an equation that is equivalent to the original one only for the values the original allows. The polynomial you are left with is defined everywhere, so it can carry roots at excluded values. Those roots have to be rejected.
Extraneous: makes a denominator zero, so it breaks the equation. Reject on algebra.
Non viable: satisfies the equation but contradicts the situation, such as a negative time or length. Reject on the context, and say which.
xx − 4 + 1x + 1 = 20x2 − 3x − 4 → List 1: 4, −1 → x2 + 2x − 24 = 0 → List 2: −6, 4 → 4 is excluded → x = −6
Download the cheat sheet (PDF)
Work in order. Part A keeps the algebra light so the checking routine becomes automatic. Part B adds the LCD. Part C adds factoring, and four of its seven problems involve a candidate that has to be thrown out. Part D mixes everything and adds word problems, where a candidate can fail for a reason that has nothing to do with the domain.
For every problem, write List 1 before you solve, state the LCD, and finish with a solution set. Answers with no solution and answers with two solutions both appear in this set, so a clean looking answer is not evidence that you are right.
Four Common Core standards touch this page. The first one names extraneous solutions directly, which is why the checking routine gets as much space here as the solving does.
| Standard | What it asks for | Where it appears above |
|---|---|---|
| HSA.REI.A.2 | Solve simple rational and radical equations in one variable, and give examples showing how extraneous solutions may arise. | The two list method, all four worked examples, Parts A through C |
| HSA.REI.A.1 | Explain each step in solving an equation, and construct a viable argument to justify a solution method. | Why extraneous solutions appear, including the table of reversible and non reversible steps |
| HSA.CED.A.1 | Create equations in one variable and use them to solve problems, including equations arising from rational functions. | Worked example 4 and Part D problems 20 through 22 and 25 |
| HSA.CED.A.3 | Interpret solutions as viable or non viable options in a modeling context. | Worked example 4 and Part D problem 21, where a true solution is discarded as a time |
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In tutoring sessions we work through the factoring, the solving, and the checking together, spending the time on whichever of those three the student finds difficult. For rational equations that is often the checking, since it is the step that is easiest to leave off when the algebra already looks finished.
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