Digital SAT Math · Advanced Math
Rational equations and expressions
A rational equation is an ordinary equation wearing a trap. Somewhere in it a denominator is zero for one or two values of , those values are never allowed, and the algebra you are about to do — multiply both sides by a common denominator — can spit those forbidden numbers back out as if they were solutions. The test knows this. The excluded value is almost always a printed choice, and “the equation has no solution” is waiting for the day every candidate fails the domain check. The whole skill is four steps with one rule that is not optional: write the excluded values before you clear anything, then throw them out again at the end.
On the test
| Domain | Advanced Math (score report) |
| What it looks like | An equation with the variable in one or more denominators; sometimes a rational expression to simplify first, then solve |
| Often asked | “What is the solution…?”, “What is the solution set…?”, “Which value of satisfies…?”, and the choice “The equation has no solution” |
| Format | Multiple choice and student-produced response; expect solution-set wording on the harder items |
| Calculator | Desmos can plot both sides and show the real intersections — it never plots an extraneous root, which is the point of the box below |
Recognition cues: a variable under a fraction bar; two or more fractions set equal or combined; the phrases solution set, no solution, for all x such that; and any item that factors a quadratic denominator into linear pieces before you touch the numerators.
What this page owns
The College Board files rational equations under “nonlinear equations in one variable,” so most free pages bury them inside a five-family nonlinear lesson. We split them out. Here the defining move is clear a denominator and then discard forbidden roots. When the same factoring is used to rewrite an expression and nothing is ever solved, that is equivalent expressions. When the equation is a plain quadratic, a radical, or a product already equal to zero, that is nonlinear equations — which keeps one teaser rational item so the domain check is not a surprise the first time it appears. When the answer is a feature of a parabola, that is quadratic functions. The polynomial work is shared; the domain-first discipline is this page’s job.
Pattern recognition
Five shapes cover nearly every item:
- Single fraction equals a constant — , or . One excluded value; one linear equation after clearing.
- Two fractions set equal — cross-multiply (which is LCD multiply with two terms), watch both excluded values.
- Sum or difference of fractions equals a third — LCD is usually the product of the linear denominators, or the factored form of a quadratic denominator already printed.
- Polynomial over linear equals a linear or a constant — clearing produces a quadratic; one root is often the excluded value.
- No solution — every candidate from the cleared equation is excluded, or the cleared equation is inconsistent after the domain cut.
Method
- Note every excluded value. Any that zeros a denominator in the original equation is out. Factor quadratic denominators if needed so you can see the linear pieces.
- Multiply through by the common denominator. Multiply every term — constants and lone variables included, not only the fractions. Distribute carefully through any parentheses that appear.
- Solve the resulting polynomial equation. Linear ones finish in one line; quadratics factor or use the formula. You now have a list of candidates, not solutions.
- Discard any candidate that was excluded. Substitute the survivors into the original equation (not the cleared one) to confirm. If the list is empty, the equation has no solution.
| What you are looking at | LCD move | What usually goes wrong |
|---|---|---|
| one fraction constant | multiply both sides by the denominator | picking the excluded value as the answer |
| two fractions equal | cross-multiply, or LCD of both | forgetting one of the two excluded values |
| sum/difference of fractions | multiply every term by the product of the linear denoms | multiplying only the fraction terms; sign error on a subtracted fraction |
| rational rational with shared quadratic denom | factor first, then multiply by the factors | keeping the root that zeros the cancelled factor |
| only candidate is excluded | discard it | reporting that candidate, or inventing a second root |
Worked example 1 — domain first, then one line
Stem. What is the solution to ?
Step 1 — excluded values. The denominator is zero at . Box it: .
Step 2 — clear. Multiply both sides by :
Step 3 — discard and check. , so it is allowed. In the original: ✓.
Trap watch. makes the left side undefined, so it cannot be a solution even though it is the number that “wants” to sit in the denominator. Reporting forgets to solve and just copies the numerator. Answer: .
Worked example 2 — clearing invents a root, so the domain kills it
Stem. What is the solution set of ?
Step 1 — excluded values. gives . Box it.
Step 2 — clear. Multiply both sides by :
Candidates: and .
Step 3 — discard. is exactly the boxed value, so it is extraneous. Only remains.
Check in the original. At : ✓. At the original is , undefined. Answer: .
Trap watch. Three wrong finishes, all common. Keeping both candidates reports . Keeping only the forbidden one reports . Throwing both out reports no solution. The cleared quadratic does have two roots; the original equation has one. Checking in will “confirm” both and teach you nothing.
Worked example 3 — the only candidate is excluded
Stem. What is the solution set of ?
Step 1 — excluded values. Both denominators vanish at . Box it.
Step 2 — clear. Multiply every term by :
The only candidate is .
Step 3 — discard. That candidate is exactly the excluded value. Nothing remains.
Check. At neither side of the original equation is defined, so is not a solution of anything you were asked. The cleared equation is true at and the original is not an equation there at all. Answer: the equation has no solution.
Trap watch. Reporting is the distractor the item exists to catch. A partial multiply that forgets the constant produces , which fails the original: on the left and on the right. And “no solution” here does not mean the cleared algebra was wrong — it means the algebra was right and the domain finished the job.
Practice
Answer before opening the explanation. Two items are student-produced response. At least three items print an extraneous root as a choice on purpose; “the equation has no solution” is a real answer on this skill, not a throwaway. Every wrong choice below is one named misconception — excluded value kept, constant term not multiplied by the LCD, minus distributed to only half a numerator, both candidates kept after a domain cut. When you miss one, log the misconception, not the item number.
Question 1 Warm-up
What is the solution to the equation 6/(x − 2) = 3?
Show the answer Choice A
Why it is right
The denominator x − 2 is zero at x = 2, so x = 2 is excluded before any algebra. Multiply both sides by x − 2 to get 6 = 3(x − 2). Divide both sides by 3: 2 = x − 2, so x = 4. The candidate 4 is not the excluded value. Checking in the original equation, 6/(4 − 2) = 6/2 = 3, which matches the right side.
Why each other choice fails
- Choice B
- This is the excluded value that makes the denominator zero. At x = 2 the left side of the original equation is undefined, so 2 cannot be a solution even though it is the number sitting in the denominator.
- Choice C
- Adds the 2 instead of solving 2 = x − 2. After clearing, 6 = 3(x − 2) simplifies to x − 2 = 2, so x is 2 more than 2, which is 4, not 8.
- Choice D
- Flips the sign when finishing x − 2 = 2, reporting −4 instead of 4. Substituting −4 into the original gives 6/(−6) = −1, which is not 3.
Question 2 Standard
What is the solution to the equation 2/(x + 3) = 4/(x + 9)?
Show the answer Choice B
Why it is right
The denominators exclude x = −3 and x = −9. Cross-multiplying (or multiplying both sides by (x + 3)(x + 9)) gives 2(x + 9) = 4(x + 3), so 2x + 18 = 4x + 12. Subtract 2x from both sides: 18 = 2x + 12. Subtract 12: 6 = 2x, so x = 3. The value 3 is not excluded. Checking: left side 2/(3 + 3) = 2/6 = 1/3, and right side 4/(3 + 9) = 4/12 = 1/3.
Why each other choice fails
- Choice A
- The excluded value from the first denominator. At x = −3 the left side is undefined, so −3 cannot solve the original equation.
- Choice C
- The excluded value from the second denominator. At x = −9 the right side is undefined. Excluded values are never solutions, even when they appear as clean integers among the choices.
- Choice D
- Stops after getting 6 = 2x and reports 6 instead of dividing by 2. Substituting x = 6 gives 2/9 on the left and 4/15 on the right, which are not equal.
Question 3 Standard
What is the solution to the equation (3x + 1)/(x − 4) = 2?
Show the answer Choice C
Why it is right
The denominator excludes x = 4. Multiply both sides by x − 4: 3x + 1 = 2(x − 4) = 2x − 8. Subtract 2x: x + 1 = −8. Subtract 1: x = −9. The candidate −9 is not excluded. Checking: (3(−9) + 1)/(−9 − 4) = (−27 + 1)/(−13) = (−26)/(−13) = 2, which matches the right side.
Why each other choice fails
- Choice A
- The excluded value that zeros the denominator. At x = 4 the left side is undefined, so 4 is not a solution of the original equation.
- Choice B
- Solves 3x + 1 = 2x + 8 or otherwise flips a sign when expanding 2(x − 4), landing on the positive counterpart of the true root. At x = 9 the left side is (27 + 1)/5 = 28/5, not 2.
- Choice D
- Sets only the numerator equal to zero, as if solving 3x + 1 = 0. That finds where the left side is zero, not where it equals 2.
Question 4 Standard Student-produced response
What is the solution to the equation 4/(x − 1) = 8/(3x + 1)?
Show the answer −3
Why it is right
The denominators exclude x = 1 and x = −1/3. Cross-multiplying gives 4(3x + 1) = 8(x − 1), so 12x + 4 = 8x − 8. Subtract 8x: 4x + 4 = −8. Subtract 4: 4x = −12, so x = −3. The value −3 is not excluded. Checking: left side 4/(−3 − 1) = 4/(−4) = −1, and right side 8/(−9 + 1) = 8/(−8) = −1.
Answers students type instead
- 1
- The excluded value from the first denominator. At x = 1 the left side is undefined, so 1 cannot be a solution even though it is a clean integer near the problem.
- 3
- Loses the sign when finishing 4x = −12, reporting the positive counterpart. Substituting x = 3 gives 4/2 = 2 on the left and 8/10 = 4/5 on the right.
- -1/3
- The excluded value from the second denominator, where 3x + 1 = 0. Excluded values solve nothing in the original equation.
Question 5 Standard
What is the solution to the equation 3/(x + 2) + 1 = 5/(x + 2)?
Show the answer Choice D
Why it is right
The shared denominator excludes x = −2. Multiply every term by x + 2: 3 + 1·(x + 2) = 5, so 3 + x + 2 = 5, hence x + 5 = 5 and x = 0. The candidate 0 is not excluded. Checking: 3/(0 + 2) + 1 = 3/2 + 1 = 5/2, and 5/(0 + 2) = 5/2.
Why each other choice fails
- Choice A
- The excluded value that zeros both denominators. At x = −2 neither fraction is defined, so −2 cannot be a solution.
- Choice B
- Copies a numerator from the equation instead of solving. After clearing, the equation is x + 5 = 5, not x = 5.
- Choice C
- Solves as if the constant 1 were never multiplied by the LCD, or subtracts incorrectly from 5 − 3. At x = 2 the left side is 3/4 + 1 = 7/4 and the right side is 5/4.
Question 6 Harder
What is the solution set of the equation (x² − x − 6)/(x − 3) = 4?
Show the answer Choice A
Why it is right
The denominator excludes x = 3. Multiply both sides by x − 3: x² − x − 6 = 4(x − 3) = 4x − 12. Bring all terms to one side: x² − x − 6 − 4x + 12 = 0, so x² − 5x + 6 = 0, which factors as (x − 2)(x − 3) = 0. Candidates are x = 2 and x = 3. Discard x = 3 because it is excluded. Only x = 2 remains, and it checks: (4 − 2 − 6)/(2 − 3) = (−4)/(−1) = 4.
Why each other choice fails
- Choice B
- Keeps the extraneous root and discards the valid one. At x = 3 the original left side is undefined (0/0 after the numerator also vanishes), so 3 is not a solution.
- Choice C
- Solves the cleared quadratic correctly and never returns to the domain restriction. Multiplying by x − 3 is legal only when x ≠ 3, so that candidate must be discarded.
- Choice D
- Assumes that once the extraneous root is thrown out nothing is left. The quadratic has two roots and only one of them is excluded; x = 2 makes the denominator −1 and satisfies the equation.
Question 7 Harder
What is the solution to the equation 5/(x − 2) − 3/(x + 2) = 4/(x² − 4)?
Show the answer Choice B
Why it is right
Factor x² − 4 = (x − 2)(x + 2). The equation excludes x = 2 and x = −2. Multiply every term by (x − 2)(x + 2): 5(x + 2) − 3(x − 2) = 4. Expand carefully: 5x + 10 − 3x + 6 = 4, because −3(x − 2) = −3x + 6. Combine: 2x + 16 = 4, so 2x = −12 and x = −6. The candidate −6 is not excluded. Checking: 5/(−8) − 3/(−4) = −5/8 + 3/4 = −5/8 + 6/8 = 1/8, and 4/(36 − 4) = 4/32 = 1/8.
Why each other choice fails
- Choice A
- Distributes the minus on the second fraction incorrectly, treating −3(x − 2) as −3x − 6 instead of −3x + 6. That produces 2x + 4 = 4, hence x = 0, which fails in the original: 5/(−2) − 3/2 = −4, while 4/(0 − 4) = −1.
- Choice C
- The excluded value from the first denominator (and a factor of the third). At x = 2 every original denominator that contains x − 2 is zero, so the equation is undefined there.
- Choice D
- The positive counterpart of the true root, from a sign slip when solving 2x = −12 or when combining constants. At x = 6 the left side is 5/4 − 3/8 = 10/8 − 3/8 = 7/8, not 4/32.
Question 8 Harder
What is the solution to the equation x/(x − 3) + 4 = 12/(x − 3)?
Show the answer Choice C
Why it is right
The shared denominator excludes x = 3. Multiply every term by x − 3, including the constant 4: x + 4(x − 3) = 12. Expand: x + 4x − 12 = 12, so 5x − 12 = 12, 5x = 24, and x = 24/5. The candidate 24/5 is not 3. Checking: (24/5)/(24/5 − 3) + 4 = (24/5)/(9/5) + 4 = 24/9 + 4 = 8/3 + 12/3 = 20/3, and 12/(24/5 − 3) = 12/(9/5) = 12 · 5/9 = 60/9 = 20/3.
Why each other choice fails
- Choice A
- The excluded value that zeros both denominators. At x = 3 the original equation is undefined, so 3 cannot be a solution.
- Choice B
- Multiplies only the fraction terms by the LCD and leaves the constant 4 alone, producing x + 4 = 12 and x = 8. The constant must become 4(x − 3). At x = 8 the left side is 8/5 + 4 = 28/5 and the right side is 12/5.
- Choice D
- Divides 12 by 5 after a partial clear, or solves 5x = 12 instead of 5x = 24. Substituting 12/5 fails the original equation the same way any wrong linear root does.
Question 9 Harder
What is the solution set of the equation (x² − 6x + 8)/(x − 2) = 3?
Show the answer Choice D
Why it is right
The denominator excludes x = 2. Multiply both sides by x − 2: x² − 6x + 8 = 3(x − 2) = 3x − 6. Bring all terms to one side: x² − 6x + 8 − 3x + 6 = 0, so x² − 9x + 14 = 0, which factors as (x − 2)(x − 7) = 0. Candidates are x = 2 and x = 7. Discard x = 2 because it is excluded. Only x = 7 remains, and it checks: (49 − 42 + 8)/(7 − 2) = 15/5 = 3.
Why each other choice fails
- Choice A
- Keeps the extraneous root that the domain forbids. At x = 2 the original left side is undefined, so 2 is not a solution even though it is a root of the cleared quadratic.
- Choice B
- Reports both roots of the cleared equation without applying the domain filter. Multiplying by x − 2 is valid only for x ≠ 2, so 2 must be discarded after solving.
- Choice C
- Throws out every candidate because one of them is excluded. The quadratic has two roots; only one is forbidden, and x = 7 survives and checks in the original equation.
Question 10 Hardest
What is the solution to the equation 4/(x − 3) + 2/(x + 1) = 5x/(x² − 2x − 3)?
Show the answer Choice A
Why it is right
Factor x² − 2x − 3 = (x − 3)(x + 1). The equation excludes x = 3 and x = −1. Multiply every term by (x − 3)(x + 1): 4(x + 1) + 2(x − 3) = 5x. Expand: 4x + 4 + 2x − 6 = 5x, so 6x − 2 = 5x. Subtract 5x: x − 2 = 0, hence x = 2. The candidate 2 is not excluded. Checking: 4/(2 − 3) + 2/(2 + 1) = 4/(−1) + 2/3 = −4 + 2/3 = −10/3, and 5(2)/(4 − 4 − 3) = 10/(−3) = −10/3.
Why each other choice fails
- Choice B
- The excluded value from the factor x − 3. At x = 3 the original equation has two undefined terms, so 3 cannot be a solution.
- Choice C
- Comes from a sign error when clearing the second fraction, treating the left side as 4(x + 1) − 2(x − 3) instead of 4(x + 1) + 2(x − 3). That yields 2x + 10 = 5x and x = 10/3, which fails the original.
- Choice D
- The excluded value from the factor x + 1. At x = −1 two of the three denominators are zero, so the original equation is undefined there.
Question 11 Hardest Student-produced response
If 1/x + 1/(x + 6) = 1/4, what is the positive value of x?
Show the answer 6
Why it is right
The denominators exclude x = 0 and x = −6. Multiply every term by the LCD 4x(x + 6): 4(x + 6) + 4x = x(x + 6). Expand: 4x + 24 + 4x = x² + 6x, so 8x + 24 = x² + 6x. Bring all terms to one side: 0 = x² − 2x − 24 = (x − 6)(x + 4). Candidates are x = 6 and x = −4. Neither is excluded. The question asks for the positive value, so x = 6. Checking: 1/6 + 1/12 = 2/12 + 1/12 = 3/12 = 1/4.
Answers students type instead
- 0
- An excluded value where the first denominator vanishes. At x = 0 the original left side is undefined, so 0 is never a solution.
- 4
- Copies the constant from the right side of the equation instead of solving. Substituting x = 4 gives 1/4 + 1/10 = 7/20, which is not 1/4.
- -4
- The other algebraic root. Both 6 and −4 satisfy the original equation, but the stem asks for the positive value of x only.
Question 12 Hardest
What is the solution set of the equation 2/(x − 1) − 3/(x + 1) = 4/(x² − 1)?
Show the answer Choice B
Why it is right
Factor x² − 1 = (x − 1)(x + 1). The equation excludes x = 1 and x = −1. Multiply every term by (x − 1)(x + 1): 2(x + 1) − 3(x − 1) = 4. Expand: 2x + 2 − 3x + 3 = 4, so −x + 5 = 4. Then −x = −1, so x = 1. The only candidate is x = 1, which is excluded. Nothing remains, so the solution set is empty. At x = 1 the original equation is undefined, which confirms that 1 is not a solution of anything that was asked.
Why each other choice fails
- Choice A
- Reports the extraneous root that the domain forbids. Clearing produces x = 1, but at x = 1 two denominators are zero, so the original equation has no value there and 1 cannot be a solution.
- Choice C
- Picks the other excluded value without solving. At x = −1 the original equation is also undefined; −1 never appears as a candidate of the cleared equation either.
- Choice D
- Lists both excluded values as if they were roots. Domain restrictions remove inputs; they do not create a solution set out of the forbidden numbers.
Common mistakes
- Picking the excluded value as the answer — it solves the cleared equation and makes the original undefined. Write the excluded list before clearing so the match is obvious at the end.
- Keeping every root of the cleared equation — solution-set items put next to where is forbidden. Domain is a filter, not optional commentary.
- Multiplying only the fraction terms by the LCD — a constant or a lone sitting next to a fraction still has to be multiplied. The partial clear produces a clean wrong linear root.
- Sign error on a subtracted fraction — times is . Writing flips the constant and lands on a tidy distractor.
- Checking in the cleared equation — every candidate satisfies it by construction. Substitute into the equation as printed.
- Calling a removable hole a solution — after cancelling a common factor , the simplified equation may hold at , but the original expression never had a value there.
- Forgetting one factor of a quadratic denominator — excludes both . Clearing by alone leaves a still-rational equation.
- Reporting “no solution” too early — a quadratic can have two candidates of which only one is excluded. Discard the forbidden one; keep the survivor.
- Reporting a solution when every candidate is excluded — the reverse of (8). Empty after the filter means no solution, not the excluded number.
- Cross-multiplying a sum — is not a proportion. Build the LCD and multiply through; cross-multiply only when one fraction equals one fraction.
- Answering a single root when the item asked for the solution set — and the reverse. Read the last words of the stem before choosing a letter.
- Entering both roots on a grid-in that asked for the positive (or the greatest) one — the other root is often the excluded value or a negative the question already ruled out.
FAQ
Why can clearing a denominator create a fake solution? Because multiplying both sides by an expression is only reversible where that expression is not zero. At a zero of a denominator the original equation is undefined, so that input was never in the conversation — but the cleared polynomial does not know that and may list it as a root.
Is “no solution” the same as ? No. Zero is a number that can solve an equation. “No solution” means the solution set is empty. On multiple choice it is a written option; a grid-in will not ask a question whose answer does not exist.
Do I always get a quadratic after clearing? No. Single-fraction and two-fraction proportions usually stay linear. Quadratics appear when a numerator is already degree 2, or when a constant multiplies a linear factor and you bring everything to one side. Degree is a signal, not a requirement.
Can I cancel before I solve? Yes, if you cancel a common factor of numerator and denominator and keep the restriction from that factor. Cancelling turns into with , which is often faster than expanding. Cancelling terms (pieces of a sum) is still illegal.
When is Desmos faster than hand clearing? When there are three different linear denominators or ugly coefficients. For a clean two-denominator item, writing the excluded list and multiplying is usually under thirty seconds and gives an exact fraction for free.
How is this different from radical extraneous roots? Same idea, different one-way step. Squaring destroys sign information; multiplying by a denominator invents roots at the zeros of that denominator. In both cases the cure is the same: test candidates in the original equation, and know why a particular candidate is likely to fail before you test it.
Grid-in format?
Enter the exact value. Fractions are safer than rounded decimals (3/2 rather than 1.5). If the question asks for the positive solution and both a positive and a negative candidate appear, enter only the positive — and still confirm it is not excluded.