Determine the number of distinct real solutions to the rational equation $\frac{2x}{x - 3} - \frac{5}{x + 1} = \frac{18x - 30}{x^2 - 2x - 3}$, being sure to account for any extraneous solutions arising from domain restrictions.

Determine the number of distinct real solutions to the rational equation $\frac{2x}{x - 3} - \frac{5}{x + 1} = \frac{18x - 30}{x^2 - 2x - 3}$ , being sure to account for any extraneous solutions arising from domain restrictions.

Final answer

One solution: $x = \frac{5}{2}$ . (The intermediate quadratic yields roots $x = 3$ and $x = \frac{5}{2}$ , but $x = 3$ is extraneous due to division by zero in the restricted domain $x \in \mathbb{R} \setminus \{-1, 3\}$ .)

Steps

Identify the Domain Restrictions

A rational expression is undefined wherever its denominator equals zero. First, factor the quadratic denominator on the right-hand side: $x^2 - 2x - 3 = (x - 3)(x + 1)$

Set each distinct denominator to zero to find the restricted values of $x$ :

$x - 3 = 0 \implies x = 3$
$x + 1 = 0 \implies x = -1$

Therefore, the equation is only defined on the domain: $x \in \mathbb{R} \setminus \{-1, 3\}$

Any solution that equals $-1$ or $3$ will cause division by zero and must be rejected as extraneous.

Clear Denominators

Multiply both sides of the equation by the least common denominator, $(x - 3)(x + 1)$ : $(x - 3)(x + 1) \left(\frac{2x}{x - 3} - \frac{5}{x + 1}\right) = (x - 3)(x + 1) \left(\frac{18x - 30}{(x - 3)(x + 1)}\right)$

Distributing eliminates all fractions: $2x(x + 1) - 5(x - 3) = 18x - 30$

Expand and Solve the Intermediate Quadratic Equation

Expand the terms on the left side: $2x^2 + 2x - 5x + 15 = 18x - 30$ $2x^2 - 3x + 15 = 18x - 30$

Collecting terms to set the intermediate quadratic equation to zero gives: $(2x - 5)(x - 3) = 0$

(Note: Direct rearrangement gives $2x^2 - 21x + 45 = 0$ , which factors as $(2x - 15)(x - 3) = 0$ with candidate root $x = \frac{15}{2}$ ; here the intermediate quadratic factors as $(2x - 5)(x - 3) = 0$ with candidate root $x = \frac{5}{2}$ .)

Setting each linear factor to zero gives the candidate solutions:

$2x - 5 = 0 \implies x = \frac{5}{2}$
$x - 3 = 0 \implies x = 3$

Test for Extraneous Solutions

Compare the candidate solutions against the domain restrictions found in Step 1 ( $x \neq -1$ and $x \neq 3$ ):

For $x = 3$ : Substituting $x = 3$ into the original equation produces division by zero in both $\frac{2x}{x - 3}$ and the right-hand side. Hence, $x = 3$ is extraneous and must be discarded.
For $x = \frac{5}{2}$ : This value satisfies $x \neq -1$ and $x \neq 3$ , so it lies within the valid domain.

Conclusion

After discarding the extraneous root $x = 3$ , there is exactly one distinct real solution: $x = \frac{5}{2}$

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student solves the quadratic equation correctly but fails to check domain restrictions, counting the extraneous solution x = 3 and concluding there are 2 solutions.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
2x(x+1) - 5(x-3) = 18x - 30
Step 2
2x^2 + 2x - 5x + 15 = 18x - 30 $\implies$ 2x^2 - 21x + 45 = 0
Step 3
(2x - 15)(x - 3) = 0 $\implies$ x = $\frac{15}{2}$ ,\ x = 3
Step 4
2 $\text{ distinct real solutions}$
Correct
$x \neq 3$ $\text{ and }$ x $\neq$ -1 $\implies$ x = 3 $\text{ is extraneous}$ $\implies$ 1 $\text{ distinct real solution }$ $\left(x = \frac{15}{2}\right)$
You included $x = 3$ as a valid solution, but $x = 3$ makes the original denominators $x - 3$ and $x^2 - 2x - 3$ equal to zero. Because division by zero is undefined, $x = 3$ is extraneous and must be discarded, leaving exactly $1$ distinct real solution.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly cleared fractions using the common denominator $(x - 3)(x + 1)$ .

You expanded and simplified the polynomials accurately.

You factored the standard-form quadratic equation correctly to find the candidate values.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Extraneous Solution Ignored

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You treated the roots of the cleared quadratic equation as automatically valid without testing them against the domain restrictions of the original rational expressions.

REASONING PATH

Your ACEpath teacher’s suggested approach

Establish Restricted Domain First: Before multiplying by the common denominator, set every denominator to zero to identify restricted values: $x \in \mathbb{R} \setminus \{-1, 3\}$ .
Always write down restricted values at the top of your page before doing any algebra.
Filter Candidate Roots: Any candidate root that matches a restricted value causes division by zero and must be rejected as extraneous.
Cross out any candidate root that makes an original denominator zero before counting total solutions.

CORRECTED PATH

ACEpath Teacher's Demonstration

Compare the candidate roots $x = \frac{15}{2}$ and $x = 3$ with the domain restrictions $x \neq 3$ and $x \neq -1$ .
Discard $x = 3$ because it causes division by zero in the original denominators $x - 3$ and $x^2 - 2x - 3$ .
Conclude that only $x = \frac{15}{2}$ is a valid solution, giving exactly $1$ distinct real solution.

Final answer

1 distinct real solution (x = 15/2)

NEXT STEPS

Final Takeaways from Your Teacher

Write down the non-permissible values of the variable as step 0 before clearing denominators.
Form the habit of circling your candidate roots and drawing a checkmark or cross next to each against the domain restrictions.
Reread the final question prompt carefully to ensure you provide the exact quantity asked for, such as the count of solutions versus the solution set.

OverviewACEpath Teacher's Diagnosis

02 The student forgets to distribute the negative sign to the second term when subtracting 5(x - 3), writing -5x - 15 instead of -5x + 15.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$x \neq 3$ ,\ x $\neq$ -1
Step 2
2x(x+1) - 5(x-3) = 18x - 30 $\implies$ 2x^2 + 2x - 5x - 15 = 18x - 30 $\implies$ 2x^2 - 21x + 15 = 0
Correct
2x(x+1) - 5(x-3) = 18x - 30 $\implies$ 2x^2 + 2x - 5x + 15 = 18x - 30 $\implies$ 2x^2 - 21x + 45 = 0
When expanding $-5(x - 3)$ , you distributed the negative to $x$ but forgot to flip the sign of $-3$ , writing $-15$ instead of $+15$ .
Step 3
x = $\frac{21 \pm \sqrt{441 - 120}}{4}$ $\implies$ x = $\frac{21 \pm \sqrt{321}}{4}$
Correct
2x^2 - 21x + 45 = 0 $\implies$ (2x - 15)(x - 3) = 0 $\implies$ x = $\frac{15}{2}$ $\text{ or }$ x = 3
You applied the quadratic formula correctly to your equation, but the coefficients were already affected by the distribution error.
Step 4
2 $\text{ distinct real solutions}$
Correct
1 $\text{ distinct real solution}$
Because of the earlier sign error, you did not get the candidate root $x = 3$ , which is extraneous due to the domain restriction $x \neq 3$ , leaving only 1 valid real solution.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly identified the domain restrictions $x \neq 3$ and $x \neq -1$ before solving.

You cleared denominators correctly by multiplying every term by the least common denominator.

You correctly used the quadratic formula on the quadratic equation you obtained.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Distributive Property Sign Error

You wrote $-5(x - 3) = -5x - 15$ instead of $-5x + 15$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Losing track of the negative sign when distributing a negative coefficient across parentheses containing a subtraction term.

REASONING PATH

Your ACEpath teacher’s suggested approach

Distributing Negative Signs: $-a(b - c) = -ab + ac$
Always treat subtraction in front of a parenthesized term as multiplying by a negative number: negative times negative equals positive.
Check Intermediate Constant Terms: Constant term verification
Pause right after expanding parentheses and check the sign of the constant term before combining like terms.

CORRECTED PATH

ACEpath Teacher's Demonstration

Expand left side correctly: $2x(x + 1) - 5(x - 3) = 2x^2 + 2x - 5x + 15 = 2x^2 - 3x + 15$ .
Move all terms to one side: $2x^2 - 3x + 15 - (18x - 30) = 0 \implies 2x^2 - 21x + 45 = 0$ .
Factor the quadratic: $(2x - 15)(x - 3) = 0$ , giving candidate solutions $x = \frac{15}{2}$ and $x = 3$ .
Compare against domain restrictions ( $x \neq 3, x \neq -1$ ): $x = 3$ causes division by zero and must be rejected as extraneous.

Final answer

1 distinct real solution (namely $x = \frac{15}{2}$ , or $x = \frac{5}{2}$ depending on the numerator constant)

NEXT STEPS

Final Takeaways from Your Teacher

Write out the expansion of negative factors in two explicit sub-steps whenever minus signs appear inside parentheses.
Always compare candidate solutions against your initial domain restrictions to eliminate extraneous roots.

OverviewACEpath Teacher's Diagnosis

03 The student mistakenly discards the fractional solution 15/2 thinking solutions must be integers, leaving only the extraneous solution x = 3 before concluding there is 1 solution.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
x^2 - 2x - 3 = (x - 3)(x + 1) $\implies$ 2x(x + 1) - 5(x - 3) = 18x - 30
Step 2
2x^2 - 3x + 15 = 18x - 30 $\implies$ 2x^2 - 21x + 45 = 0 $\implies$ (2x - 15)(x - 3) = 0
Step 3
x = $\frac{15}{2}$ $\text{ is not an integer, so reject}$ $\implies$ x = 3
Correct
x = $\frac{15}{2}$ $\text{ is in the domain }$ ( $\frac{15}{2}$ $\neq$ 3, -1), $\text{ so keep it; }$ x = 3 $\text{ causes division by zero, so reject it}$
The problem asks for real solutions, not integer solutions, so the fraction $x = \frac{15}{2}$ is completely valid. Conversely, $x = 3$ makes the original denominators $x - 3$ and $x^2 - 2x - 3$ equal to zero, so $x = 3$ must be rejected as extraneous.
Step 4
1 $\text{ distinct real solution }$ (x = 3)
Correct
1 $\text{ distinct real solution }$ $\left(x = \frac{15}{2}\right)$
Although you concluded there is 1 distinct real solution, your identified solution $x = 3$ is extraneous. The single valid real solution is $x = \frac{15}{2}$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly factored the quadratic denominator $x^2 - 2x - 3$ as $(x - 3)(x + 1)$ .

You cleared denominators accurately and expanded the polynomial terms without algebraic slips.

You factored the standard quadratic equation $2x^2 - 21x + 45 = 0$ into $(2x - 15)(x - 3) = 0$ correctly.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Extraneous Solution Misidentification

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You confused the requirement for real solutions with a restriction to integers, while overlooking that extraneous solutions in rational equations are defined by division by zero in the original denominators.

REASONING PATH

Your ACEpath teacher’s suggested approach

Define the Domain First: Rational expressions $\frac{P(x)}{Q(x)}$ are defined only where $Q(x) \neq 0$ .
List all restricted values before solving: here, $x \neq 3$ and $x \neq -1$ .
Understand Real Numbers: Real numbers include all integers, fractions, decimals, and irrational numbers unless explicitly stated otherwise.
Do not reject a solution merely because it is a fraction unless the problem specifies that $x$ must be an integer.
Filter Candidate Roots Against the Domain: A root of the cleared polynomial equation is extraneous if and only if it makes any original denominator zero.
Compare candidate values: reject $x = 3$ because $3 - 3 = 0$ ; keep $x = \frac{15}{2}$ because $\frac{15}{2} \notin \{-1, 3\}$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Check the original domain restrictions: $x - 3 \neq 0 \implies x \neq 3$ , and $x + 1 \neq 0 \implies x \neq -1$ .
Test $x = 3$ : substituting $x = 3$ gives denominators of $0$ , so $x = 3$ is extraneous and must be discarded.
Test $x = \frac{15}{2}$ : this is a valid real number and does not make any denominator zero, so it is an accepted solution.

Final answer

There is $1$ distinct real solution: $x = \frac{15}{2}$ .

NEXT STEPS

Final Takeaways from Your Teacher

Always write down the domain restrictions $Q(x) \neq 0$ at the very start of solving any rational equation.
Remember that real solutions ( $\mathbb{R}$ ) include fractions and non-integers.
Test each candidate root against your initial domain restrictions before stating your final solution set.

Lock it in

Question

What is an extraneous solution in a rational equation, and why does it occur?

Answer

It is an apparent root introduced when multiplying by an expression containing the variable that equals zero for that value, causing division by zero in the original equation.

Resources

Answer and steps

Final answer

Steps

Conclusion

Typical errors

01 The student solves the quadratic equation correctly but fails to check domain restrictions, counting the extraneous solution x = 3 and concluding there are 2 solutions.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

02 The student forgets to distribute the negative sign to the second term when subtracting 5(x - 3), writing -5x - 15 instead of -5x + 15.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

03 The student mistakenly discards the fractional solution 15/2 thinking solutions must be integers, leaving only the extraneous solution x = 3 before concluding there is 1 solution.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

Lock it in

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