Why Rational Equations Need a Domain Check
A rational equation contains one or more fractions whose numerators or denominators involve the variable. Solving such an equation involves more than clearing fractions and solving the result: the original denominators may rule out values that appear among the algebraic candidates.
For instance, multiplying an equation by an expression that is zero at a particular value can produce an equation that has that value as a root, even though the original equation is undefined there. A reliable solution therefore keeps track of the original domain from the beginning and checks candidates in the original equation at the end.
Start with the Original Restrictions
A rational expression is undefined wherever its denominator is zero. Thus, before doing any algebra, identify every value that makes any original denominator zero. These values are excluded from the domain, even if a later step cancels a factor or clears the denominators.
For example, the equation
has the restrictions \(x\ne 1\) and \(x\ne -2\). Neither value can be a solution, because at least one side of the original equation would be undefined.
The restrictions come from the equation as first written. They are not erased by multiplying both sides by a denominator, simplifying an expression, or solving a resulting polynomial.
A Reliable Solution Process
Once the restrictions are recorded, clear denominators by multiplying both sides by a common denominator that is nonzero at every value in the original domain. Then solve the resulting equation. The polynomial equation may be easier to work with, but its roots are only candidates until they have been checked against the original restrictions and equation.
Set each original denominator unequal to zero and exclude the resulting values.
Multiply both sides by a common denominator, keeping the recorded restrictions in view.
Use appropriate algebra to find all its real candidates.
Reject candidates outside the original domain and verify the remaining candidates in the original equation.
Worked Examples
Worked Example: Clearing Two Linear Denominators
Solve \(\frac{2}{x-1}=\frac{3}{x+2}\).
First identify the restrictions. The denominators are zero when \(x-1=0\) or \(x+2=0\), so \(x\ne 1\) and \(x\ne -2\).
For all values in the domain, multiply both sides by \((x-1)(x+2)\):
Cancel the factors that are nonzero on the domain:
Expand both sides, collect the \(x\)-terms, and solve:
The candidate \(7\) is not either excluded value. Check it in the original equation:
Both sides are defined and equal, so the solution is \(x=7\).
Worked Example: Keeping a Valid Root and Rejecting an Extraneous One
Solve \(\frac{x-1}{x-1}=x-2\).
The denominator on the left is zero at \(x=1\), so the original equation requires \(x\ne 1\). Multiply both sides by \(x-1\), which is nonzero for values allowed by this restriction:
Expand the right-hand side and move all terms to one side:
The transformed equation gives the candidates \(x=1\) and \(x=3\). The first candidate violates the original restriction, so it must be rejected. For the other candidate, check the original equation:
At \(x=3\), both sides are defined and equal. Therefore the solution is \(x=3\), not \(x=1\).
Worked Example: Solving a Rational Equation That Becomes a Quadratic
Solve \(\frac{1}{x}+\frac{1}{x+1}=\frac{3}{2}\).
The original denominators require \(x\ne 0\) and \(x\ne -1\). Multiply every term by \(2x(x+1)\), which is nonzero for all values allowed by those restrictions:
Cancel the corresponding nonzero factors in each term:
Expand and collect terms:
The candidates are \(x=-\frac{2}{3}\) and \(x=1\). Neither is \(0\) or \(-1\), but each still needs verification. At \(x=1\):
At \(x=-\frac{2}{3}\), the two terms are:
Their sum is \(-\frac{3}{2}+3=\frac{3}{2}\), as required. Both candidates satisfy the original equation, so the solutions are \(x=-\frac{2}{3}\) and \(x=1\).
Why Clearing Denominators Works
The algebra behind this method is an equivalence on the original domain. Suppose an equation has denominators \(q_1(x),\ldots,q_n(x)\), and let \(D(x)\) be their product. Every value in the original domain has \(D(x)\ne 0\). For such values, multiplying both sides by \(D(x)\) preserves equality in both directions: if two sides are equal, their products by \(D(x)\) are equal; and if their products by this same nonzero number are equal, the original sides are equal.
The condition that \(D(x)\ne 0\) matters. If a root of the cleared equation makes an original denominator zero, then the multiplication step was not reversible at that value: the original equation was never defined there. This is why the roots found after clearing denominators must be treated as candidates, not accepted automatically.
In the second worked example, the original left-hand side is defined only when \(x\ne 1\). The cleared equation factors as \((x-1)(x-3)=0\). The root \(x=1\) belongs to the transformed equation, but not to the original domain; \(x=3\) belongs to both and satisfies the original equation.
A Common Mistake
A common error is to solve the cleared equation and report every root without comparing those roots with the original restrictions. For example, from the second worked example, one might write:
The last implication is wrong. The original equation contains the denominator \(x-1\), so \(x=1\) is not in its domain. The correction is to keep the restriction \(x\ne 1\), reject that candidate, and check \(x=3\) in the original equation. A restriction written down at the start remains in force throughout the calculation.
How to Present a Complete Solution
A clear written solution should make the domain check visible rather than leaving it implicit. In particular, show the restrictions before clearing denominators, show the algebra that produces the candidates, and say which candidates are rejected and why. When candidates remain, substitute them into the original equation; checking only the transformed equation does not establish that they satisfy the original one.
A compact final statement can list the original restrictions and the accepted solution set. If every candidate is excluded, the equation has no solution. If no real roots arise from the transformed equation, it also has no real solution. In either case, the conclusion concerns the original equation, not merely its transformed form.
Check Your Understanding
For each question, remember to use the restrictions from the original equation.
- What values must be excluded from the domain of \(\frac{4}{x+3}=\frac{1}{x-2}\)?
- Why is a root of the cleared equation only a candidate solution?
- In \(\frac{x-1}{x-1}=x-2\), why must \(x=1\) be rejected even though it is a root of the cleared equation?
- What condition on a common denominator allows multiplication to preserve equivalence?
- What should you do with each candidate that is not excluded by the original restrictions?
