Rational Expressions

Record the domain before simplifying

A rational expression $P(x)/Q(x)$ requires $Q(x)\ne0$. For

factor the denominator:

Therefore the original domain is all real $x$ except 3 and −2. Factoring the numerator gives

The shorter formula is defined at 3, but the original one is not. If the formula is being used to describe the same function, its inherited domain must remain attached. The missing point would be $(3,6/5)$; at −2 the denominator still vanishes.

The OpenStax rational-function treatment supplies the broader connection between these restrictions and graph behaviour.

Cancel factors, not terms

Cancellation is division of a common nonzero factor in the whole numerator and denominator. It is not deletion of matching symbols across addition.

For example,

But

in general. At $x=1$, the left side is $3/6=1/2$, not $2/5$. A single counterexample refutes that proposed identity.

Factor before cancelling. If a numerator is $2x+4$, write it as $2(x+2)$. Removing $x$ from $2x+4$ would not divide the entire numerator by $x$.

Multiply and divide with all restrictions visible

Multiplication requires each original denominator to be nonzero. Division also requires the divisor expression to be nonzero.

Consider

The first denominator excludes 2. The second excludes 2 and −2. The divisor is zero at −1, which must also be excluded. Thus the original domain has $x\ne-2,-1,2$.

Now multiply by the reciprocal and factor:

The final polynomial keeps all three original exclusions. None can be recovered by looking only at its formula. At $x=-1$, the original calculation attempts division by zero; at $x=\pm2$, one of its component fractions is already undefined.

As a check, choose an allowed input such as $x=3$. The original calculation is $8\div(4/5)=10$, and the product is $(3-1)(3+2)=10$. This numerical check supports the worked example; the factor cancellation supplies the general equivalence.

Add using a least common denominator

For

the original restrictions are $x\ne1,-1$. A common denominator is $(x-1)(x+1)$. Multiply each fraction by a form of one:

The numerator is $(x+1)+2(x-1)=3x-1$. Keep the signs attached to the full parenthesised expressions. For subtraction,

The minus sign changes both terms in the second numerator. Writing $x+1-x-1$ would be a different subtraction and would incorrectly give zero.

When a denominator contains repeated factors, the least common denominator must include the highest power required. For denominators $(x-1)^2(x+2)$ and $(x-1)(x+2)^3$, it is $(x-1)^2(x+2)^3$. Taking only one copy of each factor would not produce a multiple of both denominators.

Simplify a complex fraction without dropping conditions

A complex fraction contains fractions within a numerator or denominator. Suppose

Its domain requires $x\ne0$, $y\ne0$ and $x\ne y$. Combine the numerator:

Therefore

At $x=y\ne0$, the simplified formula has a value, but the original expression has the indeterminate form $0/0$ and is undefined. An algebraic simplification on allowed inputs does not settle a limit at an excluded input; that is a separate question.

Alternatively, multiply the entire numerator and denominator of the complex fraction by $xy$. Doing so is legal on the already stated domain and gives the same result. Multiplying only selected terms would change the expression.

Keep an expression distinct from an equation

The expression

can be simplified to $x+1$ for $x\ne1$. There is nothing to solve until a condition is imposed. If the condition is that the expression equals 2, then the candidate from $x+1=2$ is $x=1$, which the original domain excludes. That equation has no solution.

By contrast, setting the same expression equal to $x+1$ produces an identity on all real inputs except 1. The formula and the equation play different roles. See Solving Rational Equations for complete solution-set reasoning.

A domain can change which sign intervals are allowed

For the original $R(x)$ above, use the simplified quotient $(x+3)/(x+2)$ to determine its sign, but retain the extra exclusion at 3.

The numerator changes sign at −3 and the denominator changes sign at −2. Testing one point in each interval gives a positive quotient on $(-\infty,-3)$, a negative quotient on $(-3,-2)$, and a positive quotient on $(-2,\infty)$. At −3 the value is zero. At −2 it is undefined.

Consequently $R(x)\ge0$ holds on

The hole at 3 must be removed even though no sign change occurs there. This is why solving a rational inequality by multiplying through by an unknown-sign denominator is unsafe. A sign chart or justified case split preserves the correct inequality.

Related topics

Use Asymptotes to interpret surviving denominator factors and cancelled points. Solving Rational Equations applies the same domain discipline to finding complete solution sets.