Rational expressions

Rational equations

ALEKS placement

Multiply every term by the LCD to clear denominators, then check against the restrictions.

What this covers

Factor all denominators and identify the LCD.
List the values that make any denominator zero.
Multiply every term by the LCD and simplify.
Solve, then discard any root on the restricted list.

Worked example

Solve 1/x + 1/2 = 3/(2x)

x=0A zero input would make an original denominator zero. Exclude it before multiplying.
2x⋅x1+2x⋅21=2x⋅2x3The least common denominator is twice the variable. Multiply every term on both sides by it.
2+x=3Cancel within each product. This step preserves the equation on the original allowed domain.
x=1Subtract two from both sides. One is a candidate solution and is not excluded.
11+21=2(1)3Substitute the candidate into the original equation, where the denominators still matter.
23=23Both sides have the same value. The allowed candidate is therefore a solution.

Another worked example

Solve 1/x + 1/2 = 0

x=0Only zero is forbidden by the original variable denominator. A negative input is not automatically forbidden.
2x⋅x1+2x⋅21=2x⋅0Multiply every term on both sides by the least common denominator.
2+x=0Simplify the three products on the allowed domain.
x=−2Subtract two to find a candidate. It is not on the excluded list.
−21+21=0Substitution into the original equation gives a true statement, so this candidate is a solution.

A common mistake

A mistake Lemma catches

From 1/x+1/2=3/(2x)

2+21=32+x=3

“Multiplying by the LCD 2x means every term is multiplied: 2x · 1/x = 2, 2x · 1/2 = x and 2x · 3/(2x) = 3. The 1/2 was left as it was, so that term never got multiplied by 2x.”

Try one

Sample problem

Solve and respect the denominator restriction.

x1=101

Practice rational equations free

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Where this fits

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