Clear denominators only after stating their nonzero restrictions.
Extended · Syllabus E2.5
Notes and key methods
List the excluded inputs before multiplying an equation by a common denominator. Multiply every term by that denominator, solve the resulting equation and check candidates in the original fraction expressions.
A value that makes an original denominator zero is never a solution, even if it satisfies a later polynomial equation. A denominator can be negative; the restriction is nonzero, rather than positive, unless another part of the question imposes positivity.
Check the conditions
Use the stated units and precision.
Show the method as well as the final answer.
Three worked examples
Example 1 · Equations with algebraic denominators
Solve \(\displaystyle \frac{6}{x-6}=\frac{7}{x-13}\) over the real numbers.
Solve the equation, including all domain checks.
Find x−6 at the valid solution.
Find the common value of the two fractions at the solution.
Find the sum of the two excluded inputs.
Show the reasoning
(a)
\(6(x-13)=7(x-6)\) gives \(\displaystyle x=-36\). It differs from both excluded values and both original fractions equal \(\displaystyle - \frac{1}{7}\).
Answer: \(x=-36,\quad x\ne 6,13\)
(b)
Exclude both denominator zeros, cross-multiply whole factors and substitute the resulting candidate into the original equation.
\(-42\)
Answer: \(-42\).
(c)
Exclude both denominator zeros, cross-multiply whole factors and substitute the resulting candidate into the original equation.
\(\displaystyle - \frac{1}{7}\)
Answer: \(\displaystyle - \frac{1}{7}\).
(d)
Exclude both denominator zeros, cross-multiply whole factors and substitute the resulting candidate into the original equation.
\(19\)
Answer: \(19\).
Example 2 · Equations with algebraic denominators
Solve \(\displaystyle \frac{2}{x-1}=\frac{3}{x-4}\) over the real numbers.
Solve the equation, including all domain checks.
Find x−1 at the valid solution.
Find the common value of the two fractions at the solution.
Find the sum of the two excluded inputs.
Show the reasoning
(a)
\(2(x-4)=3(x-1)\) gives \(\displaystyle x=-5\). It differs from both excluded values and both original fractions equal \(\displaystyle - \frac{1}{3}\).
Answer: \(x=-5,\quad x\ne 1,4\)
(b)
Exclude both denominator zeros, cross-multiply whole factors and substitute the resulting candidate into the original equation.
\(-6\)
Answer: \(-6\).
(c)
Exclude both denominator zeros, cross-multiply whole factors and substitute the resulting candidate into the original equation.
\(\displaystyle - \frac{1}{3}\)
Answer: \(\displaystyle - \frac{1}{3}\).
(d)
Exclude both denominator zeros, cross-multiply whole factors and substitute the resulting candidate into the original equation.
\(5\)
Answer: \(5\).
Example 3 · Equations with algebraic denominators
Solve \(\displaystyle \frac{3}{x-2}=\frac{4}{x-6}\) over the real numbers.
Solve the equation, including all domain checks.
Find x−2 at the valid solution.
Find the common value of the two fractions at the solution.
Find the sum of the two excluded inputs.
Show the reasoning
(a)
\(3(x-6)=4(x-2)\) gives \(\displaystyle x=-10\). It differs from both excluded values and both original fractions equal \(\displaystyle - \frac{1}{4}\).
Answer: \(x=-10,\quad x\ne 2,6\)
(b)
Exclude both denominator zeros, cross-multiply whole factors and substitute the resulting candidate into the original equation.
\(-12\)
Answer: \(-12\).
(c)
Exclude both denominator zeros, cross-multiply whole factors and substitute the resulting candidate into the original equation.
\(\displaystyle - \frac{1}{4}\)
Answer: \(\displaystyle - \frac{1}{4}\).
(d)
Exclude both denominator zeros, cross-multiply whole factors and substitute the resulting candidate into the original equation.
\(8\)
Answer: \(8\).
Common mistakes
Keep intermediate values exact until the final step.