Solving Rational Equations, SAT math lesson

Two pipes fill a tank, one alone in $x$ hours and the other in $x + 2$ hours, and together they fill $\frac{5}{12}$ of it each hour. Finding $x$ means solving an equation with the variable in denominators. The method is short. Multiply away the fractions and solve what is left. The catch is that this step can let in a number the original equation never allowed, so the last step is always a check.

See also Solving Equations and Checking Solutions in Foundations for Calculus and Solving Rational and Radical Equations in Precalculus.

Key terms

Rational equation

Definition A rational equation is an equation in which at least one side contains a rational expression with the variable in a denominator. A real number is a solution when every expression in the equation is defined at that number and both sides are equal there.

In plain words It is an equation with fractions that have $x$ in the bottom. A solution has to make the equation true and also keep every fraction meaningful.

Example The equation

\frac{3}{x} = \frac{5}{x + 4}

is rational. The number $6$ is a solution, since $\frac{3}{6} = \frac{1}{2}$ and $\frac{5}{10} = \frac{1}{2}$ .

Restriction

Definition The restrictions of a rational equation are the values of the variable that make any denominator in the original equation equal to $0$ . They are excluded before any solving starts, and no later step can bring them back. Write them down beside the equation at the beginning.

In plain words Before solving, list the numbers that would break a fraction. Those numbers can never be answers.

Example In

\frac{5}{x^2 - 9} + \frac{1}{x + 3} = \frac{2}{x}

the denominators are $x^2 - 9 = (x - 3)(x + 3)$ , $x + 3$ and $x$ . The restrictions are $x \ne 3$ , $x \ne -3$ and $x \ne 0$ .

Common denominator

Definition A common denominator of the fractions in an equation is an expression that every denominator divides evenly. The least common denominator (LCD) is the product of each distinct factor of the denominators, raised to the highest power in which it appears. Multiplying both sides of the equation by the LCD, which means multiplying every single term, leaves an equation with no fractions.

In plain words Find one expression that every bottom fits into, then multiply every term on both sides by it. Each fraction loses its denominator.

Example Solve the equation from the opening,

\frac{1}{x} + \frac{1}{x + 2} = \frac{5}{12},

with $x \ne 0$ and $x \ne -2$ . The LCD is $12x(x + 2)$ . Multiplying every term by it gives

\begin{aligned} 12(x + 2) + 12x &= 5x(x + 2) \\[0.4em] 24x + 24 &= 5x^2 + 10x \\[0.4em] 0 &= 5x^2 - 14x - 24 \\[0.4em] 0 &= (x - 4)(5x + 6). \end{aligned}

The candidates are $4$ and $-\frac{6}{5}$ . Neither is a restricted value, and both check. At $x = 4$ the left side is $\frac{1}{4} + \frac{1}{6} = \frac{5}{12}$ . At $x = -\frac{6}{5}$ it is $-\frac{5}{6} + \frac{5}{4} = \frac{5}{12}$ . In the tank story only the positive value makes sense, so the first pipe takes $4$ hours.

Candidate solution

Definition A candidate solution is a solution of the equation obtained after clearing denominators. Multiplying both sides by the LCD is a reversible step only where the LCD is not zero, that is, on the restricted domain. So on that domain the cleared equation has exactly the same solutions as the original, and every solution of the original appears among the candidates. A candidate that is a restricted value is not a solution.

In plain words Clearing fractions gives you a short list of suspects. The list contains every true solution, but it may also contain an impostor that sits on a forbidden value.

Example Solve

\frac{x^2}{x - 2} = \frac{4}{x - 2},

with $x \ne 2$ . Multiplying by $x - 2$ gives $x^2 = 4$ , so the candidates are $2$ and $-2$ . The candidate $2$ is restricted. The candidate $-2$ checks, since both sides equal $\frac{4}{-4} = -1$ . The only solution is $-2$ .

Keep exact answers exact. If a candidate is $-\frac{6}{5}$ , check and report $-\frac{6}{5}$ (or $-1.2$ , which is the same number), not a rounded value such as $-1$ . A rounded number is not a solution, and on a student-produced response question either $-6/5$ or $-1.2$ can be entered, since each uses only $4$ characters.

Extraneous solution

Definition An extraneous solution is a candidate produced by the solving process that does not satisfy the original equation. When you only clear denominators and then solve correctly, an extraneous candidate is always a value that makes some original denominator zero. If every candidate is extraneous, the equation has no solution.

In plain words It is a fake answer created by the algebra. Plug each candidate back into the original equation, not into a later line, and throw out any that break a fraction.

Example Solve

\frac{1}{x - 3} + \frac{1}{x + 3} = \frac{6}{x^2 - 9},

with $x \ne 3$ and $x \ne -3$ . The LCD is $(x - 3)(x + 3)$ , and multiplying every term by it gives $(x + 3) + (x - 3) = 6$ , so $2x = 6$ and $x = 3$ . That value is restricted, so it is extraneous. The equation has no solution.

Finding an input from an output

Definition For a rational function $f$ and a number $c$ , the inputs with $f(x) = c$ are the solutions of the rational equation $f(x) = c$ that lie in the domain of $f$ . Solve it like any rational equation, keeping the excluded values of $f$ out. If no candidate survives, the value $c$ is not an output of $f$ .

In plain words A question that gives you the output and asks for the input is a rational equation in disguise.

Example Let

f(x) = \frac{2x + 1}{x - 4},

with $x \ne 4$ . To find $x$ with $f(x) = 3$ , multiply by $x - 4$ to get $2x + 1 = 3x - 12$ , so $x = 13$ . Check $f(13) = \frac{27}{9} = 3$ . Now try $f(x) = 2$ . Clearing gives $2x + 1 = 2x - 8$ , which says $1 = -8$ , so there is no input at all. The rewritten form

f(x) = 2 + \frac{9}{x - 4}

shows why, since the fraction is never $0$ . If you graph $y = f(x)$ and $y = 2$ in the Desmos graphing calculator, the curve gets closer and closer to the line $y = 2$ far to the left and right but never meets it. That line is a horizontal asymptote.

Check yourself

What is the solution set of $\frac{x + 3}{x - 1} = \frac{4}{x - 1}$ ? (A) $\{1\}$ (B) $\{4\}$ (C) $\{-3\}$ (D) There is no solution.

Show the answer

The answer is (D). The restriction is $x \ne 1$ . Multiplying by $x - 1$ gives $x + 3 = 4$ , so the only candidate is $x = 1$ , which makes both denominators zero. It is extraneous, so nothing is left. Choice (A) skips the check. Choice (B) takes the numerator $4$ as the answer and choice (C) sets the numerator $x + 3$ equal to $0$ , but neither satisfies the equation. For example, at $x = 4$ the left side is $\frac{7}{3}$ and the right side is $\frac{4}{3}$ .

What value of $x$ satisfies $\frac{2}{x} + \frac{3}{2x} = \frac{7}{8}$ ?

Show the answer

The restriction is $x \ne 0$ and the LCD is $8x$ . Multiplying every term by $8x$ gives $16 + 12 = 7x$ , so $x = 4$ . Check by substituting,

\frac{2}{4} + \frac{3}{8} = \frac{4}{8} + \frac{3}{8} = \frac{7}{8}.

Which of the following lists all solutions of $\frac{x^2}{x + 3} = \frac{9}{x + 3}$ ? (A) $3$ only (B) $-3$ only (C) $-3$ and $3$ (D) There is no solution.

Show the answer

The answer is (A). The restriction is $x \ne -3$ . Clearing the denominator gives $x^2 = 9$ , with candidates $3$ and $-3$ . The candidate $-3$ makes the denominator zero and is rejected. The candidate $3$ checks, since both sides equal $\frac{9}{6}$ . Choice (C) keeps the extraneous candidate, choice (B) rejects the wrong one, and choice (D) throws out both.

Solving Rational Equations

Theory, part 61

Practice, part 62

Key terms

Rational equation

Restriction

Common denominator

Candidate solution

Extraneous solution

Finding an input from an output

Check yourself

How this lesson connects

Builds on