C. Polynomial and Rational Functions · Rational Functions and Equations

Solving Rational Equations Algebraically

C3.6About 75 minutes

Learning goals

I can state the restrictions on the variable before solving a rational equation.
I can clear the fractions by multiplying both sides by the lowest common denominator.
I can solve the resulting linear or quadratic equation and reject extraneous roots.
I can verify solutions by substitution and with graphing technology.

Words to know

restriction
A value the variable cannot take because it would make a denominator 0.

lowest common denominator (LCD)
The simplest expression that every denominator divides into. For 1/(x − 2) and 1/(x² − 4) it is (x − 2)(x + 2).

extraneous root
A value that solves the equation you get after clearing fractions but does not solve the original equation, usually because it is a restricted value.

cross-multiplying
For a/b = c/d (b, d ≠ 0), writing ad = bc. It is the same as multiplying both sides by bd.

From pictures to exact answers

In the last lesson you saw roots as x-intercepts and intersection points. Graphs show how many roots there are, but they usually give only decimal estimates. Algebra gives exact answers. The key move is to get rid of the fractions.

Remember from Grade 11: in the MCR3U unit Functions and Equivalent Expressions you added and subtracted rational expressions using a common denominator and stated restrictions. Here the same skills solve equations.

Four steps. Step 1 comes first because once the fractions are gone, you can no longer see which values were forbidden.

The method

Factor every denominator and write the restrictions (the values that make any denominator 0).
Multiply both sides by the LCD. Every fraction cancels, leaving a polynomial equation. If there is exactly one fraction on each side, this is the same as cross-multiplying.
Solve the polynomial equation (it is usually linear or quadratic).
Check each answer against the restrictions. Reject any restricted value: it is an extraneous root. Substitute the others into the original equation, or check them with technology.

Why do extraneous roots appear? Multiplying both sides by an expression like x − 3 is fine when x − 3 ≠ 0. But the new equation also 'allows' x = 3, so x = 3 can sneak in as a false answer.

Worked example

Example 1: One fraction on each side

Solve 3/(x + 2) = 5/(x − 4).

Restrictions: x ≠ −2, x ≠ 4.
Cross-multiply: 3(x − 4) = 5(x + 2).
3x − 12 = 5x + 10, so −2x = 22 and x = −11.
−11 is not restricted. Check: 3/(−9) = −1/3 and 5/(−15) = −1/3 ✓.

x = −11

Worked example

Example 2: An extraneous root

Solve x/(x − 3) − 2 = 3/(x − 3).

Restriction: x ≠ 3.
Multiply every term by x − 3: x − 2(x − 3) = 3.
x − 2x + 6 = 3, so −x = −3 and x = 3.
But x = 3 is restricted: both fractions are undefined there. It is extraneous.

No solution. The graph of the left side minus the right side is the line y = −1 with a hole at x = 3 (see the diagram below): it never touches the x-axis.

Worked example

Example 3: A quadratic appears

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

Restrictions: x ≠ 0, x ≠ −1. LCD: x(x + 1).
Multiply every term by x(x + 1): 2(x + 1) + 3x = 2x(x + 1).
5x + 2 = 2x² + 2x, so 2x² − 3x − 2 = 0.
Factor: (2x + 1)(x − 2) = 0, so x = −1/2 or x = 2.
Check x = 2: 1 + 1 = 2 ✓. Check x = −1/2: 2/(−1/2) = −4 and 3/(1/2) = 6, and −4 + 6 = 2 ✓.

x = −1/2 or x = 2

Worked example

Example 4: A variable outside the fraction

Solve x + 6/x = 5.

Restriction: x ≠ 0.
Multiply by x: x² + 6 = 5x, so x² − 5x + 6 = 0.
(x − 2)(x − 3) = 0, so x = 2 or x = 3.
Check: 2 + 3 = 5 ✓ and 3 + 2 = 5 ✓.

x = 2 or x = 3

Worked example

Example 5: Factor the denominators first

Solve x/(x − 2) + 1/(x + 2) = 8/(x² − 4).

Factor: x² − 4 = (x − 2)(x + 2). Restrictions: x ≠ ±2. LCD: (x − 2)(x + 2).
Multiply every term by the LCD: x(x + 2) + (x − 2) = 8.
x² + 3x − 10 = 0, so (x + 5)(x − 2) = 0, and x = −5 or x = 2.
x = 2 is restricted, so it is extraneous.
Check x = −5: (−5)/(−7) + 1/(−3) = 5/7 − 1/3 = 8/21, and 8/(25 − 4) = 8/21 ✓.

x = −5 (x = 2 is extraneous).

Graph the left side minus the right side. Left: Example 2 gives y = −1 with a hole, so there is no root. Right: Example 5 gives y = (x + 5)/(x + 2) with a hole at x = 2. The real root x = −5 is an x-intercept; the extraneous value x = 2 is only a hole.

Verifying with technology

To check an answer with a graphing tool, move everything to one side and graph y = (left side) − (right side). Real roots are the x-intercepts. A restricted value never shows up as an x-intercept. Or graph each side separately and look at the x-coordinates of the intersection points. A CAS 'solve' command will also list the roots; most CAS tools reject extraneous roots for you, but you should still be able to explain why.

Just for fun

Extraneous roots are like party guests who RSVP 'yes' and then arrive at the wrong address. Your algebra invited them, but the original equation never lets them in.

Hands-on activity

Mistake hunt

You need: paper, a pencil, and a family member or friend to read the 'student work' aloud.

Each solution below has one mistake. Find it, explain it, and fix the answer.

Solve 4/x = 2/(x + 3). 'Student': 4(x + 3) = 2x, so 4x + 3 = 2x, so x = −3/2.
Solve 1/(x − 1) + 1 = 2/(x − 1). 'Student': 1 + (x − 1) = 2, so x = 2. Then they write 'x = 2 and x = 1'.
Solve (x + 4)/(x − 4) = 3. 'Student': x + 4 = 3x − 4, so −2x = −8, and x = 4.

Then make up your own wrong solution for your partner to fix.

What to look for: (1) The 4 was not distributed: 4(x + 3) = 4x + 12, so 4x + 12 = 2x and x = −6 (check: 4/(−6) = −2/3 and 2/(−3) = −2/3 ✓). (2) Only x = 2 came out of the algebra, and x = 1 is a restricted value; the answer is x = 2 alone (check: 1 + 1 = 2/1 ✓). (3) The algebra is right, but x = 4 is restricted, so it is extraneous: no solution.

Your turn

Solve 4/(x − 1) = 2/(x + 3).

State the restrictions for 5/(x² − 9) + 1/(x + 3) = 2.

Solve 6/x − 1 = 1/x.

Solve x − 8/x = 2.

Solve (x + 1)/(x − 2) = 3/(x − 2).

Solve 1/x + 1/(2x) = 3/4.

Solve 3/(x − 1) + 2/(x + 1) = 2.

More practice: warm-up

Solve 6/x = 3.

Solve 10/(x + 1) = 2.

State the restrictions for 1/(x − 5) + 2/x = 1.

What is the LCD of 2/(x − 1) and 3/(x + 4)?

Solve 1/x = x/4.

More practice: on your own

Solve (x − 2)/(x + 3) = 2/3.

Solve 2/(x − 3) − 1/x = 1. Give exact answers.

Solve x/(x + 2) − 2/(x − 2) = (x² + 4)/(x² − 4).

Solve 2/(x + 1) + 1 = 4/(x² + x).

A student solved x/(x − 1) = 1/(x − 1) + 2 and wrote 'x = 1'. Explain the error and give the correct answer.

Challenge

Solve (x + 2)/(x − 1) − (x − 2)/(x + 1) = 16/(x² − 1).

Solve x/(x − 1) + (x − 1)/x = 5/2.

Show that 1/(x − 2) + k/(x + 2) = 4/(x² − 4) has no solution for every k ≠ −1. What happens when k = −1?

Check your understanding

Quiz

What should you do first when solving 2/(x − 4) = 5/x?

What is the solution of 3/(x + 2) = 5/(x − 4)?

What is the solution of x/(x − 3) − 2 = 3/(x − 3)?

What is an extraneous root?

What are the solutions of x + 6/x = 5?

Clearing the fractions in x/(x − 2) + 1/(x + 2) = 8/(x² − 4) gives x² + 3x − 10 = 0. What is the solution of the original equation?

For parents and teachers

Big idea (C3.6): to solve a rational equation, state the restrictions, multiply by the lowest common denominator to clear the fractions, solve, and then check. Any answer that makes a denominator zero is extraneous and must be rejected.

How to help: Before your teen starts multiplying, ask, 'What can x not be?' After they finish, ask, 'Did you check each answer in the original equation?' Those two questions prevent most lost marks.

Common difficulties:

Multiplying only some of the terms by the LCD (every term, including whole numbers, must be multiplied).
Forgetting to distribute: 4(x + 3) is 4x + 12, not 4x + 3.
Keeping an extraneous root, or throwing away a good root by mistake.
Not factoring x² − 4 or x² + x, which hides the LCD.