Solving Rational Equations and Inequalities

Solving Rational Equations and Inequalities — Moosa Academy

Clearing the fractions turns a rational equation into an ordinary one, which is the easy part. The hard part comes at the end: multiplying by an expression containing the variable can invent solutions that were never allowed. Every answer has to be checked against the denominators before it is accepted.

Concept Two strategies, side by side

Solving an equation	Solving an inequality
1. Find the LCM of the denominators	1. Note the excluded values (denominator $= 0$ )
2. Multiply both sides by the LCM to clear fractions	2. Solve the related equation (replace $<$ with $=$ )
3. Simplify and solve what remains	3. Mark all critical points on a number line
4. Check — reject any root that zeroes a denominator	4. Test one value from each interval

Note that an inequality is never solved by cross multiplying. The sign of the denominator is unknown, and multiplying by a negative quantity would flip the inequality without warning. Testing intervals sidesteps that trap entirely.

Theorem Why extraneous roots appear

Multiplying both sides by an expression that contains the variable is only reversible when that expression is non-zero — so the cleared equation can have roots the original never had.

Suppose the LCM is $(x+3)(x+5)$ . At $x = -5$ that factor is zero, and multiplying by zero makes any two sides agree. The cleared polynomial happily reports $x = -5$ as a root, but the original expression is undefined there. Such a value is called an extraneous root and must be discarded.

Example An equation with a rejected root

Solve $\dfrac{x}{x+5} + \dfrac{1}{x+3} = \dfrac{10}{(x+3)(x+5)}$ .

LCM of the denominators:

(x+3)(x+5)

Multiply through:

x(x+3) + 1(x+5) = 10

Expand:

x^{2} + 3x + x + 5 = 10

Collect:

x^{2} + 4x - 5 = 0

Factor:

(x - 1)(x + 5) = 0

Possible roots:

x = 1

and

x = -5

Check

x = -5

: it zeroes

(x+5)

— rejected

Check

x = 1

: both sides equal

\dfrac{5}{12}

— valid

$x = 1$

Two roots came out of the algebra and only one survived. Skipping the final check would have produced a wrong answer that looked perfectly reasonable.

Example A motion problem

A boat travels at $6$ mi/h in still water. It covers $10$ miles in total — $5$ miles downstream and $5$ miles back upstream — taking $3$ hours altogether. Find the speed $v$ of the current.

Time

=

distance

\div

speed

Downstream the current helps: speed

= 6 + v

, time

= \dfrac{5}{6+v}

Upstream the current hinders: speed

= 6 - v

, time

= \dfrac{5}{6-v}

Equation:

\dfrac{5}{6+v} + \dfrac{5}{6-v} = 3

LCM:

(6+v)(6-v)

Multiply through:

5(6-v) + 5(6+v) = 3(36 - v^{2})

Expand:

60 = 108 - 3v^{2}

Rearrange:

3v^{2} = 48

, so

v^{2} = 16

Roots:

v = 4

and

v = -4

A current speed cannot be negative — reject

v = -4

The current flows at $4$ mi/h

This time the rejected root broke no denominator; it was ruled out by the situation itself. Both kinds of check matter.

Example A rational inequality

Solve $\dfrac{4}{x-2} < \dfrac{7}{x+1}$ .

Excluded values:

x = 2

and

x = -1

Related equation:

\dfrac{4}{x-2} = \dfrac{7}{x+1}

Cross multiply:

4(x+1) = 7(x-2)

Expand:

4x + 4 = 7x - 14

Solve:

3x = 18

, so

x = 6

Critical points:

-1

2

6

Interval	Test value	Result	Works?
$x < -1$	$x = -3$	$-\dfrac{4}{5} < -\dfrac{7}{2}$	No
$-1 < x < 2$	$x = 0$	$-2 < 7$	Yes
$2 < x < 6$	$x = 4$	$2 < \dfrac{7}{5}$	No
$x > 6$	$x = 8$	$\dfrac{2}{3} < \dfrac{7}{9}$	Yes

Solution: $-1 < x < 2$ or $x > 6$ . The answer is two separate stretches of the number line — something cross multiplying could never have revealed.

Summary

Clear a rational equation by multiplying both sides by the LCM of the denominators.
Always check the roots — any value that zeroes a denominator is extraneous and must be rejected.
A word problem can rule out a root on physical grounds too, such as a negative speed.
Never cross multiply an inequality; the sign of the denominator is unknown.
Instead, find the excluded values and the roots of the related equation — these are the critical points.
Test one value from each interval; the solution may be several separate intervals.