Simple Rational Equation

Solve: $\frac{5}{x} = \frac{15}{6}$

Simplify the right side if possible: $\frac{15}{6} = \frac{5}{2}$ = $\frac{5}{x} = \frac{5}{2}$

Cross multiply: $5 \cdot 2 = 5 \cdot x$ = $10 = 5x$

Solve for x: $x = \frac{10}{5}$ = $x = 2$

Check: Does $x = 2$ make any denominator zero?: Denominators: $x = 2$ and $6$. Neither is zero when $x = 2$. = Valid solution

Answer: $x = 2$

Using the LCD Method

Solve: $\frac{3}{x} + \frac{1}{2} = \frac{5}{x}$

Identify all denominators: Denominators: $x$, $2$, $x$ = LCD = $2x$

Multiply every term by the LCD: $2x \cdot \frac{3}{x} + 2x \cdot \frac{1}{2} = 2x \cdot \frac{5}{x}$ = $6 + x = 10$

Solve the resulting equation: $x = 10 - 6$ = $x = 4$

Check for extraneous solutions: When $x = 4$: denominators are $4$ and $2$. Neither is zero. = Valid solution

Answer: $x = 4$

Equation with Variable in Multiple Denominators

Solve: $\frac{2}{x-3} - \frac{1}{x+2} = \frac{5}{(x-3)(x+2)}$

Find the LCD: Denominators: $(x-3)$, $(x+2)$, $(x-3)(x+2)$ = LCD = $(x-3)(x+2)$

Multiply every term by LCD: $(x-3)(x+2) \cdot \frac{2}{x-3} - (x-3)(x+2) \cdot \frac{1}{x+2} = (x-3)(x+2) \cdot \frac{5}{(x-3)(x+2)}$ = $2(x+2) - 1(x-3) = 5$

Distribute and simplify: $2x + 4 - x + 3 = 5$ $x + 7 = 5$ = $x = -2$

Check for extraneous solutions: When $x = -2$: $(x+2) = 0$. This makes the denominator zero! = EXTRANEOUS - reject this solution

Answer: No solution (the only candidate $x = -2$ is extraneous)

Quadratic Result

Solve: $\frac{x}{x-2} + \frac{1}{x+1} = \frac{3}{(x-2)(x+1)}$

Find the LCD: Denominators: $(x-2)$, $(x+1)$, $(x-2)(x+1)$ = LCD = $(x-2)(x+1)$

Multiply every term by LCD: $x(x+1) + 1(x-2) = 3$ = $x^2 + x + x - 2 = 3$

Simplify to standard form: $x^2 + 2x - 2 = 3$ $x^2 + 2x - 5 = 0$ = Quadratic equation

Use the quadratic formula: $x = \frac{-2 \pm \sqrt{4 + 20}}{2} = \frac{-2 \pm \sqrt{24}}{2} = \frac{-2 \pm 2\sqrt{6}}{2}$ = $x = -1 \pm \sqrt{6}$

Check both solutions: $x = -1 + \sqrt{6} \approx 1.45$ (valid) $x = -1 - \sqrt{6} \approx -3.45$ (valid) Neither makes $(x-2)$ or $(x+1)$ equal to zero. = Both solutions are valid

Answer: $x = -1 + \sqrt{6}$ or $x = -1 - \sqrt{6}$

Mistake: Forgetting to check for extraneous solutions

Why: When you multiply by an expression containing the variable, you might introduce solutions that don't work in the original equation.

Correct: Always substitute your answer back into the original equation to verify it doesn't make any denominator zero.

Mistake: Not multiplying ALL terms by the LCD

Why: Students often forget to multiply terms that are already integers or simple fractions.

Correct: Every term on both sides of the equation must be multiplied by the LCD, including constants like $3$ or $\frac{1}{2}$.

Mistake: Using the wrong LCD

Why: Finding the LCD of algebraic expressions is harder than finding LCD of numbers.

Correct: Factor all denominators first, then include each factor the maximum number of times it appears in any one denominator.

Mistake: Sign errors when distributing

Why: After clearing denominators, students often make errors with negative signs during distribution.

Correct: Be extra careful with subtraction: $-(x-3) = -x + 3$, not $-x - 3$.

Work Rate Problems

When two workers or machines complete a task together, their combined rate is the sum of individual rates.

Pipe A fills a pool in 6 hours. Pipe B fills it in 4 hours. Together: $\frac{1}{6} + \frac{1}{4} = \frac{1}{t}$ gives $t = 2.4$ hours.

Average Speed Problems

Finding average speed for a round trip requires rational equations when speeds differ.

Drive 60 km at 30 km/h, return at 60 km/h. Average speed is NOT 45 km/h! Solve $\frac{120}{v} = \frac{60}{30} + \frac{60}{60}$ to get $v = 40$ km/h.

Electrical Circuits

Parallel resistors combine using the formula $\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2}$.

Two resistors of 6 ohms and 3 ohms in parallel: $\frac{1}{R_T} = \frac{1}{6} + \frac{1}{3} = \frac{1}{2}$, so $R_T = 2$ ohms.

Q: What is an extraneous solution?

A: An extraneous solution is a value that satisfies the simplified equation but makes one of the original denominators equal to zero. It appears during the solving process but is not a valid solution to the original equation.

Q: Why do we multiply by the LCD?

A: Multiplying by the LCD eliminates all denominators, converting the rational equation into a polynomial equation that's easier to solve. Think of it as clearing fractions.

Q: Can a rational equation have no solution?

A: Yes! If all solutions found are extraneous (they make a denominator zero), then the equation has no solution.

Q: How do I find the LCD with variable expressions?

A: First factor all denominators completely. The LCD includes each unique factor raised to the highest power it appears in any denominator.