Solving Rational Equations

➗ Solving Rational Equations

Part 1 of 5 — Restrictions & the LCD

Topics in This Part

🔑 Key Concept: A rational equation has a variable in a denominator. Before we solve anything, we must answer two questions: what values are forbidden (they'd divide by zero), and what single denominator (the LCD) clears all the fractions. Part 1 builds those two tools.

What Is a Rational Equation?

A rational equation is an equation containing at least one rational expression — a fraction with a polynomial (and a variable) in the denominator.

$\frac{3}{x} = \frac{6}{x+2}, \qquad \frac{x}{x-1} + 2 = \frac{1}{x-1}, \qquad \frac{5}{x^2-9} = \frac{1}{x-3}$

The big idea behind solving them is simple:

🔑 Strategy: Multiply every term by the least common denominator (LCD). This clears all the fractions and leaves an ordinary linear or quadratic equation you already know how to solve.

But there's a catch that makes rational equations special: dividing by zero is undefined, so some $x$-values are off-limits. We deal with those first.

Excluded Values (Restrictions)

A denominator can never equal zero. Any $x$ that makes a denominator $0$ is an excluded value (a restriction) — it can never be a solution, even if it survives the algebra.

To find restrictions: set each denominator equal to $0$ and solve.

⚠️ Write the restrictions down before you solve. If your final answer equals a restricted value, you must throw it out — it's called an extraneous solution (Part 4).

Concept Check 🎯

Find the Restriction 🧮

Enter the value of $x$ that is excluded from each expression.

1) $\dfrac{7}{x - 8}$ → $x \ne \,?$ 2) $\dfrac{x+1}{x + 3}$ → $x \ne \,?$ 3) $\dfrac{10}{2x - 6}$ → $x \ne \,?$

Finding the LCD

The least common denominator is the smallest expression every denominator divides into evenly. To build it:

1. Factor every denominator completely.
2. Take each distinct factor, using the highest power it appears with anywhere.
3. Multiply those factors together.

Examples

💡 If a factor like $(x-3)$ already lives inside another denominator's factorization, you do not repeat it. The LCD takes each factor just once at its highest power.

Concept Check 🎯

➗ Solving Rational Equations

Part 2 of 5 — The Core Method: Clear the Denominators

🔑 The Idea: Multiply every term by the LCD. Each denominator cancels, the fractions vanish, and you're left with a plain polynomial equation to solve.

The Four Steps

To solve any rational equation:

1. State restrictions — set each denominator to $0$.
2. Multiply every term by the LCD and cancel.
3. Solve the resulting linear or quadratic equation.
4. Check each answer against the restrictions; discard any that are excluded.

Worked Example: $\dfrac{3}{x} = \dfrac{6}{x+2}$

Restrictions: $x \ne 0$ and $x \ne -2$. LCD $= x(x+2)$.

Multiply both sides by $x(x+2)$:

$x(x+2)\cdot \frac{3}{x} = x(x+2)\cdot \frac{6}{x+2}$

Cancel: the left loses its $x$, the right loses its $(x+2)$:

$3(x+2) = 6x \;\Rightarrow\; 3x + 6 = 6x \;\Rightarrow\; 6 = 3x \;\Rightarrow\; x = 2$

✅ Check: $x = 2$ is not restricted. $\dfrac{3}{2} = \dfrac{6}{4} = \dfrac{3}{2}$ ✓

Worked Example: $\dfrac{x}{4} + \dfrac{1}{2} = \dfrac{5}{4}$

Even when the denominators are just numbers, the method is the same. LCD $= 4$ (no restrictions, since no variable is in a denominator):

$4\cdot\frac{x}{4} + 4\cdot\frac{1}{2} = 4\cdot\frac{5}{4}$

$x + 2 = 5 \;\Rightarrow\; x = 3$

Worked Example: $\dfrac{2}{x} + 3 = \dfrac{8}{x}$

Restriction: $x \ne 0$. LCD $= x$. Multiply every term by $x$ — including the lone $3$:

$x\cdot\frac{2}{x} + x\cdot 3 = x\cdot\frac{8}{x} \;\Rightarrow\; 2 + 3x = 8 \;\Rightarrow\; 3x = 6 \;\Rightarrow\; x = 2$

⚠️ Don't forget the whole-number terms. Every term gets multiplied by the LCD — the $3$ became $3x$, not $3$.

Concept Check 🎯

Order the Steps 🔽

You're solving $\dfrac{4}{x} = \dfrac{6}{x+1}$. Choose what happens at each stage.

Solve It 🧮

Multiply by the LCD and solve.

1) $\dfrac{6}{x} = 3 \;\Rightarrow\; x = \,?$ 2) $\dfrac{x}{5} + 1 = \dfrac{7}{5} \;\Rightarrow\; x = \,?$ 3) $\dfrac{3}{x} + 4 = \dfrac{15}{x} \;\Rightarrow\; x = \,?$

➗ Solving Rational Equations

Part 3 of 5 — Factored Denominators & Quadratics

🔑 Level up: When denominators factor (like $x^2 - 9$), build the LCD from the factors. Clearing fractions often produces a quadratic — solve it by factoring, then check both roots.

Proportions: Cross-Multiplication

When the equation is one fraction = one fraction, you can cross-multiply — a shortcut for clearing both denominators at once:

$\frac{a}{b} = \frac{c}{d} \quad\Longrightarrow\quad a\cdot d = b\cdot c$

Example: $\dfrac{x+3}{4} = \dfrac{x-1}{2}$

Cross-multiply:

$2(x+3) = 4(x-1) \;\Rightarrow\; 2x + 6 = 4x - 4 \;\Rightarrow\; 10 = 2x \;\Rightarrow\; x = 5$

💡 Cross-multiplication only works for a single fraction equal to a single fraction. If there are three or more terms, go back to the full LCD method.

Concept Check 🎯

When Denominators Factor

Worked Example: $\dfrac{1}{x-3} + \dfrac{1}{x+3} = \dfrac{6}{x^2-9}$

Factor first: $x^2 - 9 = (x-3)(x+3)$.

Restrictions: $x \ne 3$ and $x \ne -3$. LCD $= (x-3)(x+3)$.

Multiply every term by $(x-3)(x+3)$ and cancel:

$(x+3) + (x-3) = 6$

$2x = 6 \;\Rightarrow\; x = 3$

⚠️ Stop! $x = 3$ is a restriction — it makes the original denominators zero. So $x = 3$ is extraneous and must be rejected. This equation has no solution. (Much more on this in Part 4.)

When Clearing Gives a Quadratic

Worked Example: $\dfrac{x}{x+2} = \dfrac{3}{x}$

Restrictions: $x \ne 0$ and $x \ne -2$. LCD $= x(x+2)$.

Cross-multiply (it's a single fraction = single fraction):

$x\cdot x = 3(x+2) \;\Rightarrow\; x^2 = 3x + 6 \;\Rightarrow\; x^2 - 3x - 6 = 0$

This one doesn't factor nicely — but $\dfrac{x}{x+2} = \dfrac{3}{x}$ with nicer numbers does. Try instead $\dfrac{x}{x+3} = \dfrac{4}{x}$:

$x^2 = 4(x+3) \;\Rightarrow\; x^2 - 4x - 12 = 0 \;\Rightarrow\; (x-6)(x+2) = 0$

$x = 6 \quad\text{or}\quad x = -2$

✅ Both pass the restrictions ($x \ne 0, -3$), so both are valid: $x = 6$ and $x = -2$.

Build It Right 🔽

You're solving $\dfrac{2}{x-1} + \dfrac{3}{x} = \dfrac{5}{x}$. Choose correctly at each stage.

Solve It 🧮

1) Cross-multiply: $\dfrac{x}{6} = \dfrac{3}{2} \;\Rightarrow\; x = \,?$ 2) $\dfrac{x-2}{3} = \dfrac{x+4}{5} \;\Rightarrow\; x = \,?$ 3) $\dfrac{x}{x+3} = \dfrac{4}{x}$ leads to $x^2 - 4x - 12 = 0$. Enter the larger root. $x = \,?$

➗ Solving Rational Equations

Part 4 of 5 — Extraneous Solutions & Applications

🔑 The Most Important Check: Multiplying by an expression containing $x$ can introduce a fake solution — one that makes a denominator zero. Always test every answer against the restrictions. If it's excluded, reject it.

Why Extraneous Solutions Happen

When you multiply both sides by the LCD, you might be multiplying by zero (if $x$ equals a restricted value). Multiplying an equation by zero can turn a false statement into a true one — creating a solution that doesn't actually work in the original.

Worked Example: $\dfrac{x}{x-4} = \dfrac{4}{x-4} + 3$

Restriction: $x \ne 4$. LCD $= x - 4$. Multiply every term:

$x = 4 + 3(x-4) \;\Rightarrow\; x = 4 + 3x - 12 \;\Rightarrow\; x = 3x - 8$

$8 = 2x \;\Rightarrow\; x = 4$

⚠️ Reject it! $x = 4$ is restricted (it makes $x - 4 = 0$). It's extraneous, so this equation has no solution.

Concept Check 🎯

Valid, Extraneous, or No Solution? 🔽

For each, the algebra has been done — decide the outcome.

Application: Work-Rate Problems

Rational equations model combined rates. If one worker finishes a job in $a$ hours and another in $b$ hours, their rates are $\frac{1}{a}$ and $\frac{1}{b}$ jobs per hour. Working together for $t$ hours:

$\frac{t}{a} + \frac{t}{b} = 1 \quad(\text{one completed job})$

Example: Pipes Filling a Pool

Pipe A fills a pool in $6$ hours; pipe B in $3$ hours. How long together?

$\frac{t}{6} + \frac{t}{3} = 1$

Multiply by the LCD $6$: $\;t + 2t = 6 \;\Rightarrow\; 3t = 6 \;\Rightarrow\; t = 2$ hours.

💡 Sanity check: Together they must be faster than either alone. $2$ hours is less than both $6$ and $3$ ✓.

Work-Rate Practice 🧮

1) Alone, Maya paints a room in $4$ hours and Leo in $4$ hours. Together: $\dfrac{t}{4} + \dfrac{t}{4} = 1 \Rightarrow t = \,?$ hours. 2) A pipe fills a tank in $2$ hours, another in $6$ hours. Together $\dfrac{t}{2} + \dfrac{t}{6} = 1 \Rightarrow t = \,?$ hours. 3) Solve $\dfrac{x}{x-1} = \dfrac{1}{x-1} + 2$. If the only root is extraneous, enter $0$ for "no solution." $x = \,?$

➗ Solving Rational Equations

Part 5 of 5 — Mixed Practice & Mastery Check

You can now (1) find restrictions, (2) build the LCD, (3) clear fractions and solve, and (4) catch extraneous solutions. Let's put it all together.

Quick Reference

⚠️ The two most common mistakes: (1) forgetting to multiply a whole-number term by the LCD, and (2) keeping a root that violates a restriction. Always check.

Mixed Practice 🎯

Mixed Drill 🧮

1) $\dfrac{8}{x} = 2 \Rightarrow x = \,?$ 2) $\dfrac{x}{3} - 1 = \dfrac{x}{6} \Rightarrow x = \,?$ 3) $\dfrac{x+2}{5} = \dfrac{x-2}{3} \Rightarrow x = \,?$

Exit Quiz ✅

Answer all three to finish the lesson.