Learn on PengiSaxon Algebra 1Chapter 10: Systems and Problem Solving

Lesson 99: Solving Rational Equations

New Concept A rational equation is an equation containing at least one rational expression. What’s next Next, you’ll master two key methods for solving these equations and learn how to spot tricky “extraneous” solutions.

Section 1

📘 Solving Rational Equations

New Concept

A rational equation is an equation containing at least one rational expression.

What’s next

Next, you’ll master two key methods for solving these equations and learn how to spot tricky “extraneous” solutions.

Section 2

Rational equation

Property

A rational equation is an equation containing at least one rational expression. If it's a proportion, you can solve it using cross products: if ab=cd\frac{a}{b} = \frac{c}{d}, then ad=bcad = bc.

Explanation

Think of these as fraction puzzles where 'x' is hiding! When two fractions are equal, you can cross-multiply to get rid of the denominators and turn it into a regular equation. It’s a neat trick to un-complicate things and find the hidden value of x.

Examples

To solve 7x=4x3\frac{7}{x} = \frac{4}{x-3}, cross-multiply to get 7(x3)=4x7(x-3) = 4x, so x=7x=7.
To solve x5=2x+3\frac{x}{5} = \frac{2}{x+3}, cross-multiply to get x(x+3)=10x(x+3) = 10, so x2+3x10=0x^2+3x-10=0, which means x=2x=2 or x=5x=-5.

Section 3

Extraneous solution

Property

An extraneous solution is a solution that is acquired through the solving process, but makes a denominator in the original equation equal to 0.

Explanation

Watch out for these impostor solutions! They seem to solve the equation, but they are mathematical outlaws. Plugging them back into the original equation makes a denominator zero, which breaks the fundamental rules of fractions. Always check your answers to expose these fakes!

Examples

Solve x+3x4=2x+1x4\frac{x+3}{x-4} = \frac{2x+1}{x-4}. Cross-multiplying gives x+3=2x+1x+3=2x+1, so x=2x=2. This solution is valid.
Solve x5x5=2x9x5\frac{x-5}{x-5} = \frac{2x-9}{x-5}. This simplifies to 1=2x91 = 2x-9, so x=5x=5. But x=5x=5 makes the denominator 0, so it's an extraneous solution.

Section 4

Example Card: Checking for Extraneous Solutions

Sometimes an equation has a 'disguised' trap, let's learn how to find it. This example demonstrates how to solve rational equations using cross products, a key idea from this lesson, and how to identify extraneous solutions.

Example Problem

Solve the equation x+3x5=x+73x15\frac{x+3}{x-5} = \frac{x+7}{3x-15}.

Section 5

Using the LCD to solve equations

Property

If a rational equation includes a sum or difference, find the least common denominator (LCD) of all the terms and multiply the entire equation by it to solve.

Explanation

Fractions in an equation can be messy. Multiplying every single term by the LCD is like using a magic wand to make all the denominators disappear! This leaves you with a much friendlier, fraction-free equation to solve. It’s the ultimate cleanup tool for rational equations.

Examples

Solve 5x+12=3\frac{5}{x} + \frac{1}{2} = 3. The LCD is 2x2x. Multiply by 2x2x to get 10+x=6x10 + x = 6x, so x=2x=2.
Solve 4x11x=2x\frac{4}{x-1} - \frac{1}{x} = \frac{2}{x}. The LCD is x(x1)x(x-1). This becomes 4x(x1)=2(x1)4x - (x-1) = 2(x-1), which simplifies to x=1x=-1.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 10: Systems and Problem Solving

  1. Lesson 1

    Lesson 91: Solving Absolute-Value Inequalities

  2. Lesson 2

    Lesson 92: Simplifying Complex Fractions

  3. Lesson 3

    Lesson 93: Dividing Polynomials

  4. Lesson 4

    Lesson 94: Solving Multi-Step Absolute-Value Equations

  5. Lesson 5

    Lesson 95: Combining Rational Expressions with Unlike Denominators

  6. Lesson 6

    Lesson 96: Graphing Quadratic Functions

  7. Lesson 7

    Lesson 97: Graphing Linear Inequalities

  8. Lesson 8

    Lesson 98: Solving Quadratic Equations by Factoring

  9. Lesson 9Current

    Lesson 99: Solving Rational Equations

  10. Lesson 10

    Lesson 100: Solving Quadratic Equations by Graphing

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

📘 Solving Rational Equations

New Concept

A rational equation is an equation containing at least one rational expression.

What’s next

Next, you’ll master two key methods for solving these equations and learn how to spot tricky “extraneous” solutions.

Section 2

Rational equation

Property

A rational equation is an equation containing at least one rational expression. If it's a proportion, you can solve it using cross products: if ab=cd\frac{a}{b} = \frac{c}{d}, then ad=bcad = bc.

Explanation

Think of these as fraction puzzles where 'x' is hiding! When two fractions are equal, you can cross-multiply to get rid of the denominators and turn it into a regular equation. It’s a neat trick to un-complicate things and find the hidden value of x.

Examples

To solve 7x=4x3\frac{7}{x} = \frac{4}{x-3}, cross-multiply to get 7(x3)=4x7(x-3) = 4x, so x=7x=7.
To solve x5=2x+3\frac{x}{5} = \frac{2}{x+3}, cross-multiply to get x(x+3)=10x(x+3) = 10, so x2+3x10=0x^2+3x-10=0, which means x=2x=2 or x=5x=-5.

Section 3

Extraneous solution

Property

An extraneous solution is a solution that is acquired through the solving process, but makes a denominator in the original equation equal to 0.

Explanation

Watch out for these impostor solutions! They seem to solve the equation, but they are mathematical outlaws. Plugging them back into the original equation makes a denominator zero, which breaks the fundamental rules of fractions. Always check your answers to expose these fakes!

Examples

Solve x+3x4=2x+1x4\frac{x+3}{x-4} = \frac{2x+1}{x-4}. Cross-multiplying gives x+3=2x+1x+3=2x+1, so x=2x=2. This solution is valid.
Solve x5x5=2x9x5\frac{x-5}{x-5} = \frac{2x-9}{x-5}. This simplifies to 1=2x91 = 2x-9, so x=5x=5. But x=5x=5 makes the denominator 0, so it's an extraneous solution.

Section 4

Example Card: Checking for Extraneous Solutions

Sometimes an equation has a 'disguised' trap, let's learn how to find it. This example demonstrates how to solve rational equations using cross products, a key idea from this lesson, and how to identify extraneous solutions.

Example Problem

Solve the equation x+3x5=x+73x15\frac{x+3}{x-5} = \frac{x+7}{3x-15}.

Section 5

Using the LCD to solve equations

Property

If a rational equation includes a sum or difference, find the least common denominator (LCD) of all the terms and multiply the entire equation by it to solve.

Explanation

Fractions in an equation can be messy. Multiplying every single term by the LCD is like using a magic wand to make all the denominators disappear! This leaves you with a much friendlier, fraction-free equation to solve. It’s the ultimate cleanup tool for rational equations.

Examples

Solve 5x+12=3\frac{5}{x} + \frac{1}{2} = 3. The LCD is 2x2x. Multiply by 2x2x to get 10+x=6x10 + x = 6x, so x=2x=2.
Solve 4x11x=2x\frac{4}{x-1} - \frac{1}{x} = \frac{2}{x}. The LCD is x(x1)x(x-1). This becomes 4x(x1)=2(x1)4x - (x-1) = 2(x-1), which simplifies to x=1x=-1.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 10: Systems and Problem Solving

  1. Lesson 1

    Lesson 91: Solving Absolute-Value Inequalities

  2. Lesson 2

    Lesson 92: Simplifying Complex Fractions

  3. Lesson 3

    Lesson 93: Dividing Polynomials

  4. Lesson 4

    Lesson 94: Solving Multi-Step Absolute-Value Equations

  5. Lesson 5

    Lesson 95: Combining Rational Expressions with Unlike Denominators

  6. Lesson 6

    Lesson 96: Graphing Quadratic Functions

  7. Lesson 7

    Lesson 97: Graphing Linear Inequalities

  8. Lesson 8

    Lesson 98: Solving Quadratic Equations by Factoring

  9. Lesson 9Current

    Lesson 99: Solving Rational Equations

  10. Lesson 10

    Lesson 100: Solving Quadratic Equations by Graphing