Learn on PengiSaxon Algebra 2Chapter 4: Lessons 31-40, Investigation 4

Lesson 37: Adding and Subtracting Rational Expressions

In this Grade 10 Saxon Algebra 2 lesson, students learn to add and subtract rational expressions by combining like denominators, finding the least common denominator (LCD) for unlike denominators, and simplifying results by factoring and canceling common terms. The lesson also covers identifying values of x for which a rational expression is undefined by setting the denominator equal to zero. Students work through multi-step problems involving polynomial denominators such as quadratics and higher-degree expressions.

Section 1

πŸ“˜ Adding and Subtracting Rational Expressions

New Concept

A rational expression is a ratio of polynomials: f(x)g(x)\frac{f(x)}{g(x)}, for polynomials f(x)f(x) and g(x)g(x), where g(x)β‰ 0g(x) \neq 0 and is of degree β‰₯1\geq 1.

What’s next

Next, you’ll practice adding and subtracting these expressions by finding a common denominator, just like with numerical fractions.

Section 2

Combining with like denominators

When adding or subtracting rational expressions with the same denominator, you simply combine the numerators and place the result over the common denominator. The rule is AC+BC=A+BC\frac{A}{C} + \frac{B}{C} = \frac{A+B}{C}. Be careful to distribute the negative sign when subtracting numerators that are polynomials.

54y+74y=5+74y=124y=3y\frac{5}{4y} + \frac{7}{4y} = \frac{5+7}{4y} = \frac{12}{4y} = \frac{3}{y}
zz2βˆ’16βˆ’4z2βˆ’16=zβˆ’4(zβˆ’4)(z+4)=1z+4\frac{z}{z^2 - 16} - \frac{4}{z^2 - 16} = \frac{z-4}{(z-4)(z+4)} = \frac{1}{z+4}
3aa2+aβˆ’6βˆ’aβˆ’4a2+aβˆ’6=3aβˆ’(aβˆ’4)(a+3)(aβˆ’2)=2a+4(a+3)(aβˆ’2)\frac{3a}{a^2+a-6} - \frac{a-4}{a^2+a-6} = \frac{3a-(a-4)}{(a+3)(a-2)} = \frac{2a+4}{(a+3)(a-2)}

Think of it like adding pizza slices! If you have 18\frac{1}{8} of a pizza and someone gives you 38\frac{3}{8} more, you just add the numerators to find out how many slices you have in total. The size of the slices, which is the denominator, stays the same. The same simple logic applies to algebraic fractions!

Section 3

Least common denominator

The least common denominator (LCD) is the least common multiple (LCM) of the denominators of the fractions. To find the LCD of polynomial denominators, first factor each polynomial completely. Then, construct the LCD by taking the highest power of each unique factor that appears in any of the denominators and multiplying them together.

For 14x3\frac{1}{4x^3} and 16x2\frac{1}{6x^2}, the factors are 22,x32^2, x^3 and 2,3,x22, 3, x^2. The LCD is 22β‹…3β‹…x3=12x32^2 \cdot 3 \cdot x^3 = 12x^3.
For 1x2βˆ’4\frac{1}{x^2-4} and 1x2+xβˆ’6\frac{1}{x^2+x-6}, factor to 1(xβˆ’2)(x+2)\frac{1}{(x-2)(x+2)} and 1(x+3)(xβˆ’2)\frac{1}{(x+3)(x-2)}. The LCD is (xβˆ’2)(x+2)(x+3)(x-2)(x+2)(x+3).

Think of the LCD as finding a 'common language' for different fractions before they can be combined. By breaking down each denominator into its prime factors, you can see all the essential 'building blocks.' The LCD is the smallest and most efficient denominator that includes every single one of these blocks, ensuring all fractions can be converted smoothly.

Section 4

Undefined values of a rational expression

A rational expression in the form f(x)g(x)\frac{f(x)}{g(x)} is undefined for any value of the variable that causes the denominator, g(x)g(x), to equal zero. This is because division by zero is not a defined operation in mathematics. To find these values, you must set the denominator equal to zero and solve for the variable.

The expression x+24xβˆ’12\frac{x+2}{4x-12} is undefined when 4xβˆ’12=04x-12=0, which happens when x=3x=3.
The expression 3xx2βˆ’25\frac{3x}{x^2-25} is undefined when x2βˆ’25=0x^2-25=0. Factoring gives (xβˆ’5)(x+5)=0(x-5)(x+5)=0, so it is undefined at x=5x=5 and x=βˆ’5x=-5.

Remember the most important rule in fractions: you can never, ever divide by zero! It's a mathematical impossibility. Finding the 'undefined values' is like finding the secret trap doors in your equation. By setting the denominator to zero, you are identifying the specific inputs that would break the math, helping you steer clear of trouble.

Book overview

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Chapter 4: Lessons 31-40, Investigation 4

  1. Lesson 1

    Lesson 31: Multiplying and Dividing Rational Expressions

  2. Lesson 2

    Lesson 32: Solving Linear Systems with Matrix Inverses (Exploration: Exploring Matrix Inverses)

  3. Lesson 3

    Lesson 33: Applying Counting Principles

  4. Lesson 4

    Lesson 34: Graphing Linear Equations II

  5. Lesson 5

    Lesson 35: Solving Quadratic Equations I

  6. Lesson 6

    Lesson 36: Using Parallel and Perpendicular Lines

  7. Lesson 7Current

    Lesson 37: Adding and Subtracting Rational Expressions

  8. Lesson 8

    Lesson 38: Dividing Polynomials Using Long Division

  9. Lesson 9

    Lesson 39: Graphing Linear Inequalities in Two Variables

  10. Lesson 10

    Lesson 40: Simplifying Radical Expressions

  11. Lesson 11

    Investigation 4: Understanding Cryptography

Lesson overview

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Section 1

πŸ“˜ Adding and Subtracting Rational Expressions

New Concept

A rational expression is a ratio of polynomials: f(x)g(x)\frac{f(x)}{g(x)}, for polynomials f(x)f(x) and g(x)g(x), where g(x)β‰ 0g(x) \neq 0 and is of degree β‰₯1\geq 1.

What’s next

Next, you’ll practice adding and subtracting these expressions by finding a common denominator, just like with numerical fractions.

Section 2

Combining with like denominators

When adding or subtracting rational expressions with the same denominator, you simply combine the numerators and place the result over the common denominator. The rule is AC+BC=A+BC\frac{A}{C} + \frac{B}{C} = \frac{A+B}{C}. Be careful to distribute the negative sign when subtracting numerators that are polynomials.

54y+74y=5+74y=124y=3y\frac{5}{4y} + \frac{7}{4y} = \frac{5+7}{4y} = \frac{12}{4y} = \frac{3}{y}
zz2βˆ’16βˆ’4z2βˆ’16=zβˆ’4(zβˆ’4)(z+4)=1z+4\frac{z}{z^2 - 16} - \frac{4}{z^2 - 16} = \frac{z-4}{(z-4)(z+4)} = \frac{1}{z+4}
3aa2+aβˆ’6βˆ’aβˆ’4a2+aβˆ’6=3aβˆ’(aβˆ’4)(a+3)(aβˆ’2)=2a+4(a+3)(aβˆ’2)\frac{3a}{a^2+a-6} - \frac{a-4}{a^2+a-6} = \frac{3a-(a-4)}{(a+3)(a-2)} = \frac{2a+4}{(a+3)(a-2)}

Think of it like adding pizza slices! If you have 18\frac{1}{8} of a pizza and someone gives you 38\frac{3}{8} more, you just add the numerators to find out how many slices you have in total. The size of the slices, which is the denominator, stays the same. The same simple logic applies to algebraic fractions!

Section 3

Least common denominator

The least common denominator (LCD) is the least common multiple (LCM) of the denominators of the fractions. To find the LCD of polynomial denominators, first factor each polynomial completely. Then, construct the LCD by taking the highest power of each unique factor that appears in any of the denominators and multiplying them together.

For 14x3\frac{1}{4x^3} and 16x2\frac{1}{6x^2}, the factors are 22,x32^2, x^3 and 2,3,x22, 3, x^2. The LCD is 22β‹…3β‹…x3=12x32^2 \cdot 3 \cdot x^3 = 12x^3.
For 1x2βˆ’4\frac{1}{x^2-4} and 1x2+xβˆ’6\frac{1}{x^2+x-6}, factor to 1(xβˆ’2)(x+2)\frac{1}{(x-2)(x+2)} and 1(x+3)(xβˆ’2)\frac{1}{(x+3)(x-2)}. The LCD is (xβˆ’2)(x+2)(x+3)(x-2)(x+2)(x+3).

Think of the LCD as finding a 'common language' for different fractions before they can be combined. By breaking down each denominator into its prime factors, you can see all the essential 'building blocks.' The LCD is the smallest and most efficient denominator that includes every single one of these blocks, ensuring all fractions can be converted smoothly.

Section 4

Undefined values of a rational expression

A rational expression in the form f(x)g(x)\frac{f(x)}{g(x)} is undefined for any value of the variable that causes the denominator, g(x)g(x), to equal zero. This is because division by zero is not a defined operation in mathematics. To find these values, you must set the denominator equal to zero and solve for the variable.

The expression x+24xβˆ’12\frac{x+2}{4x-12} is undefined when 4xβˆ’12=04x-12=0, which happens when x=3x=3.
The expression 3xx2βˆ’25\frac{3x}{x^2-25} is undefined when x2βˆ’25=0x^2-25=0. Factoring gives (xβˆ’5)(x+5)=0(x-5)(x+5)=0, so it is undefined at x=5x=5 and x=βˆ’5x=-5.

Remember the most important rule in fractions: you can never, ever divide by zero! It's a mathematical impossibility. Finding the 'undefined values' is like finding the secret trap doors in your equation. By setting the denominator to zero, you are identifying the specific inputs that would break the math, helping you steer clear of trouble.

Book overview

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

Continue this chapter

Chapter 4: Lessons 31-40, Investigation 4

  1. Lesson 1

    Lesson 31: Multiplying and Dividing Rational Expressions

  2. Lesson 2

    Lesson 32: Solving Linear Systems with Matrix Inverses (Exploration: Exploring Matrix Inverses)

  3. Lesson 3

    Lesson 33: Applying Counting Principles

  4. Lesson 4

    Lesson 34: Graphing Linear Equations II

  5. Lesson 5

    Lesson 35: Solving Quadratic Equations I

  6. Lesson 6

    Lesson 36: Using Parallel and Perpendicular Lines

  7. Lesson 7Current

    Lesson 37: Adding and Subtracting Rational Expressions

  8. Lesson 8

    Lesson 38: Dividing Polynomials Using Long Division

  9. Lesson 9

    Lesson 39: Graphing Linear Inequalities in Two Variables

  10. Lesson 10

    Lesson 40: Simplifying Radical Expressions

  11. Lesson 11

    Investigation 4: Understanding Cryptography