Learn on PengiYoshiwara Elementary AlgebraChapter 4: Applications of Linear Equations

Lesson 1: The Distributive Law

New Concept Master the distributive law, $a(b+c)=ab+ac$, to simplify expressions containing parentheses. This skill is essential for solving more complex linear equations and writing algebraic models for applied problems.

Section 1

πŸ“˜ The Distributive Law

New Concept

Master the distributive law, a(b+c)=ab+aca(b+c)=ab+ac, to simplify expressions containing parentheses. This skill is essential for solving more complex linear equations and writing algebraic models for applied problems.

What’s next

Next, you’ll work through interactive examples and practice cards to apply this law when solving equations and tackling real-world problems.

Section 2

Products

Property

To simplify a product such as 3(2a)3(2a), we multiply the numerical factors to find 3(2a)=(3β‹…2)a=6a3(2a) = (3 \cdot 2)a = 6a. This calculation is an application of the associative law for multiplication. We can see why the product is reasonable by recalling that multiplication is repeated addition, so that 3(2a)=2a+2a+2a=6a3(2a) = 2a + 2a + 2a = 6a.

Examples

  • To simplify 4(5b)4(5b), you multiply the numerical factors: (4β‹…5)b=20b(4 \cdot 5)b = 20b.
  • To simplify βˆ’6(2y)-6(2y), you multiply the numbers: (βˆ’6β‹…2)y=βˆ’12y(-6 \cdot 2)y = -12y.
  • To simplify βˆ’3(βˆ’8z)-3(-8z), you multiply the coefficients: (βˆ’3β‹…βˆ’8)z=24z(-3 \cdot -8)z = 24z.

Explanation

When a number is outside parentheses with a single term inside, just multiply the numbers together. Think of it as a shortcut for adding the term multiple times. It's all about grouping the numbers first to simplify.

Section 3

The Distributive Law

Property

If aa, bb, and cc are any numbers, then

a(b+c)=ab+aca(b + c) = ab + ac

If the terms inside parentheses are not like terms, we have no choice but to use the distributive law to simplify the expression.

Examples

  • To simplify 5(x+4)5(x+4), distribute the 5 to each term: 5(x)+5(4)=5x+205(x) + 5(4) = 5x + 20.
  • To simplify βˆ’3(2yβˆ’1)-3(2y-1), multiply each inner term by βˆ’3-3: βˆ’3(2y)βˆ’3(βˆ’1)=βˆ’6y+3-3(2y) - 3(-1) = -6y + 3.
  • To simplify (aβˆ’5)6(a-5)6, distribute the 6 from the right: a(6)βˆ’5(6)=6aβˆ’30a(6) - 5(6) = 6a - 30.

Explanation

The distributive law lets you multiply a number outside parentheses by each term inside. It's like sharing the outside number with every term in the group through multiplication. This is essential when you can't combine the terms inside first.

Section 4

Solving Linear Equations

Property

Steps for Solving Linear Equations.

  1. Use the distributive law to remove any parentheses.
  2. Combine like terms on each side of the equation.
  3. By adding or subtracting the same quantity on both sides of the equation, get all the variable terms on one side and all the constant terms on the other.
  4. Divide both sides by the coefficient of the variable to obtain an equation of the form x=ax = a.

Examples

  • Solve 3(xβˆ’4)=93(x-4) = 9. First, distribute the 3 to get 3xβˆ’12=93x - 12 = 9. Add 12 to both sides to get 3x=213x = 21. Finally, divide by 3 to find x=7x = 7.
  • Solve 5(y+1)=2yβˆ’45(y+1) = 2y - 4. Distribute to get 5y+5=2yβˆ’45y+5 = 2y-4. Subtract 2y2y from both sides, then subtract 5 from both sides to get 3y=βˆ’93y = -9. Divide by 3 to find y=βˆ’3y=-3.
  • Solve 25βˆ’4x=2xβˆ’5(2βˆ’x)25 - 4x = 2x - 5(2-x). Distribute to get 25βˆ’4x=2xβˆ’10+5x25 - 4x = 2x - 10 + 5x. Combine like terms to get 25βˆ’4x=7xβˆ’1025 - 4x = 7x - 10. Add 4x4x to both sides, then add 10 to both sides to get 35=11x35 = 11x, so x=3511x = \frac{35}{11}.

Explanation

When an equation has parentheses, first use the distributive law to clear them. After that, tidy up by combining like terms on each side. This simplifies the equation, making it easier to isolate the variable and find your solution.

Section 5

Simplifying Algebraic Expressions

Property

To build and simplify an algebraic expression from a description, first translate the words into variables and operations. If the expression involves multiplying a quantity in parentheses, use the distributive law to expand it. Finally, combine any like terms to get the simplest form. For example, for a rectangle with perimeter P=2l+2wP = 2l + 2w, if the length ll is 2wβˆ’32w-3, we substitute to get P=2(2wβˆ’3)+2wP = 2(2w - 3) + 2w.

Examples

  • A rectangle's length is 5 feet more than twice its width ww. The perimeter is P=2(2w+5)+2wP = 2(2w+5) + 2w. Simplifying gives P=4w+10+2w=6w+10P = 4w + 10 + 2w = 6w + 10.
  • A large pizza costs 5 dollars more than a medium pizza, mm. If you buy 3 large pizzas, the total cost is 3(m+5)3(m+5). Using the distributive law, this simplifies to 3m+153m + 15.
  • Ben is twice as old as his sister, ss. In 4 years, Ben's age will be 2s+42s+4, and his sister's age will be s+4s+4. The sum of their ages in 4 years will be (2s+4)+(s+4)=3s+8(2s+4) + (s+4) = 3s + 8.

Explanation

The distributive law is a key tool for turning word problems into math. It helps you set up a formula based on a description and then simplify it into a clean, usable expression for solving problems.

Book overview

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Chapter 4: Applications of Linear Equations

  1. Lesson 1Current

    Lesson 1: The Distributive Law

  2. Lesson 2

    Lesson 2: Systems of Linear Equations

  3. Lesson 3

    Lesson 3: Algebraic Solution of Systems

  4. Lesson 4

    Lesson 4: Problem Solving with Systems

  5. Lesson 5

    Lesson 5: Point-Slope Formula

  6. Lesson 6

    Lesson 4.6: Chapter Summary and Review

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

πŸ“˜ The Distributive Law

New Concept

Master the distributive law, a(b+c)=ab+aca(b+c)=ab+ac, to simplify expressions containing parentheses. This skill is essential for solving more complex linear equations and writing algebraic models for applied problems.

What’s next

Next, you’ll work through interactive examples and practice cards to apply this law when solving equations and tackling real-world problems.

Section 2

Products

Property

To simplify a product such as 3(2a)3(2a), we multiply the numerical factors to find 3(2a)=(3β‹…2)a=6a3(2a) = (3 \cdot 2)a = 6a. This calculation is an application of the associative law for multiplication. We can see why the product is reasonable by recalling that multiplication is repeated addition, so that 3(2a)=2a+2a+2a=6a3(2a) = 2a + 2a + 2a = 6a.

Examples

  • To simplify 4(5b)4(5b), you multiply the numerical factors: (4β‹…5)b=20b(4 \cdot 5)b = 20b.
  • To simplify βˆ’6(2y)-6(2y), you multiply the numbers: (βˆ’6β‹…2)y=βˆ’12y(-6 \cdot 2)y = -12y.
  • To simplify βˆ’3(βˆ’8z)-3(-8z), you multiply the coefficients: (βˆ’3β‹…βˆ’8)z=24z(-3 \cdot -8)z = 24z.

Explanation

When a number is outside parentheses with a single term inside, just multiply the numbers together. Think of it as a shortcut for adding the term multiple times. It's all about grouping the numbers first to simplify.

Section 3

The Distributive Law

Property

If aa, bb, and cc are any numbers, then

a(b+c)=ab+aca(b + c) = ab + ac

If the terms inside parentheses are not like terms, we have no choice but to use the distributive law to simplify the expression.

Examples

  • To simplify 5(x+4)5(x+4), distribute the 5 to each term: 5(x)+5(4)=5x+205(x) + 5(4) = 5x + 20.
  • To simplify βˆ’3(2yβˆ’1)-3(2y-1), multiply each inner term by βˆ’3-3: βˆ’3(2y)βˆ’3(βˆ’1)=βˆ’6y+3-3(2y) - 3(-1) = -6y + 3.
  • To simplify (aβˆ’5)6(a-5)6, distribute the 6 from the right: a(6)βˆ’5(6)=6aβˆ’30a(6) - 5(6) = 6a - 30.

Explanation

The distributive law lets you multiply a number outside parentheses by each term inside. It's like sharing the outside number with every term in the group through multiplication. This is essential when you can't combine the terms inside first.

Section 4

Solving Linear Equations

Property

Steps for Solving Linear Equations.

  1. Use the distributive law to remove any parentheses.
  2. Combine like terms on each side of the equation.
  3. By adding or subtracting the same quantity on both sides of the equation, get all the variable terms on one side and all the constant terms on the other.
  4. Divide both sides by the coefficient of the variable to obtain an equation of the form x=ax = a.

Examples

  • Solve 3(xβˆ’4)=93(x-4) = 9. First, distribute the 3 to get 3xβˆ’12=93x - 12 = 9. Add 12 to both sides to get 3x=213x = 21. Finally, divide by 3 to find x=7x = 7.
  • Solve 5(y+1)=2yβˆ’45(y+1) = 2y - 4. Distribute to get 5y+5=2yβˆ’45y+5 = 2y-4. Subtract 2y2y from both sides, then subtract 5 from both sides to get 3y=βˆ’93y = -9. Divide by 3 to find y=βˆ’3y=-3.
  • Solve 25βˆ’4x=2xβˆ’5(2βˆ’x)25 - 4x = 2x - 5(2-x). Distribute to get 25βˆ’4x=2xβˆ’10+5x25 - 4x = 2x - 10 + 5x. Combine like terms to get 25βˆ’4x=7xβˆ’1025 - 4x = 7x - 10. Add 4x4x to both sides, then add 10 to both sides to get 35=11x35 = 11x, so x=3511x = \frac{35}{11}.

Explanation

When an equation has parentheses, first use the distributive law to clear them. After that, tidy up by combining like terms on each side. This simplifies the equation, making it easier to isolate the variable and find your solution.

Section 5

Simplifying Algebraic Expressions

Property

To build and simplify an algebraic expression from a description, first translate the words into variables and operations. If the expression involves multiplying a quantity in parentheses, use the distributive law to expand it. Finally, combine any like terms to get the simplest form. For example, for a rectangle with perimeter P=2l+2wP = 2l + 2w, if the length ll is 2wβˆ’32w-3, we substitute to get P=2(2wβˆ’3)+2wP = 2(2w - 3) + 2w.

Examples

  • A rectangle's length is 5 feet more than twice its width ww. The perimeter is P=2(2w+5)+2wP = 2(2w+5) + 2w. Simplifying gives P=4w+10+2w=6w+10P = 4w + 10 + 2w = 6w + 10.
  • A large pizza costs 5 dollars more than a medium pizza, mm. If you buy 3 large pizzas, the total cost is 3(m+5)3(m+5). Using the distributive law, this simplifies to 3m+153m + 15.
  • Ben is twice as old as his sister, ss. In 4 years, Ben's age will be 2s+42s+4, and his sister's age will be s+4s+4. The sum of their ages in 4 years will be (2s+4)+(s+4)=3s+8(2s+4) + (s+4) = 3s + 8.

Explanation

The distributive law is a key tool for turning word problems into math. It helps you set up a formula based on a description and then simplify it into a clean, usable expression for solving problems.

Book overview

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

Continue this chapter

Chapter 4: Applications of Linear Equations

  1. Lesson 1Current

    Lesson 1: The Distributive Law

  2. Lesson 2

    Lesson 2: Systems of Linear Equations

  3. Lesson 3

    Lesson 3: Algebraic Solution of Systems

  4. Lesson 4

    Lesson 4: Problem Solving with Systems

  5. Lesson 5

    Lesson 5: Point-Slope Formula

  6. Lesson 6

    Lesson 4.6: Chapter Summary and Review