Learn on PengiBig Ideas Math, Advanced 2Chapter 1: Equations

Lesson 4: Rewriting Equations and Formulas

In this Grade 7 lesson from Big Ideas Math Advanced 2, Chapter 1, students learn how to rewrite literal equations and geometric formulas by solving for one variable in terms of the other variables. They apply properties of equality to isolate specific variables in formulas for perimeter, area, volume, surface area, and temperature conversion, including the Fahrenheit-to-Celsius formula. By the end of the lesson, students can rearrange multi-variable equations such as the cone surface area formula or a linear equation like 2y + 5x = 6 to express any target variable.

Section 1

Solve a formula for a variable

Property

To solve a formula for a specific variable means to get that variable by itself with a coefficient of 1 on one side of the equation and all the other variables and constants on the other side. This process is also called solving a literal equation.

Examples

  • To solve the perimeter formula for a rectangle, P=2L+2WP = 2L + 2W, for LL, first subtract 2W2W to get P2W=2LP - 2W = 2L. Then divide by 2: L=P2W2L = \frac{P - 2W}{2}.
  • To solve the simple interest formula I=PrtI = Prt for the principal PP, you divide both sides by the product of rate and time: P=IrtP = \frac{I}{rt}.

Section 2

Solving Formulas with Fractions

Property

To solve a formula for a specific variable when fractions are present, first clear the fractions by multiplying the entire equation by the denominator or LCD. Then, use inverse operations to isolate the desired variable on one side of the equation.

Examples

Section 3

Solving Formulas with Fractions

Property

When solving a formula that includes a fraction, like the area of a triangle A=12bhA = \frac{1}{2}bh, you can first clear the fraction by multiplying both sides of the equation by the denominator.
Then, isolate the desired variable. To solve for hh, we get 2A=bh2A = bh, which simplifies to h=2Abh = \frac{2A}{b}.

Examples

  • Solve the formula A=12bhA = \frac{1}{2}bh for bb. First, multiply by 2 to get 2A=bh2A = bh. Then, divide by hh to get b=2Ahb = \frac{2A}{h}.
  • The formula for the average of two numbers is M=a+b2M = \frac{a+b}{2}. Solve for aa. Multiply by 2 to get 2M=a+b2M = a+b. Then subtract bb to get a=2Mba = 2M - b.
  • Solve the formula for the area of a trapezoid, A=12h(b1+b2)A = \frac{1}{2}h(b_1 + b_2), for hh. Multiply by 2: 2A=h(b1+b2)2A = h(b_1 + b_2). Then divide by (b1+b2)(b_1+b_2) to get h=2Ab1+b2h = \frac{2A}{b_1 + b_2}.

Explanation

To get a variable out of a fraction, first 'free' it by multiplying everything by the denominator. This clears the fraction and makes it much easier to isolate the variable you're looking for using standard algebra steps.

Book overview

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Chapter 1: Equations

  1. Lesson 1

    Section 1.1: Solving Simple Equations

  2. Lesson 2

    Section 1.2: Solving Multi-Step Equations

  3. Lesson 3

    Section 1.3: Solving Equations with Variables on Both Sides

  4. Lesson 4Current

    Lesson 4: Rewriting Equations and Formulas

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Solve a formula for a variable

Property

To solve a formula for a specific variable means to get that variable by itself with a coefficient of 1 on one side of the equation and all the other variables and constants on the other side. This process is also called solving a literal equation.

Examples

  • To solve the perimeter formula for a rectangle, P=2L+2WP = 2L + 2W, for LL, first subtract 2W2W to get P2W=2LP - 2W = 2L. Then divide by 2: L=P2W2L = \frac{P - 2W}{2}.
  • To solve the simple interest formula I=PrtI = Prt for the principal PP, you divide both sides by the product of rate and time: P=IrtP = \frac{I}{rt}.

Section 2

Solving Formulas with Fractions

Property

To solve a formula for a specific variable when fractions are present, first clear the fractions by multiplying the entire equation by the denominator or LCD. Then, use inverse operations to isolate the desired variable on one side of the equation.

Examples

Section 3

Solving Formulas with Fractions

Property

When solving a formula that includes a fraction, like the area of a triangle A=12bhA = \frac{1}{2}bh, you can first clear the fraction by multiplying both sides of the equation by the denominator.
Then, isolate the desired variable. To solve for hh, we get 2A=bh2A = bh, which simplifies to h=2Abh = \frac{2A}{b}.

Examples

  • Solve the formula A=12bhA = \frac{1}{2}bh for bb. First, multiply by 2 to get 2A=bh2A = bh. Then, divide by hh to get b=2Ahb = \frac{2A}{h}.
  • The formula for the average of two numbers is M=a+b2M = \frac{a+b}{2}. Solve for aa. Multiply by 2 to get 2M=a+b2M = a+b. Then subtract bb to get a=2Mba = 2M - b.
  • Solve the formula for the area of a trapezoid, A=12h(b1+b2)A = \frac{1}{2}h(b_1 + b_2), for hh. Multiply by 2: 2A=h(b1+b2)2A = h(b_1 + b_2). Then divide by (b1+b2)(b_1+b_2) to get h=2Ab1+b2h = \frac{2A}{b_1 + b_2}.

Explanation

To get a variable out of a fraction, first 'free' it by multiplying everything by the denominator. This clears the fraction and makes it much easier to isolate the variable you're looking for using standard algebra steps.

Book overview

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

Continue this chapter

Chapter 1: Equations

  1. Lesson 1

    Section 1.1: Solving Simple Equations

  2. Lesson 2

    Section 1.2: Solving Multi-Step Equations

  3. Lesson 3

    Section 1.3: Solving Equations with Variables on Both Sides

  4. Lesson 4Current

    Lesson 4: Rewriting Equations and Formulas