Learn on PengiCalifornia Reveal Math, Algebra 1Unit 1: Using Expressions and Equations

1-5 Using Formulas

In this Grade 9 lesson from California Reveal Math, Algebra 1, students learn how to solve formulas for a specific variable by applying algebraic properties such as factoring and dividing by a quantity, using examples like the trapezoid area formula and multi-variable equations. The lesson also introduces dimensional analysis as a method for converting units of measurement. Students practice applying rearranged formulas to real-world contexts, including calculating geometric dimensions and modeling costs with linear equations.

Section 1

Factoring a Variable from Multiple Terms to Isolate It

Property

When the variable you want to isolate appears in more than one term, factor it out first, then divide both sides by the remaining expression.

If pp appears in two terms, rewrite as:

ap+bp=p(a+b)ap + bp = p(a + b)

Section 2

Solving Formulas with Fractions

Property

To solve a formula for a specific variable, first clear any fractions by multiplying the entire equation by the LCD. Next, gather all terms containing the desired variable on one side of the equation. If there are multiple terms with the variable, factor the variable out, and then divide both sides by the remaining factor to isolate it.

Examples

  • Solve the formula S=a1rS = \frac{a}{1-r} for rr. Multiply by 1r1-r to get
    S(1r)=aS(1-r) = a
    . Distribute SS to get
    SSr=aS - Sr = a
    . Then,
    Sr=aS-Sr = a - S
    , so
    r=aSSr = \frac{a-S}{-S}
    or
    r=SaSr = \frac{S-a}{S}
    .
  • Solve 1f=1u+1v\frac{1}{f} = \frac{1}{u} + \frac{1}{v} for uu. The LCD is fuvfuv. Multiply by the LCD to get
    uv=fv+fuuv = fv + fu
    . Move terms with uu to one side:
    uvfu=fvuv - fu = fv
    . Factor out uu:
    u(vf)=fvu(v-f) = fv
    . The solution is
    u=fvvfu = \frac{fv}{v-f}
    .
  • Solve h=2Ab1+b2h = \frac{2A}{b_1 + b_2} for b1b_1. Multiply by b1+b2b_1+b_2 to get
    h(b1+b2)=2Ah(b_1+b_2) = 2A
    . Distribute hh:
    hb1+hb2=2Ahb_1 + hb_2 = 2A
    . Then
    hb1=2Ahb2hb_1 = 2A - hb_2
    , and
    b1=2Ahb2hb_1 = \frac{2A - hb_2}{h}
    .

Explanation

Rearranging formulas is like solving a puzzle. First, get rid of fractions. Then, herd all the pieces with your target variable to one side. If it appears in multiple terms, factor it out to isolate it.

Section 3

Dimensional Analysis for Unit Conversion

Property

Dimensional analysis is a method to convert units by multiplying the original quantity by a series of unit fractions.
A unit fraction is a ratio of equivalent measurements that equals 1 (e.g., 1 foot12 inches=1\frac{1 \ \operatorname{foot}}{12 \ \operatorname{inches}} = 1).
This process cancels out unwanted units until only the desired units remain. This method relies on two key properties:

  • For any nonzero number nn, nn=1\frac{n}{n} = 1.
  • Any number nn multiplied by 1 is still nn: n1=nn \cdot 1 = n.

Examples

  • To convert 7000 pounds to tons, use the unit fraction relating tons and pounds: 7000 lbs×1 ton2000 lbs=70002000 tons=3.5 tons7000 \ \operatorname{lbs} \times \frac{1 \ \operatorname{ton}}{2000 \ \operatorname{lbs}} = \frac{7000}{2000} \ \operatorname{tons} = 3.5 \ \operatorname{tons}.
  • To convert 2.5 hours into seconds, use a chain of unit fractions: 2.5 hours×60 minutes1 hour×60 seconds1 minute=2.5×60×60 seconds=9000 seconds2.5 \ \operatorname{hours} \times \frac{60 \ \operatorname{minutes}}{1 \ \operatorname{hour}} \times \frac{60 \ \operatorname{seconds}}{1 \ \operatorname{minute}} = 2.5 \times 60 \times 60 \ \operatorname{seconds} = 9000 \ \operatorname{seconds}.
  • How many kilograms are in 55 pounds? Use the conversion 1 lb = 0.454 kg: 55 lbs×0.454 kg1 lb=55×0.454 kg=24.97 kg55 \ \operatorname{lbs} \times \frac{0.454 \ \operatorname{kg}}{1 \ \operatorname{lb}} = 55 \times 0.454 \ \operatorname{kg} = 24.97 \ \operatorname{kg}.

Explanation

This is a powerful technique for converting any unit. You create a chain of fractions where the unit names you want to eliminate cancel out from the numerator and denominator, leaving you with only the units you need for your answer.

Book overview

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Continue this chapter

Unit 1: Using Expressions and Equations

  1. Lesson 1

    1-1 Properties of Real Numbers

  2. Lesson 2

    1-2 Descriptive Modeling with Expressions

  3. Lesson 3

    1-3 Solving Equations in One Variable

  4. Lesson 4

    1-4 Absolute Value Expressions and Equations

  5. Lesson 5Current

    1-5 Using Formulas

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Factoring a Variable from Multiple Terms to Isolate It

Property

When the variable you want to isolate appears in more than one term, factor it out first, then divide both sides by the remaining expression.

If pp appears in two terms, rewrite as:

ap+bp=p(a+b)ap + bp = p(a + b)

Section 2

Solving Formulas with Fractions

Property

To solve a formula for a specific variable, first clear any fractions by multiplying the entire equation by the LCD. Next, gather all terms containing the desired variable on one side of the equation. If there are multiple terms with the variable, factor the variable out, and then divide both sides by the remaining factor to isolate it.

Examples

  • Solve the formula S=a1rS = \frac{a}{1-r} for rr. Multiply by 1r1-r to get
    S(1r)=aS(1-r) = a
    . Distribute SS to get
    SSr=aS - Sr = a
    . Then,
    Sr=aS-Sr = a - S
    , so
    r=aSSr = \frac{a-S}{-S}
    or
    r=SaSr = \frac{S-a}{S}
    .
  • Solve 1f=1u+1v\frac{1}{f} = \frac{1}{u} + \frac{1}{v} for uu. The LCD is fuvfuv. Multiply by the LCD to get
    uv=fv+fuuv = fv + fu
    . Move terms with uu to one side:
    uvfu=fvuv - fu = fv
    . Factor out uu:
    u(vf)=fvu(v-f) = fv
    . The solution is
    u=fvvfu = \frac{fv}{v-f}
    .
  • Solve h=2Ab1+b2h = \frac{2A}{b_1 + b_2} for b1b_1. Multiply by b1+b2b_1+b_2 to get
    h(b1+b2)=2Ah(b_1+b_2) = 2A
    . Distribute hh:
    hb1+hb2=2Ahb_1 + hb_2 = 2A
    . Then
    hb1=2Ahb2hb_1 = 2A - hb_2
    , and
    b1=2Ahb2hb_1 = \frac{2A - hb_2}{h}
    .

Explanation

Rearranging formulas is like solving a puzzle. First, get rid of fractions. Then, herd all the pieces with your target variable to one side. If it appears in multiple terms, factor it out to isolate it.

Section 3

Dimensional Analysis for Unit Conversion

Property

Dimensional analysis is a method to convert units by multiplying the original quantity by a series of unit fractions.
A unit fraction is a ratio of equivalent measurements that equals 1 (e.g., 1 foot12 inches=1\frac{1 \ \operatorname{foot}}{12 \ \operatorname{inches}} = 1).
This process cancels out unwanted units until only the desired units remain. This method relies on two key properties:

  • For any nonzero number nn, nn=1\frac{n}{n} = 1.
  • Any number nn multiplied by 1 is still nn: n1=nn \cdot 1 = n.

Examples

  • To convert 7000 pounds to tons, use the unit fraction relating tons and pounds: 7000 lbs×1 ton2000 lbs=70002000 tons=3.5 tons7000 \ \operatorname{lbs} \times \frac{1 \ \operatorname{ton}}{2000 \ \operatorname{lbs}} = \frac{7000}{2000} \ \operatorname{tons} = 3.5 \ \operatorname{tons}.
  • To convert 2.5 hours into seconds, use a chain of unit fractions: 2.5 hours×60 minutes1 hour×60 seconds1 minute=2.5×60×60 seconds=9000 seconds2.5 \ \operatorname{hours} \times \frac{60 \ \operatorname{minutes}}{1 \ \operatorname{hour}} \times \frac{60 \ \operatorname{seconds}}{1 \ \operatorname{minute}} = 2.5 \times 60 \times 60 \ \operatorname{seconds} = 9000 \ \operatorname{seconds}.
  • How many kilograms are in 55 pounds? Use the conversion 1 lb = 0.454 kg: 55 lbs×0.454 kg1 lb=55×0.454 kg=24.97 kg55 \ \operatorname{lbs} \times \frac{0.454 \ \operatorname{kg}}{1 \ \operatorname{lb}} = 55 \times 0.454 \ \operatorname{kg} = 24.97 \ \operatorname{kg}.

Explanation

This is a powerful technique for converting any unit. You create a chain of fractions where the unit names you want to eliminate cancel out from the numerator and denominator, leaving you with only the units you need for your answer.

Book overview

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

Continue this chapter

Unit 1: Using Expressions and Equations

  1. Lesson 1

    1-1 Properties of Real Numbers

  2. Lesson 2

    1-2 Descriptive Modeling with Expressions

  3. Lesson 3

    1-3 Solving Equations in One Variable

  4. Lesson 4

    1-4 Absolute Value Expressions and Equations

  5. Lesson 5Current

    1-5 Using Formulas