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Solving Literal Equations for a Variable7-3

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Lesson PresentationLesson Presentation

Problem of the DayProblem of the Day

Lesson QuizzesLesson Quizzes


Warm UpSolve.

1. x + 14 = 27

2. 50 = 2x – 16

3. = 12

4. 6x = 72

13

33

48x4

12


Problem of the Day

Sandra has some pencils. Kylie has twice as many pencils as Sandra. Together they have 42 pencils. Write and solve an equation to find how many pencils each girl has.x + 2x = 42; Sandra: 14, Kylie: 28


MA.8.A.4.1 Solve literal equations for a specified variable.

Sunshine State Standards


Vocabularyliteral equation


A literal equation is an equation with two or more variables.


Solve each equation for the given variable.

A = lw for l

Additional Example 1A: Solving for Variables in Formulas

A = lw Divide both sides by w to isolate l.



I = prt for t

Additional Example 1B: Solving for Variables in Formulas

I = prt Divide both sides by pr to isolate t.



Check It Out: Example 1A

A = ℓw for w.

Aℓ

= ℓwℓ

Aℓ

= w



Check It Out: Example 1B

I = Prt for P.

Irt

= Prtrt

Irt

= P


The area of a rectangular table is 30 ft2. What is the width of the table if the length is 7.5 feet?

Additional Example 2: Application

A = lw

Divide both sides by l to isolate 1.

First solve the area formula for w because you want to find the width. Then use the given values to find w.


The area of a rectangular table is 30 ft2. What is the width of the table if the length is 7.5 feet?

Additional Example 2 Continued

w = 4

The width of the table is 4 feet.


The formula V = Bh gives the volume V of a rectangular prism with base area B and height h. The volume of a rectangular prism is 96 cm3. What is the height of the prism if base area is 12 cm?

Check It Out: Example 2

V = Bh

VB

= BhB

VB

= h

h =

9612

h = 8

VB

=

The height of the prism is 8 cm.


Additional Example 3A: Solving Literal Equations with Addition or Subtraction

Solve each equation for the given variable.c = d – e for d

Locate d in the equation.

+e + e

c + e = d

Add e to both sides of the equation.

c = d – e


Additional Example 3B: Solving Literal Equations with Addition or Subtraction

Solve each equation for the given variable.4x = y + 1 – 3x for y

Locate y in the equation.

–1 – 1

7x – 1 = y

Subtract 1 from both sides of the equation.

4x = y + 1 – 3x

4x – 1 = y – 3xAdd 3x to both sides of the

equation. +3x + 3x


Check It Out: Example 3A

Solve each equation for the given variable.3x – 2y = y + z for z

–y –y

3x – 3y = z

3x – 2y = y + z

z = 3x – 3y


Check It Out: Example 3B

Solve each equation for the given variable.4t + 2s = 2t + r – s for r

+s +s

4t + 3s = 2t + r

4t + 2s = 2t + r – s

2t + 3s = r–2t –2t


Additional Example 4: Solving Literal Equations for a Variable

The equation t = m + 10e gives the test score t for a student who answers m multiple-choice questions and e essay questions correctly. Solve this equation for e.

t = m + 10e Locate e in the equation.

Since m is added to 10e, subtract m from both sides.

t = m + 10e

–m –mt – m = 10e

t – m = 10e 10 10t – m = e 10

Since e is multiplied 10, divide both sides by 10.


Check It Out! Example 4

Solve 2k - 4d = r for k

2k – 4d = r

+ 4d +4d k = r + 4d2

2k2

= r + 4d2


Standard Lesson Quiz

Lesson Quizzes

Lesson Quiz for Student Response Systems


Lesson Quiz

1. Solve 3Y = F for Y.

2. Solve g = hk for h.

3. Solve 2b + 3c = d for b.

4. Solve 6 + 2m – n = 3n + 9 for m.

5. What would be the length of a rectangle if its

perimeter was 72 meters and its width was 16

meters?

20 m


1. Solve for D.

A.

B. D = V

C. D = mV

D.



2. Solve for m.

A.

B. D = Vm

C. DV = m

D.


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