Solving for a Variable - Dolfanescobar's Weblog | Middle ... · PDF fileSolving for a Variable Going Deeper Essential question: ... - 7 to x = -10? Divide -6 by 2 to get -3, then subtract

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1-6

Standards for Mathematical Content

A-CED.1.2 Create equations in two or more variables to represent relationships between quantities; . . . *

A-CED.1.4 Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations. . . . *

A-REI.1.1 Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method.

A-REI.2.3 Solve linear equations ... in one variable, including equations with coefficients represented by letters.

Vocabularyliteral equation

PrerequisitesSolving linear equations

Math BackgroundIn this lesson students will extend their skill at solving specific linear equations to solving entire classes of linear equations by replacing the numbers in an equation with letters. For instance, the equation 3x = 12 has the general form ax = b, which also represents other equations like 2x = -4 and -5x = 17. The general forms of equations are called literal equations. (The word literal refers to letter.) Restrictions must be placed on the letters used for the coefficients of a variable in a literal equation. One common restriction is that the coefficient of a variable cannot be 0; otherwise, the variable term drops out of the equation. Students will then extend this knowledge to solving and rewriting formulas.

When solving literal equations and formulas, students will follow the same steps as when solving specific equations. Once they have solved a literal equation, they should recognize that it represents the solutions of all members of that class of equations. For instance, x = b __ a   gives the solution of all equations of the form ax = b. Students just need to substitute the values of a and b into x = b __ a   and simplify to obtain the solution of a specific equation like 3x = 12. This is the idea behind the quadratic formula, which gives the solution of the literal equation ax2 + bx + c = 0. It is also the idea behind calculator or computer programs that allow you to enter an equation or inequality having a particular form and obtain the solution automatically.

Show the following graphic organizer to help students see that solving literal equations uses the same steps as solving non-literal linear equations. Leave the steps on the right side blank until the appropriate part of the lesson.

INTRODUCE

Solving for a VariableGoing DeeperEssential question: How do you solve literal equations and rewrite formulas?

4x + 3 = 15

4x =

ax + b = c

ax  = 

4x _____    = 12 ____        ax ____  = c - b _____   

x =       x  =

-3 -3 -b

a a

-b

c - b

c - b ____ a   

12

4

3

4

Chapter 1 33 Lesson 6

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1b. Choose one of the other specific equations from part A. Show that the solution of

the literal equation gives the solution of the specific equation.

When you solve a literal equation, you use properties of equality and other properties

to isolate the variable. The result is not a number, but rather an expression involving the

letters that represent the coefficients and constants.

Solving a Literal Equation and Evaluating Its Solution

Solve the literal equation a(x + b) = c where a ≠ 0. Then use the literal equation’s solution to obtain the solution of the specific equation 2(x + 7) = -6.

A Solve a(x + b) = c for x. Use the properties of equality to justify your solution steps.

a(x + b) = c Original equation

x + b = __ Property of Equality

x = __ - Property of Equality

B Obtain the solution of 2(x + 7) = -6 from the literal equation’s solution by letting

a = 2, b = 7, and c = -6.

x = __ - Write the literal equation’s solution.

x = ___ - Substitute 2 for a, 7 for b, and -6 for c.

x = Simplify.

REFLECT

2a. When solving a(x + b) = c, why do you divide by a before you subtract b?

2b. Write an equation that has the form a(x + b) = c. Find the solution of your equation

using the literal equation’s solution.

E X A M P L E2A-REI.2.3

c

cb

a

a

The order of operations for a(x + b) tells you that b is added to x before the

sum is multiplied by a. When undoing these operations, you perform the inverse

operations in reverse order: dividing by a before subtracting b.

For -2x + 5 = 11, x = c - b ____ a = 11 - 5 _____ -2

= -3.

For 4x + 3 = -1, x = c - b ____ a = -1 - 3 _____ 4 = -1.

Answers will vary. Sample answer: -3(x + 4) = 6; x = 6 ___ -3

- 4 = -6

c

-6

b

7

a

2

-10

Division

Subtraction


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Name Class Date

Understanding Literal Equations

A For each equation given below, solve the equation by writing two equivalent

equations: one where the x-term is isolated and then one where x is isolated.

3x + 1 = 7 -2x + 5 = 11 4x + 3 = -1

B Identify the two properties of equality that you used in part A. List them in the

order that you used them.

Each equation in part A has the general form ax + b = c where a ≠ 0. Solve this

literal equation for x using the properties of equality that you identified in part B.

ax + b = c Write the literal equation.

ax = Subtract b from both sides.

x = - ______

Divide both sides by a.

Show that the solution of the literal equation gives the same solution of 3x + 1 = 7

as you found in part A. Recognize that when a = 3, b = 1, and c = 7, the literal

equation ax + b = c gives the specific equation 3x + 1 = 7.

x = - ______

Write the literal equation’s solution.

x = - ______

Substitute 3 for a, 1 for b, and 7 for c.

x = Simplify.

REFLECT

1a. Why must the restriction a ≠ 0 be placed on the literal equation ax + b = c ?

E X P L O R E1

C

D

Solving for a VariableGoing DeeperEssential question: How do you solve literal equations and rewrite formulas?

A literal equation is an equation in which the coefficients and

constants have been replaced by letters. In the following Explore,

you will see how a literal equation can be used to represent specific

equations having the same form.

1-6

-

-

A-REI.2.3

Subtraction Property of Equality; Division Property of Equality

3x = 6

x = 2

-2x = 6

x = -3

4x = -4

x = -1

cc

c

7

bb

b

1

a

a

3

2

The equation would have no variable term if a = 0.


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Notes


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Questioning Strategies• Could you use other letters besides a, b, and c to

write the literal equation? If so, give an example.

Yes; you could write the literal equation as mx + n = p, for instance.

• How does the solution of the literal equation

differ from the solution of a specific equation of

that form? Unlike a specific equation, the literal equation’s solution is not a number; it is an expression that involves the letters representing the coefficients and constants in the literal equation.

• How is the solution of a literal equation useful? It provides a rule for calculating the solution of any specific equation having the form of the literal equation; there is no need to go through the steps of solving the specific equation.

Teaching StrategyAfter completing the Explore, have students work

in pairs. One student should write an equation of

the form ax + b = c, share it with his or her partner,

and then solve it on his or her own. The partner

should determine the solution of the equation from

the general solution x = c - b _____ a . The partners should

then compare their two solutions to make sure

they are the same. (If not, they should check each

other’s work to see where a mistake was made.)

The partners can then switch roles and repeat the

process.

Questioning Strategies• In part A, is it possible to use the Multiplication

Property of Equality rather than the Subtraction

Property of Equality as the first step of solving

a(x + b) = c? If so, how? Yes; you could multiply both sides by the reciprocal of a.

• In part A, is it possible to use the Addition

Property of Equality rather than the Division

Property of Equality as the second step of solving

a(x + b) = c? If so, how? Yes; you could add the opposite of b to both sides.

• In part B, what mental calculations do you

perform in going from x = -6 ___

2 - 7 to x = -10?

Divide -6 by 2 to get -3, then subtract 7 from -3 to get -10.

Avoid Common ErrorsBecause the letters in the solution of a literal

equation appear in a different order than they do

in the equation, students may substitute the wrong

values for the letters. To slow down the process of

substituting values for so many letters, have them

work with one letter at a time, as shown below

when using x = c __ a - b to solve 2(x + 7) = -6.

EXTRA EXAMPLESolve the literal equation x - a _____

b = c where b ≠ 0.

Then use the literal equation’s solution to obtain

the solution of x - 3 _____

2 = -4. x = bc + a;

x = 2(-4) + 3 = -5

Questioning Strategies• In part A, what would you get if you solve for

w instead of h? w = A __ lh

• How can you tell whether a rewritten formula

is correct? Choose values for the variables that make the original formula a true statement. Then substitute those values in the rewritten formula to see if it is a true statement.

EXTRA EXAMPLE Solve the formula for the given variable.

A. The formula I = prt gives the simple interest on

a savings account with principal p at a decimal

rate r for t years. Solve for r to find the rate of a

savings account with a given interest, principal,

and time.

r = I __ pt

B. The formula A = 1 __ 2

d 1 d

2 gives the area of a

rhombus with diagonal lengths d 1 and d

2 . Solve

for d 1 to find the length of one diagonal with

a given area and known length of the other

diagonal.

d 1 = 2A __ d 2

TEACH

EXPLORE1

EXAMPLE2

EXAMPLE3

Step Result

Identify the value of a. 2

Substitute the value of a. x = c __ 2 - b

Identify the value of b. 7

Substitute the value of b. x = c __ 2 - 7

Identify the value of c. -6

Substitute the value of c. x = -6 ___ 2 - 7

continued


x

x

x

78

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REFLECT

3a. The formula T = p + sp gives the total cost of an item with price p and sales tax s,

expressed as a decimal. Describe a situation in which you would want to solve the

formula for s.

3b. What is true about the restrictions on the value of a variable in a formula that

might not be true of other literal equations?

Writing and Rearranging a Formula

The flower garden at the right is made up of a square and an isosceles triangle. Write a formula for the perimeter P in terms of x, and then solve for x to find a formula for the side length of the square in terms of P.

A Write a formula for the perimeter of each shape. Use only the

sides of the square and the triangle that form the outer edges

of the figure.

Perimeter of square =

Perimeter of triangle =

B Combine the formulas. P =

C Solve the formula for x.

P =

P = x ( + )

P = x ( ____ )

P ( ____ ) = x ( ____ ) ( ____ )

____ = x

E X A M P L E4A-CED.1.4

Write the combined formula.

Distributive Property

Find the sum. Write the result as an improper fraction.

Multiplication Property of Equality

Simplify.

You know the total cost and the item price, and want to find the sales tax rate.

If the variable represents a measurable quantity such as length or volume, then

the value cannot be negative or 0.

3x

7 _ 4 x

3x + 7 _ 4 x

3x + 7 _ 4 x

7 _ 4 3

19

19

4

19

4 19

4 19

4

4P


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2c. Another way to solve a(x + b) = c is to start by using the Distributive Property.

Show and justify the solution steps using this method.

2d. When you start solving a(x + b) = c by dividing by a, you get x = c _ a

- b. When you

start solving a(x + b) = c by distributing a, you get x = c - ab _____ a . Use the fact that you

can rewrite c - ab _____ a as the difference of two fractions to show that the two solutions

are equivalent.

Solving a Formula for a Variable

Solve the formula for the given variable. Justify each step in your solution.

A The formula V = lwh gives the volume of a rectangular prism with length l, width w, and height h. Solve the formula for h to find the height of a rectangular

prism with a given volume, length, and width.

V = lwh Original equation

V ______ = lwh ______

h = ______ Simplify.

B The formula E = 1 _ 2 k x 2 gives the potential energy E of a spring with spring

constant k that has been stretched by length x. Solve the formula for k to find the

constant of a spring with a given potential energy and stretch.

E = 1 __ 2

kx 2 Original equation

· E = · 1 __ 2

kx 2

2E = kx 2

2E ____ = kx 2 ____

k = ______ Simplify.

E X A M P L E3A-CED.1.4

a(x + b) = c Write the literal equation.

ax + ab = c Distributive Property

ax = c - ab Subtraction Property of Equality

x = c - ab _____ a Division Property of Equality

c - ab _____ a = c _ a - ab __ a = c _ a - b

lw

lw

V

lwDivision Property of Equality

Multiplication Property of Equality

Division Property of Equality

Simplify.

2 2

x 2 x 2

2E

x 2


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Notes


x

x

x

x

x14

x14

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continued

Teaching StrategiesStudents may become confused when working with

formulas with three or more variables. You may

wish to begin with pairs of formulas with only two

variables, such as formulas for converting units of

measurement.

Converting feet and inches: i = 12f; f = 1 __ 12

i

Converting Fahrenheit and Celsius temperatures:

F = 9 __ 5

C - 32; C = 5 __ 9

(F + 32)

Questioning Strategies• In part A, why does the formula have the term 3x

instead of 4x? The perimeter of the figure only includes the outside lengths. The fourth side of the square, represented by the dashed line, is not part of the perimeter of the figure.

EXTRA EXAMPLEThe figure shown below is made up of a square and

a rectangle. Write a formula for the perimeter P in

terms of x, and then solve for x to find a formula for

the side length of the square in terms of P.

P = 4x + 1 __ 2 x; 2P ___ 9 = x

Essential QuestionHow do you solve literal equations and rewrite formulas?To solve literal equations and rewrite formulas, you must isolate the given variable on one side of the equation or formula, using the same steps and properties as when solving specific equations. You must also pay attention to any restrictions on the coefficient(s) of the variables.

SummarizeHave students write a journal entry in which they

summarize the steps for solving a literal equation

and then use the literal solution to obtain the

solution of a specific equation. Have students

give their own examples and challenge them

to determine appropriate restrictions on the

coefficient(s) of the variable in the equation.

Where skills are taught

Where skills are practiced

EXAMPLE2 EXS. 1–3

EXAMPLE3 EXS. 4, 5

EXAMPLE4 EXS. 6, 7

EXAMPLE3

EXAMPLE4

CLOSE

PRACTICE

Highlighting the Standards

The journal entry provides a connection to

Mathematical Practice Standard 6 (Attend

to precision). Placing a restriction on the

coefficient(s) of the variable in a literal

equation is an essential part of generalizing a

class of equations because it guarantees that

the literal equation will have a variable term.

Highlighting the Standards

Reflect Question 4b, rewriting a formula for

the area of a square in terms of its perimeter,

provides a connection to Mathematical

Practice Standard 4 (Model with mathematics).

If the figure shown in EXAMPLE4 models a

piece of land and the perimeter is the length

of a fence enclosing the land, then the area of

the square piece can be found in terms of the

perimeter of the entire piece of land.


x

x

12

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6. An electrician sent Bonnie an invoice in the amount of a dollars for 6 hours

of work that was done on Saturday. The electrician charges a weekend fee f in

addition to an hourly rate r. Bonnie knows what the weekend fee is. Write a

formula Bonnie can use to find r, the rate the electrician charges per hour.

7. The swimming pool below is made up of a square and two semicircles. Write a

formula for the perimeter P in terms of x, and then solve for x to find a formula

for the side length of the square in terms of P.

4. Formula for the surface area of a rectangular

prism: SA = 2(lw + hw + hl), for w

5. Formula for the area of a

trapezoid: A = 1 __ 2

(a + b)h, for b

Solve each formula for the indicated variable.

SA = 2(lw + hw + hl )

SA - 2hl = 2lw + 2hw + 2hl - 2hl

SA - 2hl = 2lw + 2hw

SA - 2hl = w(2l + 2h)

SA - 2hl ______ 2l + 2h

= w(2l + 2h) ________

2l + 2h

SA - 2hl ______ 2l + 2h

= w

A = 1 _ 2 (a + b)h

A __ h =

1 _ 2 (a + b)h

_______ h

A __ h = 1 _

2 (a + b)

2 · A __ h = 2 · 1 _

2 (a + b)

2A __ h = a + b

2A __ h - a = a + b - a

2A __ h - a = b

a = f + 6r

a - f = 6r

P = 2x + 2π ( 1 _ 2 x )

P = 2x + πx

P = x ( 2 + π )

P ____ 2 + π

= x ( 2 + π )

______ 2 + π

P ____ 2 + π

= x

a - f ____ 6 = 6r __

6

a - f ____ 6 = r


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REFLECT

4a. What are the restrictions on the values of P and x? Explain.

4b. How could you write a formula for the area of the square in terms of P?

P R A C T I C E

1. Show and justify the steps for solving x + a = b. Then use the literal equation’s

solution to obtain the solution of x + 2 = -4.

2. Show and justify the steps for solving ax = b where a ≠ 0. Then use the literal

equation’s solution to obtain the solution of 3x = -15.

3. Show and justify the steps for solving ax = bx + c where a ≠ b. Then use the literal

equation’s solution to obtain the solution of 2x = x + 7.

P > 0, x > 0; the variables represent lengths, which cannot be negative or 0.

x + a = b Original equation

x = b - a Subtraction Property of Equality

Substitute 2 for a and -4 for b to get the solution of x + a = b: x = 4 - (-2) = 6.

ax = b Original equation

x = b _ a Division Property of Equality

Substitute 3 for a and -15 for b to get the solution of 3x = -15: x = -15 ___ 3 = -5.

The area is A = x 2 . Substitute 4P __ 19

for x to get ( 4P __ 19

) 2 = 16 P 2 ____

361 . The formula is A = 16 P 2 ____

361 .

ax = bx + c Original equation

ax - bx = c Subtraction Property of Equality

(a - b)x = c Distributive Property

x = c ____ a - b

Division Property of Equality

Substitute 2 for a, 1 for b, and 7 for c to get the solution of

2x = x + 7: x = 7 ____ 2 - 1

= 7.


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Notes


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Assign these pages to help your students practice

and apply important lesson concepts. For

additional exercises, see the Student Edition.

Answers

Additional Practice

1. r = C _

2π 2. m =

y − b _ x

3. c = d _ 4

4. n = 8 + 6m

5. p = q − 5r

_ 2

6. x = −10 − z _ y

7. b = a _ c 8. j = h − 4

_ k

9. a. p = c − 215

_ 5

10. a. b = 2A _ h

b. 17 b. 32 mm

Problem Solving

1. r = d _ t 2. 9.1 m/s

3. 8.5 m/s 4. 0.4 m/s

5. B 6. F

7. D 8. J

ADDITIONAL PRACTICE AND PROBLEM SOLVING


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Problem Solving


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1-6Name Class Date

Additional Practice


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Notes


Documents

Solving for a Variable - Dolfanescobar's Weblog | Middle ... · PDF fileSolving for a Variable Going Deeper Essential question: ... - 7 to x = -10? Divide -6 by 2 to get -3, then subtract