ACTIVITY
Solving More Linear Equations
ACTIVITY 3
All Things Being Equal
Lesson 3-1 Rewriting Equations to Solve for a Variable
Learning Targets:
Solve formulas and equations that have more than one variable for a
• specified
variable.
a formula that has been solved for a specified variable to determine
• anUseunknown
quantity.
A formula usually contains more than one variable. You can solve a formula
for a specific variable using the same procedures you use for solving an
equation containing one variable.
In Activity 3, students extend properties
of equality and operations to solve
equations and formulas for a specified
variable. They will also begin to write
and solve absolute value equations.
My Notes
Lesson 3-1
PLAN
MATH TERMS
A formula is a literal equation, that
is, an equation written using
symbols that describes the
relationship between different
quantities.
Example A
The figure shown is made of a triangle and a square.
The area of the square is 36 square centimeters.
A formula for the area A of the figure is A = 3h + 36.
© 2017 College Board. All rights reserved.
© 2017 College Board. All rights reserved.
Step 2:
Example A
Check Your Understanding
Example B
#3
Check Your Understanding
Lesson Practice
Bell-Ringer Activity
Isolate the term that contains h by subtracting
36 from both sides.
3h + 36 = A
3h + 36 − 36 = A − 36
3h = A − 36
Pacing: 1 class period
Chunking the Lesson
TEACH
h
Solve the formula for h, the height of the triangle.
Step 1:
Guided
Activity Standards Focus
SUGGESTED LEARNING STRATEGIES: Identify a Subtask, Close
Reading, Work Backward, Create Representations, Discussion Groups
A formula describes how two or more quantities are related. Formulas are
important in many disciplines; geometry, physics, economics, sports, and
medicine are just a few examples of fields in which formulas are widely used.
3
POINTS OF INTEGRATION
Original equation
Subtraction Property of Equality
Simplify.
Isolate h by dividing both sides by 3.
3h = A − 36
Original equation
3h = A − 36
3
3
Division Property of Equality
Algebra and Geometry
When you use algebra to solve for
a variable in a geometry formula
that includes more than one
variable, you apply the same
reasoning you use to solve
equations with one variable. You
use inverse operations and
properties of equality.
A − 36 = h
Simplify.
3
Solution: h = A −36 , or h = A − 12
3
3
Ask students to name some of the formulas
they have learned in previous mathematics
or science classes and to explain what each
variable in the formulas represents.
Developing Math Language
This lesson contains the vocabulary term
formula. Use a familiar formula, such as
distance = rate × time to illustrate the
definition of the word. You may also want
to have students complete the Verbal and
Visual Word Association graphic organizer,
which allows them to describe uses of the
word formula that are familiar to them.
Have students add this term to their math
notebooks. Include the term on the class
Interactive Word Wall.
Try These A
POINT OF INTEGRATION
a. The pathways at a park form a rectangle with two diagonals. Each
diagonal has length l yards. A formula for the total distance D, in yards,
of all the pathways is D = 2l + 524. Solve this formula for l.
l = D − 524 , or l = D − 262
2
2
b. The formula for the circumference of a circle is C = 2πr. Solve this
formula for r.
r= C
2π
Activity 3 • Solving More Linear Equations
37
Common Core State Standards for Activity 3
HSA-SSE.A.1.B Interpret complicated expressions by viewing one or more of their parts as a single
entity.*
HSA-CED.A.4
Rearrange formulas to highlight a quantity of interest using the same reasoning as in
solving equations.
HSA-REI.A.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.
Algebra and Geometry
Have students name other familiar
geometry formulas that include two or
more variables. Include formulas that
use π, and remind students that π is not
a variable, but an irrational number. For
one or two of the formulas, work with
students to draw a diagram and label
the variables used in the formula.
Discuss the steps needed to solve for
one of the variables.
Example A Activating Prior Knowledge,
Identify a Subtask, Note Taking In this
example, students solve a formula for a
given variable. In Step 2 of the example,
students divide both sides of the equation
by 3 to isolate the variable. Be sure that
they understand that the entire expression
on the right side needs to be divided by 3,
not just the constant −36 or the variable A.
Remind students that an expression such
as A − 36 can also be written as A − 36 .
3
3
3
Activity 3 • Solving More Linear Equations
37
Debrief students’ answers to these items
to ensure that they can identify and
understand relationships between
variables in an equation with more than
one variable. Ask students to explain the
difference between solving an equation
for a variable and finding the solution of
an equation, as they did earlier in this
activity.
Lesson 3-1
Rewriting Equations to Solve for a Variable
ACTIVITY 3
continued
My Notes
Check Your Understanding
1. In the equation in Example A, 3h + 36 = A, which part of the
equation represents the area of the square? Which part represents the
area of the triangle? Explain.
2. Describe the similarities and differences between solving an equation
containing one variable and solving an equation for a variable.
Answers
1. 36 represents the area of the square.
The expression 3h represents the area
of the triangle. The area of the square
is 36 cm2, so its side length is 6 cm.
The triangle has a base of 6 cm, so its
area is 1 (6)h, or 3h cm2.
2
2. Answers may vary. Similarities: In
both cases, properties are used to
isolate a variable on one side of the
equation. Differences: The solution of
an equation in one variable is usually
a number (unless the equation has no
or infinitely many solutions), but
when a formula is solved for a
variable, the result is still an equation
containing more than one variable.
Example B Shared Reading, Marking
the Text, Close Reading, Work
Backward, Create Representations
In this example, students solve formulas
for a given variable and then use the
resulting equations to solve problems.
Suggest that in each step students circle
or highlight the variable they are solving
for. This may help them avoid confusion
and focus on the variable they want to
isolate. The last step in the example
involves applying the Symmetric
Property of Equality. Remind students
that this property states that if a = b,
then b = a. This property allows you to
switch the sides of an equation.
CONNECT TO PHYSICS
An object’s velocity is its speed in a
particular direction. Its acceleration
is the rate of change in velocity. If
an object has a positive
acceleration, it is speeding up; if it
has a negative acceleration, it is
slowing down.
Example B
The equation v = v0 + at gives the velocity in meters per second of an
object after t seconds, where v0 is the object’s initial velocity in meters per
second and a is its acceleration in meters per second squared.
a. Solve the equation for a.
b. Determine the acceleration for an object whose velocity after 15 seconds
is 25 meters per second and whose initial velocity was 15 meters per
second.
a.
Original equation
v = v0 + at
v − v0 = v0 − v0 + at
READING MATH
Sometimes a variable may include
a subscript. A subscript is a small
number or letter written to the
lower right of a variable. For
example, the variable v0 has the
subscript 0. The subscript 0 is often
read “nought.” (Nought is another
word for 0.) A variable with a
subscript of 0 usually indicates an
initial value. So, v0 indicates the
initial value of the velocity, or the
velocity when the time t = 0.
Subtraction Property of Equality:
Subtract v0 from both sides.
v − v0 = at
Combine like terms.
v − v0
Division Property of Equality: Divide
at
=
t
t
both sides by t.
v − v0
Simplify.
=a
t
v − v0
a=
Symmetric Property of Equality
t
b. To determine the acceleration for an object whose velocity after
15 seconds is 25 meters per second and whose initial velocity was
15 meters per second, substitute 25 for v, 15 for v0, and 15 for t.
a = 25 − 15 = 10 = 2 m/s2
15
15 3
Try These B
The equation t = 13p + 108 can be used to estimate the cooking time t in
minutes for a stuffed turkey that weighs p pounds. Solve the equation for p.
Then find the weight of a turkey that requires 285 minutes to cook.
p = t − 108 ; about 13.6 pounds
13
TEACHER to TEACHER
CLASSROOM-TESTED TIP
Be sure to point out the Reading
Math signal box on this page. It is
important for students to realize that
a variable with a subscript such as v0
represents a single variable, rather
than two separate variables or a
variable multiplied by a number.
38 SpringBoard® Integrated Mathematics I, Unit 1 • Linear Equations and Inequalities
Differentiating Instruction
Extend learning for students who are
fluent in solving an equation for a
variable by having them choose a
formula from their science textbook
and then solve the formula for each
variable.
38
SpringBoard® Integrated Mathematics I, Unit 1 • Linear Equations and Inequalities
© 2017 College Board. All rights reserved.
Check Your Understanding
© 2017 College Board. All rights reserved.
ACTIVITY 3 Continued
ACTIVITY 3 Continued
Lesson 3-1
Rewriting Equations to Solve for a Variable
ACTIVITY 3
continued
3. Reason abstractly. Solve for the indicated variable in each formula.
Name
Distance
Pressure
Kinetic
energy
Formula
My Notes
Solve for
d = rt, where d is the distance an object
r r = dt
travels, r is the average rate of speed, and
t is the time traveled
p = F , where p is the pressure on a surface,
A
F F = pA
F is the force applied, and A is the area of
the surface
k = 1 mv2, where k is the kinetic energy of
2
an object, m is its mass, and v is its
velocity
m m = 2k2
v
Gravitational U = mgh, where U is the gravitational
energy
energy of an object, m is its mass, g is the
h h= U
mg
acceleration due to gravity, and h is the
object’s height
Boyle’s Law p1V1 = p2V2, where p1 and V1 are the initial
pV
pressure and volume of a gas and p2 and V2
V V = p1 1
are the final pressure and volume of the gas 2 2
2
when the temperature is kept constant
ACADEMIC VOCABULARY
The word kinetic means related to
motion. It comes from a Greek
word that means to move.
Short-cycle formative assessment items
for Lesson 3-1 are also available in the
Assessment section on SpringBoard
Digital.
© 2017 College Board. All rights reserved.
© 2017 College Board. All rights reserved.
5. Why do you think being able to solve an equation for a variable would
be useful in certain situations?
Refer back to the graphic organizer the
class created when unpacking
Embedded Assessment 2. Ask students
to use the graphic organizer to identify
the concepts or skills they learned in this
lesson.
LESSON 3-1 PRACTICE
Solve each equation for the indicated variable.
6. V = Bh, for h
7. P = W , for W
t
W
8. P = , for t
9. lw + B = A, for w
t
10. Reason quantitatively. In baseball, the equation E = 9R gives a
I
pitcher’s earned run average E, where R is the number of earned runs
the player allowed and I is the number of innings pitched.
a. Solve the equation for I. State a property or provide an explanation
for each step.
b. Last season, a pitcher had an earned run average of 2.80 and allowed
70 earned runs. How many innings did the pitcher pitch last season?
ADAPT
Activity 3 • Solving More Linear Equations
s
w +1
5. Being able to solve an equation for a
variable allows you to write a formula
for any quantity. For example, if you
know that the perimeter of a
rectangle is given by P = 2(l + w),
you can write a formula for the
length of a rectangle l = P − w or
2
the width of a rectangle w = P − l .
2
4. c =
(
(
)
)
Verify that students understand
concepts related to formulas by
reviewing their answers to the Check
Your Understanding items. Ask students
to explain the steps they used to solve
the equation in Item 4 for the variable c.
Students’ answers to the Lesson Practice
items will provide a formative
assessment of their understanding of
solving an equation or formula for a
specified variable, and of students’
ability to apply their learning.
4. Solve the equation w + i = cs for c.
LESSON 3-1 PRACTICE
Check Your Understanding
ASSESS
Check Your Understanding
Answers
3 Work Backward, Create
Representations, Discussion
Groups In this item, students solve
formulas for a given variable. Monitor
group discussions of this item to ensure
that all members are participating. Point
out that students do not need to
understand the scientific vocabulary
used in the table when working with the
formulas, but encourage them to ask for
clarification of unfamiliar terms within
their group discussions.
39
6. h = V
B
7. W = Pt
8. t = W
P
9. w = A − B
I
Original equation
10. a. E = 9R
I
IE = 9R
Multiply both sides by I.
9
R
I=
Divide both sides by E.
E
b. 225 innings
Check students’ answers to the Lesson
Practice to ensure that they understand
how to solve equations for an indicated
variable. If students have not yet
mastered this skill, suggest substituting
numbers for each of the variables in the
equation with the exception of the
variable they are solving for. Students
can then solve the equation, recording
each step. They can use these steps as a
model to go back and solve the
equation.
See the Activity Practice on page 45 and
the Additional Unit Practice in the
Teacher Resources on SpringBoard
Digital for additional problems for this
lesson.
You may wish to use the Teacher
Assessment Builder on SpringBoard
Digital to create custom assessments or
additional practice.
Activity 3 • Solving More Linear Equations
39