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Copyright © 2010 Hawkes Learning Systems.
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Hawkes Learning Systems:
College Algebra
Section 2.1b: Applications of Linear Equations
in One Variable
HAWKES LEARNING SYSTEMS
Copyright © 2010 Hawkes Learning Systems.
All rights reserved.
math courseware specialists
Objectives
o Solving Formulas for one variable.
o Distance and Interest problems.
HAWKES LEARNING SYSTEMS
math courseware specialists
Copyright © 2010 Hawkes Learning Systems.
All rights reserved.
Solving Formulas for One Variable
o One common task in applied mathematics is to solve
a given equation in two or more variables for one of
the variables.
o Solving for a variable means to transform the
equation into an equivalent one in which the
specified variable is isolated on one side.
o This is accomplished by the same methods we have
already used to solve equations in Section 2.1.
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Example: Solving Formulas
for One Variable
Solve the following equations for the specified variable.
P  2l  2w. Solve for w.
P  2l  2 w
P  2l  2 w
P  2l
w
2
P  2l
w
2
Step 1: subtract 2l from both
sides of the equation.
Step 2: divide by 2 on both sides
of the equation.
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Example: Solving Formulas
for One Variable
S  2 r 2  2 rh. Solve for h .
S  2 r  2 rh
2
S  2 r 2
h
2 r
S  2 r 2
h
2 r
HAWKES LEARNING SYSTEMS
Copyright © 2010 Hawkes Learning Systems.
All rights reserved.
math courseware specialists
Example: Solving Formulas
for One Variable
a) v = v0 + at ; Solve for a.
b)
; Solve for F.
c) A = P + Prt ; Solve for P.
HAWKES LEARNING SYSTEMS
math courseware specialists
Copyright © 2010 Hawkes Learning Systems.
All rights reserved.
Distance and Interest Problems
Good examples of linear equations arise from
certain distance and simple interest problems.
The basic distance formula is d  rt where d is
distance traveled at r rate for time t .
The simple interest formula is I  Prt where I is
the interest earned on principal P invested at
rate r for time t .
HAWKES LEARNING SYSTEMS
math courseware specialists
Copyright © 2010 Hawkes Learning Systems.
All rights reserved.
Example: Interest Problems
Sarah invested $8,000 in a global technology mutual fund which
had an annual return rate of 14%. What amount did she earn on
her investment after 1 year?
I  Prt
Given : P = $8000 r = 14% = 0.14 and t = 1
I = (8000)(0.14)(1)
I = 1120
She earned $1,120 in interest on her investment.
HAWKES LEARNING SYSTEMS
math courseware specialists
Copyright © 2010 Hawkes Learning Systems.
All rights reserved.
Example: Other Applications
a) The subtotal of items Jill bought is $18.40. If the
sales tax rate is 7%, what is the total she will pay?
b) Find three consecutive integers whose sum is 288.
c) Find three consecutive odd integers whose sum is
165.
HAWKES LEARNING SYSTEMS
math courseware specialists
Copyright © 2010 Hawkes Learning Systems.
All rights reserved.
Example: Distance Problems
Two trucks leave a warehouse at the same time. One travels due
west at an average speed of 61 miles per hour, and the other
travels due east at an average speed of 53 miles per hour. After
how many hours will the two trucks be 456 miles apart?
d1  d 2  456
Given.
r1  61
r2  53
d1  d 2  rt
1 1  r2t2
456  61t  53t
Plug values in.
456  114t
Combine like terms, and solve.
t  4 hours