Download General Methodology for Solving Equations

Section 2.4
Pre-Activity
Preparation
General Methodology for Solving Equations
Catering
A catering company charges $26.95 per guest with an additional set up
fee of $500. How many guests can be invited if the budget is $5000?
Construction
The cost of painting a 3 bedroom house includes indoor paint at $19.25
per gallon and outdoor paint at $27.99 per gallon. If 12 gallons of
indoor paint are required, how many gallons of outdoor paint may be
purchased, if the total budget for paint is $600?
Forensic Science
The approximate time of death* can be determined by body temperature with the formula:
T=
B-N
1.5
T = time since death
B = body temperature at time of death
N = body temperature at present
If a body is found 3 hours after death, what should the temperature be if the victim was in good health at
the time of death?
* The formula allows us to calculate the time since death; to find the time of death, we simply need to know the current time: if
T = 5 hours and it is 12:00 p.m. now, we know the time of death was 7:00 a.m. (5 hours before the current time).
Repair
A repair company charges $35 per hour plus a travel fee of $50. How many hours did the repair person
work if the total charge was $190?
Each of the above situations can be figured out through trial and error, or “in your head,” but a quicker, more
direct method is to use algebra. Each of the above situations can be modeled into an algebraic equation and
solved for the unknown quantity. This section provides you with the algebra skills to solve each of the these
situations.
Learning Objectives
• Be able to clear an equation of fractions or decimals
• Find the solution to any given algebraic equation in one variable
• Solve for a variable in a formula in terms of the other variables
Terminology
New Terms
Previously Used
clear parentheses
solve
formula
substitute
LCD
variable
to
Learn
clear fractions
143
Chapter 2 — Solving Equations
144
Building Mathematical Language
Solving Equations: x = a
One of the main objectives of algebra courses is for students to become proficient in solving equations.
Solve equations by applying the properties of equations and real numbers in successive steps until the
desired variable is isolated on one side of the equal sign and its coefficient is 1. That is, the equation is
written in the form of x = a. For algebraic equations, a will be a constant.
For formulas, a can be an expression in terms
of the other variables, such as in the equation:
for the unknown side
of a right triangle
a = c2 - b2
Clearing Fractions
c
b
a
Equations with fractions can be manipulated so that the fractions divide or multiply out, leaving all
coefficients and constants as whole numbers. Using the multiplication property of equality, multiply both
sides of an equation (with fractions) by the least common denominator (LCD) of the fractions to clear
them.
Properties
and
Principles
In order to successfully work with the equations in this section, you should remember
and be able to use the Addition, Multiplication, and Distributive Properties of Equality, all of which have
already been introduced.
Multiplication Property Review
if a = b then a : c = b : c
if x = 3, then 2x = 6
Remember that dividing is also accepted under this property, because division is multiplication by the
reciprocal. Therefore:
if a = b then
a b
=
c c
if 5x = 15, then
5 x 15
=
5
5
5 x 15
=
5
5
x=3
3
Section 2.4 — General Methodology for Solving Equations
145
Methodologies
Clearing Fractions from an Equation
The example equations in this methodology have fractions as constants. Before isolating the variable, the
fractions can be cleared by multiplying both sides of the equation by the LCD.
►
Example 1: Isolate the variable in the equation:
1
4
+ 3x =
2
5
►
Example 2: Isolate the variable in the equation:
1
2
+ 2x =
3
7
Steps in the Methodology
Step 1
Find the LCD
of all the
fractions in
the equation.
Use the methodology
for finding the LCD
from Foundations of
Mathematics.
Example 1
Try It!
Example 2
The fractions are
1
4
and .
2
5
The LCM of 2 and 5 is
their product: 10.
2 × 5 = 10
Step 2
Multiply both
sides of the
equation by
the LCD.
Step 3
Distribute the
LCD to each
term.
Apply the multiplication
property of equations to
multiply both sides of the
equation by the LCD.
Apply the Distributive
Property twice, once
on the left side of the
equation and once on the
right side of the equation.
The result is that each
term in the equation is
multiplied by the LCD.
1

4
10  + 3 x  = 10  
2

5
10 :
1
4
+ 10 : 3 x = 10 :
2
5
continued on next page
Chapter 2 — Solving Equations
146
Steps in the Methodology
Step 4
Simplify
so that the
fractions
clear to whole
numbers.
Notice that the fractions
clear by design; you chose
the LCD for that purpose.
Example 1
5
10 :
Example 2
2
1
4
+ 10 : 3 x = 10 :
2
5
5 + 30x = 8
or
10
40
+ 30 x =
2
5
5 + 30x = 8
Step 5
Validate.
Because each term was
multiplied by the LCD
(10, in this case) to arrive
at the final expression,
dividing each term by the
LCD should result in the
initial expression.
5 30 x 8
+
=
10 10 10
5
1
3
30 x
8
4
+
= 5
2
1
10
10
10
1
4
+ 3x =
2
5 
Methodologies
Congratulations! Putting together what you have learned from Section 1.1 through Section 2.4, you have
practiced each step of the General Methodology for Solving Equations:
Step 1
Determine the variable.
Step 2 Clear parentheses.
Step 3 Combine like terms.
Step 4 Clear fractions.
Step 5
Isolate the chosen variable.
Step 6
Multiply or divide to make the chosen variable coefficient equal to 1.
Step 7
Validate.
The remaining models in this section will make use of this general methodology to solve a variety of different
equations.
Section 2.4 — General Methodology for Solving Equations
147
Model 1
Solve: 5x – (2x + 3) = x – 4
Step 1
Determine the variable.
x is the variable; it appears twice on the left side and once on
the right side.
Step 2
Clear parentheses.
5x – 1(2x + 3) = x – 4
5x – 2x – 3 = x – 4
Step 3
Combine like terms.
3x – 3 = x – 4
Step 4
Clear fractions.
No fractions; omit this step.
Step 5
Isolate the variable.
3x – 3 – x = x – x – 4
2x – 3 = –4
2x – 3 + 3 = –4 + 3
2x = –1
Step 6
Divide.
2 x -1
=
2
2
1
Answer: x = 2
Step 7
Validate.
5 x – (2 x + 3) = x – 4
 1   1  ?  1
5  −  −  2  −  + 3 =  −  − 4
 2   2   2
5  2  
9
− −  −  + 3 =? −
2  2  
2
5 4? 9
− − =−
2 2
2
9
9
− =−

2
2
???
Why can we do this?
???
Why can we do this?
Sometimes it minimizes the number of steps required to keep numbers as
fractions rather than write them as whole numbers.
2
is equal to –1, but notice that 2 is the common denominator,
2
and each term on the left will eventually need to be written as a fraction so that it can be combined and
Here, inside the parentheses, the expression compared to the fraction on the right.
It actually adds a step to rewrite -
2
2
6
as –1 rather than mentally combine - with + .
2
2
2
Chapter 2 — Solving Equations
148
Model 2
Solve:
3
( x - 6) = 2( x -1)
4
Step 1
Determine the variable.
The variable to solve for is x and it appears on both sides of
the equal sign.
Step 2
Clear parentheses.
3
( x - 6) = 2( x -1)
4
3
18
x - = 2x - 2
4
4
Step 3
Combine like terms.
No terms to combine.
Step 4
Clear fractions.
LCD = 4
18 
3
4  x −  = 4(2 x − 2)
4
4
?
1
4 :3
1
4 : 18
x−
1
1
4
4
3 x − 18 = 8 x − 8
= 8x − 8
Step 5
Isolate the variable.
3 x - 8 x -18 = 8 x - 8 x - 8
-5 x -18 +18 = -8 +18
-5 x = 10
Step 6
Divide.
−5 x 10
=
−5 −5
Answer: x = –2
Step 7
Validate.
3
[(-2) - 6]=? 2 [(-2) -1]
4
2 ?
3
(- 8 ) = 2(-3)
1
4
-6 = -6 
Model 3
Solve for w: A = 2l + 2w
Step 1
Step 2
Step 3
Step 4
Determine the variable.
Clear parentheses.
Combine like terms.
Clear fractions.
w is the variable; it appears once on the right side.
No parentheses; omit this step.
No like terms to combine; omit this step.
No fractions; omit this step.
Section 2.4 — General Methodology for Solving Equations
Step 5
Isolate the variable.
Step 6
Divide.
Step 7
Validate.
A - 2l = 2l - 2l + 2 w
A - 2l = 2 wA - 2l = 2 w
2
2
A - 2l 2 w
=
2
2
A - 2l
Answer: w =
2
149
(Symmetric Property of Equality)
Validate by choosing values for A and l: Let A = 100, l = 30
100 = 2(30) + 2 w
Substitute values into
Substitute
in our final
40 = 2values
w
the original equation:
equation (from Step 6):
w = 20
100 = 2(30) + 2 w
100 - 2(30)
w=
100 = 60 + 2 w
2
100 - 60 = 2 w
100 - 60 40
w=
=
40 = 2 w
2
2
compare
w = 20 
w = 20
Model 4
Solve: 1.2x + 0.5 = x – 0.3(x + 2)
Step 1
Determine the variable.
x is the variable; it appears once on the left side and twice on
the right side.
Step 2
Clear parentheses.
1.2x + 0.5 = x – 0.3x – 0.6
Step 3
Combine like terms.
1.2x + 0.5 = 0.7x – 0.6
Step 4
Clear fractions.
No fractions; omit this step.
Step 5
Isolate the variable.
1.2x – 0.7x + 0.5 = 0.7x – 0.7x – 0.6
0.5x + 0.5 – 0.5 = –0.6 – 0.5
0.5x = –1.1
Step 6
Divide.
Step 7
Validate.
0.5 x -1.1
=
0.5
0.5
-1.1
x=
0.5
Answer: x = −2.2
?
1.2(-2.2) + 0.5 = (-2.2) - 0.3 [ (-2.2) + 2]
? -2.2 - 0.3(0.2)
-2.64 + 0.5 =
? -2.2 - 0.06
-2.14 =
-2.14 = -2.14 
Chapter 2 — Solving Equations
150
Technique
An Option for Step 4: Clearing Decimals
Sometimes it is helpful or preferable to clear the decimals in an equation. Remember that
we can convert decimal numbers to fractional numbers because numbers represented
as decimals are fractions with denominators that are powers of 10 (10, 100, 1000, . . .).
Mentally converting decimals to decimal fractions, we can determine what power of 10
the equation should be multiplied by to clear the decimals.
0.7 x = 0.14
7
14
Convert to decimal fractions :
x=
10
100
Clear the decimals:
The LCD is 100, so we need to multiply each term in the original equation by 100:
100(0.77 x) = 100(0.14)
70 x = 14
Model 5
Solve: – 0.3 + 0.6a = 0.12 – a
Step 1
Determine the variable.
a is the variable; it appears once on the left side and once on
the right side.
Step 2
Clear parentheses.
No parentheses; omit this step.
Step 3
Combine like terms.
No like terms to combine; omit this step.
Step 4
Clear decimals as in the
technique above.
To clear decimals, each term must be multiplied by 100:
-30 + 60a = 12 -100a
Step 5
Isolate the variable.
–30 + 60a + 100a = 12 – 100a + 100a
160a – 30 + 30 = 12 + 30
160a = 42
Step 6
Divide.
160a 42
=
160 160
42
a=
160
Answer: a = 0.2625
Step 7
Validate.
? 0.12 – 0.2625
– 0.3 + 0.6(0.2625) =
? 0.12 – 0.2625
– 0.3 + 0.1575 =
– 0.1425 = – 0.1425 
Section 2.4 — General Methodology for Solving Equations
151
Addressing Common Errors
Issue
Forgetting
to multiply a
term by the
LCD when
clearing
fractions
Incorrect
Process
x -2
+1 = 2
3
x -2
3:
+1 = 3 : 2
3
x - 2 +1 = 6
x -1 = 6
Resolution
Use parentheses
and distributive
property when
multiplying both
sides of the
equation by the
LCD.
x =7
Not
validating in
the original
equation
In the above example,
validate x = 7:
x – 2 + 1= 6
(7) – 2 + 1 = 6
5+1=6
6=6 
Any error might
be repeated in
the validation
if you do not
use the original
equation. Using
the original
equation will find
the error.
Correct
Process
x−2
+1 = 2
3
 x−2 
3
+ 1 = 3 (2 )
 3

x−2
3:
+ 3 : 1 = 3 (2 )
3
x−2+3= 6
x +1 = 6
x +1 −1 = 6 −1
x=5
Validation
x−2
+1= 2
3
(5) − 2
+ 1 =? 2
3
3 ?
+1= 2
3
?2
1+1=
2=2 
Validate x = 7
x -2
+1 = 2
3
(7 ) - 2
+1 = 2
3
5
+1 = 2
3
8
!2
3
2 =error
2
An
has been
made; rework the
problem.
Adding a
term when
it should
have been
subtracted
or
subtracting
when
addition is
called for
x – 6 = –6
x = –12
3x + 2 = 4
3x = 6
x=2
Even though it
looks right, take
the time to write
in the step.
x–6=–6
x–6+6=–6+6
x=0
3x + 2 = 4
3x + 2 – 2 = 4 – 2
3x = 2
3x 2
=
3 3
2
x=
3
? –6
0–6=
–6 = –6 
2
?
3  + 2 = 4
3
?
2 + 2 =4
4=4 
Chapter 2 — Solving Equations
152
Issue
Dividing
incorrectly
when
solving
for the
unknown
Incorrect
Process
6x = 3
x=2
Resolution
Even though it
looks right, take
the time to write
in the step.
x
=4
2
x =2
Solving for
the wrong
variable in a
formula
Solve for h:
1
A = bh
2
2A = bh
bh = 2A
2A
b=
h
Correct
Process
6x 3
=
6
6
1
x=
2
x
=4
2
x
2: =2: 4
2
x =8
After solving the
equation, review
Step 1 and make
sure the correct
variable has been
isolated. The
math processes
you use may be
correct, but the
answer is not if
you have solved
for the wrong
variable.
Validation
1 ?
=3
2
6?
=3
2
3=3 
6:
x
=4
2
8?
=4
2

4=4 
1
A = bh
2
2A = bh
bh = 2A
2A
h=
b
Preparation Inventory
Before proceeding, you should have an understanding of each of the following:
When and how to clear fractions in an equation
How to apply the methodology of solving equations to solve an equation or formula for a given
variable
Section 2.4
Activity
Methodology for Solving Equations
Performance Criteria
• Clear an equation of fractions
– appropriate common denominator is used
– all fractions are cleared, leaving whole number
coefficients and constants
– an equivalent equation is formed
• Solve equations
– application of the methodology
• Solve formulas (literal equations) for a given
variable
– application of the methodology
– correct variable is solved in terms of the other
variables
– all variables are accounted for in the final
answer
– accuracy of computation
– appropriate validation
Critical Thinking Questions
1. If the solution to an equation is in the form x = a, what are three types of numbers that a can be?
2. Can x = 0 be a solution for an equation? Illustrate your answer with an example.
153
Chapter 2 — Solving Equations
154
3. Why should you clear fractions before solving an equation?
4. When validating an equation, why is it important use the original equation?
5. How does solving for a variable in a formula work if there aren’t any numbers to combine?
6. What are three differences between solving an equation and solving for a variable in a formula?
Tips
for
Success
• Clearing fractions (or decimals) can make it easier to work through a problem, but answers may still
contain fractions or decimals
• Report the final answer in the same form (fraction or decimal) as the original equation
• In a formula, it might be helpful to highlight the chosen variable
Section 2.4 — General Methodology for Solving Equations
155
Demonstrate Your Understanding
1. Solve for x:
Problem
a)
2x + 3 = 11
b)
3x + 6 = 10 – x
c)
2(3x + 5) = x – 35
d)
–3(4x – 7) = 3(x + 3)
e)
–2(x + 1) – 6 = 7x – 3(x + 4)
f)
23 – (3x + 7) = 2(x + 9) + 3
Worked Solution
Validation
Chapter 2 — Solving Equations
156
2. Clear all fractions or decimals before solving for x:
Problem
a)
3
4
( x + 6) = x - 2
4
5
b)
2
1
x 4
x- = +
5
10 10 5
c)
1
( x + 6) + 3 = 5( x + 2)
2
d) 1.3x + 0.5 = x + 0.2(x + 2)
e)
0.05 x - 0.1(0.2 x + 0.3) = 0.01x - 0.4
Worked Solution
Validation
Section 2.4 — General Methodology for Solving Equations
157
3. Solve for the given variable:
Problem
Worked Solution
a) Solve for w: A = 2l +2w
b) Solve for t: I = Prt
c) Solve for t: D = rt
d) Solve for b: x = a +
e) Solve for h: V = lwh
b
2
Validation
Chapter 2 — Solving Equations
158
Identify
and
Correct
the
Errors
In the second column, identify the error(s) in the worked solution or validate its answer. If the worked solution
is incorrect, solve the problem correctly in the third column and validate your answer.
Worked Solution
1)
2)
2
x 4
x −1 = +
3
6 3
2

 x 4
3  x − 1 = 6  + 
3

6 3
2x − 3 = x + 4
2x − 3 + 3 = x + 4 + 3
2x = x + 7
2x − x = x + 7 = x
x=7
2 x − 5 = x − 10
2 x = x − 15
x = −15
2(−15) − 5 = (−15) − 10
−30 + 5 = −15 − 10
−25 = −25
3)
Solve for r:
mv 2
F=
r
r : F = mv 2
r = mv 2 F
4)
2x
1
-1 = x 3
2
2x
1
3:
-1 = x - 3 :
2
3
2x -1 = x - 6
x =5
Identify Errors
or Validate
Correct Process
Validation