Transcript Module1Topic1Notes

Module 1 ~ Topic 1
Solving Equations
Table of Contents
 Slides 6-14: Solving Linear Equations
 Slides 15-29 : Practice Questions
Audio/Video and Interactive Sites
Slides 2: Algebra Cheat Sheet
Slides 3: Graphing Calculator Use Guide
Slide 5: Video
Slide 7: Gizmos
Algebra Cheat Sheet
You may want to download and print this
sheet for reference throughout the course.
Special Instructions

This module includes graphing calculator work.

Refer to the website, TI-83/84 calculator instructions,
for resources and instructions on how to use your
calculator to obtain the results we do throughout the
lessons. I suggest that you bookmark this site if you
haven’t done so already.

Take careful notes and following along with every
example in each lesson. I encourage you to ask me
questions and think deeply as you’re studying these
concepts!
Topic #1: Solving Linear Equations

Many real-life phenomena can be described by
linear functions. It is important to learn how to
solve equations involving such functions to provide
us with information about these phenomena.
Definition: A linear equation in one
variable is an equation that can be written
in the form ax + b = c, where a, b and c
are real numbers, and a ≠ 0.
(The letter x is often used as the
variable, but it is not required to be the
variable.)
Video Break!!!!!
Click on this link to watch videos on solving equations.
Recall General Rules
Order of Operations: PEMDAS
Multiplication/Division is done in order, left to right
Addition/Subtraction is done in order, left to right
Solving Equations
First, you must know what you are solving for so you can isolate it.
To do that:
Take care of any exponents/FOIL or distribution/simplification
Get common denominators if necessary
Combine like terms on each side of the equal sign
Addition/Subtraction across = is done next to isolate the variable
Multiplication/Division across =is done last and the variable should now
be isolated
Gizmos
Gizmo: Modeling 1-Step
Equations A
Gizmo: Modeling 2-Step
Equations
Gizmo: Modeling 1-Step
Equations B
Gizmo: Solving 2-Step Equations
Gizmo: Solving Formulas
How to Solve Equations
As with any journey, you need to know where you want to
end up before you start off, or you will never get there!
Ex: Solve
4 x  8  16
1) Figure out what you are solving for.
In this case we want to end up with x = #
4 x  8  16
2) Move everything that is NOT x to the
other side. When solving equations, do all
addition/subtraction first. Then all multiplication/
division last.
Make sure you are where you wanted to end up,
x = some number.
8
8
______
4 x  _______
24
4
4
x6
x?
Example 1
Solving an equation in one variable algebraically means to find the value of the
variable that makes the mathematical statement true using appropriate algebraic
operations.
Example 1:
Solve the equation 5x - 4 = 0 .
5x – 4 = 0
+ 4 +4
5x
= 4
This means
4
5
5x – 4 = 0 true.
5x
5
= 4
5
4
x =
5
makes
One of the great things about solving equations is that we can always check our
solutions! We do this by plugging the value(s) we get for our variable into the
original equation and verify that we have a true statement.
Example 1:
Solve the equation 5x - 4 = 0 .
Check:
5x – 4 = 0
+ 4 +4
5x
= 4
5x
5
= 4
5
4
x =
5
We simplify and get 4-4, which is 0.
This is what our original equation stated.
This is true.
Original
Equation
4
5 x  4  5   4  4  4  0
5
Now, plug in the value
of x that you just found
Example 2
Example 1: Solve the equation 5(h – 2) = -4(3-h).
Check:
Solution: We have a couple
of choices as to how to solve
this equation. You may notice
that it is not in the form
introduced earlier, but it still a
linear equation in one
variable, since it can be
written in the form described
above.
Our goal is to isolate the
variable h, so that it appears
on one side of the equation,
and its value appears on the
other. We will need to
distribute on both sides of
the equation before we can
do that.
5(h - 2) = -4(3 – h)
5h – 10 = -12 + 4h
+10
5h
+10
= -2 + 4h
5h = -2 + 4h
-4h
h = -2
- 4h
5(h - 2) = -4(3 – h)
5((-2) – 2) = -4(3 – (-2))
5( - 4) = -4 ( 3 + 2)
-20 = - 4 (5)
-20 = -20
Since both sides of the equation
yield the same result, we know
that our answer is correct!
Example 1
Example 1: Solve the equation
5
1
1
1
m
 m
3
2
9
10
Remember: Solving Equations
First, you must know what you are solving for so you can isolate it.
To do that:
Take care of any exponents/FOIL or distribution/simplification
Get common denominators if necessary
Combine like terms on each side of the equal sign
Addition/Subtraction across = is done next to isolate the variable
Multiplication/Division across =is done last and the variable should now be
isolated
Example 1: Solution
5
1
1
1
m
 m
3
2
9
10
1
1
5
1
90 m    90 m  
2
10 
3
9
5 
1
1 
1
90 m   90   90 m   90 
3 
2
9 
 10 
150m  45  10m  9
-45
-45
150m  10m  54
-10m -10m
140m  54
140
m
140
54
27

140
70
Check:
5
m
1

1
m
1
?
3
2 9
10
5  27  1
9 1
9 7
2
1
           
3  70  2
14 2
14 14
14
7
1  27  1
3
1
3
7
10
1
           
9  70  10
70 10
70 70
70
7
We find that both sides of the equation give us the same result
when we plug our answer in, which means that we obtained the
correct result!
Practice Examples
Example 2: Solve the equation 3p + 2 = 0 .
Example 3: Solve the equation -7m – 1 = 0.
Example 4: Solve the equation 14z – 28 = 0.
Solutions on next slide. Solve these on your own first.
Practice Examples Answers
Example 2: Solve the equation 3p + 2 = 0 .
2
p
3
Example 3: Solve the equation -7m – 1 = 0.
 2
 3
220
Check: 3    2  0
00
Check:
1
m
7
Example 4: Solve the equation 14z – 28 = 0.
z2
 1
 7    1  0
 7
11  0
00
Check:
14(2)  28  0
28  28  0
00
More Practice Examples
Example 5: Solve the equation
4n  8  6n  19 .
x 1
  4 x  5.
Example 6: Solve the equation
3 2
Solve these on your own first. Solutions on next slide.
More Practice Examples - Answers
Example 5: Solve the equation
4n  8  6n  19 .
27
n
2
x 1
  4 x  5.
Example 6: Solve the equation
3 2
123
x
26
4n  8  6n  19
8
8
 4n  6n  27
 6n  6n
2n  27
n
27
2
x 1
  4 x  5
3 2
x 1
  4 x  20
3 2
 x 1
6    6 4 x  20
3 2
2 x  3  24 x  120
26 x  3  120
26 x  123
123
x
26
More Practice Examples
Example 7: Solve the equation
.
0.05 x  0.10(200  x)  0.45 x
Example 8: Solve the equation
.
4y  5  2y  4  6y  3
Solve these on your own first. Solutions on next slide.
More Practice Examples - Answers
Example 7: Solve the equation
0.05 x  0.10(200  x)  0.45 x
36.45  x
0.05 x  0.10(200  x)  0.45 x
.
0.05  20  0.1x  0.45 x
20.05  0.1x  0.45 x
20.05  0.55 x
__
36. 45  x
36.45  x
Example 8: Solve the equation
4y  5  2y  4  6y  3
3  y
.
4y  5  2y  4  6y  3
2y  5  7  6y
2 y  12  6 y
 12  4 y
3 y
More Practice Examples
Example 9: Solve the equation for m.
y  mx  b .
Example 10: Solve the equation A  b1  b2 h for b2.
2
1
Solve these on your own first. Solutions on next slide.
More Practice Examples
Example 9: Solve the equation for m.
y  mx  b .
y  b  mx
y b
m
x
y b
m
x
1
Example 10: Solve the equation A  b1  b2 h
2
2A
b2 
 b1
h
y  mx  b
b
b
for b2.
1
b1  b2 h
2
1

2( A)  2 b1  b2 h 
2

2 A  b1  b2 h
A
2A
 b1  b2
h
2A
 b1  b2
h
More Practice Examples
Example 11: Solve the equation for x.
No Solution
3
3
 4 x   8 x   12 x
4
5
3
3
 4 x   8 x   12 x
4
5
3 3
 4x    4x
4 5
 4x
 4x
3
3
 0
4
5
3 3

4 5
0
3 3
 , so the answer is No Solution
Obviously,
4 5
Word Problem Examples
Example:
The relationship between ºC and ºF can be represented by the equation
9
F  c  32
5
where F is the number of degrees Fahrenheit, and C is the number of
degrees Celsius.
a. Solve the above equation for C.
b. Convert 98oF into degrees Celsius
Solution a):
We want to isolate C on one side of the equation. So, we apply the following operations on our
original equation.
9
F  c  32
5 -32
-32
9
F  32  c
5
5
F  32  5  9 c 
9
95 
5
F  32   c
9
or
5
C  ( F  32)
9
Now we have an equation that allows us to compute degrees Celsius if we knew
degrees Fahrenheit.
Solution b):
We plug 98 in for F and solve for C.
9
F  c  32
5
98 
9
C  32
5
-32
-32
66 
9
C
5
9 
5(66)  5 c 
5 
330  9c
9
9
330
c
9
or
c
330
110

 33.67 0
9
3
So, 98ºF is approximately
36.67ºC.
Example:
When you buy a new car, they say that the value of the car depreciates as
soon as you drive it off the lot! Accountants use the following equation to
measure depreciation of assets:
D
CS
L
where …
D is the depreciation of the asset per year,
C is the initial cost of the asset,
S is the salvage value, and
L is the asset’s estimated life.
a. What is the salvage value of a machine that cost a company $40,000 initially, has
an annual depreciation of $3000, and an estimated life of 10 years?
b. Solve the original equation for S, the salvage value, in general.
Solution a):
We plug 30,000 in for C, 2000 for D, and 10 for L. We then solve for S.
CS
D
L
40,000  S
3000 
10
 40,000  S 
10(3000)  10

10


30,000  40,000  S
 40,000  40,000
 10,000   S
(1)( 10,000)  (1)(  S )
10,000  S
So, the salvage value for
the machine is $10,000.
Solution b):
We want to isolate the variable S, treating all of the other letters in the equation as constants.
CS
D
L
C S 
( L)( D )  ( L)

 L 
LD  C  S
-C
-C
LD  C  S
(1)( LD  C )  (1)(  S )
 LD  C  S
or
S  C  LD
This equation allows us to
calculate the salvage value for
any asset, given the initial cost,
estimated life, and depreciation
value.