Download Chapter 2 Forming Equivalent Equations The Golden Rule (of

Forming Equivalent Equations
Chapter 2
Linear Equation and Inequalities
Michael Giessing
[email protected]
An equation can be rewriten into an equivalent form
by:
1. Simplify either side
2. Using the Golden Rule of Algebra
3. Interchanging sides
University of Utah
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The Golden Rule (of Algebra)
Do unto one side as thou hath done unto the
other.
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Examples
Simplify the left hand side.
x(4 − x) = 2x →
4x − x2 = 2x
Added 4 to both sides.
x(4 − x) = 2x →
x(4 − x) + 4 = 2x + 4
Interchangesd sides.
x(4 − x) = 2x →
2x = x(4 − x)
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The Golden Rule and Addition
You can think of an equation as a scale. If you do
something to one side you must do the same thing to
the other side to keep things in balance. You must do
all operations to entire expressions!
You can add anything to both sides. This includes
negative numbers and variables
y+π
y + π + (−3)
2x + 3
2x + x
=
=
=
=
y2 →
y 2 + (−3)
x→
−x + x
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The Golden Rule and Multiplication
You can mulitiply both side by any number. You must
multiply the entire side by the number not just some
of the terms on that side.
2l + 6 = 4l → 30(2l + 6) = 30 × 4l
You can multiply by any multiplicative inverse.
Which is the same thing as saying you can divide by
any number.
4
1 4
1
= 5n → × = × 5n
n
4 n 4
There is a number that has no multiplicative inverse.
What is it?
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The Golden Rule and Multiplication by a variable
Never divide by Zero
Zero does not have a multiplicative inverse. To see
this try to find Zero’s inverse.(Remember that an
inverse is the number x such that ax = 1. For most
numbers x = 1/a)
0 ∗ x = 1; x =?
Unsual things happen if you divide by zero.
You can alway multiply both sides by a variable.
x2 + y 2 = 4 → x(x2 + y 2 ) = 4x
You can divide by x as long as x 6= 0. It is easy to
make this mistake. Be cautious whenever you divide
by a variable.
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The Golden Rule and Function
We will use the Golden Rule with more than just addition and multiplication. We will use it with many other
operations called functions. Always remember to apply the operation to the entire left hand side(LHS) and
right hand side( RHS).
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Word Problems
1. What do you need to know to solve the problem?
Write this down in english.
2. Assing numbers to the the known parts. Assign a
letter to the unknown parts
3. Translate this into an algebraic equation or
inequality.
4. Solve.
5. Make sure that your solution answers the original
question
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Example
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Example Continued
You purchased my new bike for $1500.00. The bike
shop was having 30% off sale the day you purchased
the bike. What was the original price of the bike?
1. The price you paid for the bike is 30% off the
original price. Restated, the price you paid for the
bike is the orignal price minus 30% of the
original price.
2. 1500.00 =x-.3x
1500.00=.7x
Simplifying the RHS
1500.00/.7=.7x/.7 Golden Rule of Algebra
3.
2142.86=x
Simplifying both sides
x=2142.86
Interchanging the sides
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The answer $2142.86 answers the question and seems
reasonable. Lets check it to make sure everything
worked. We need to see if 30% off of $2142.86 is
$1500.00.
2142.86 − .3 × 2142.86 = 1500.00
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Geometry Formulas
Useful Formulas
Business
Shape
Square
Rectangle
Circle
Triangle
Selling price = Cost + Mark up
Mark up = Mark up rate × Cost
Selling price = List price - Discount
Rates in Mixtures
Area
A = s2
A = lw
A = πr 2
A = bh/2
Perimeter
P = 4s
P = 2l + 2w
P = 2πr
P =a+b+c
first rate × Amount + second rate × Amount
= Final Rate × Amount
I expect you to learn all of these.
Rate problems
Distance = rate × time
I expect you to know all of these.
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3-D Geometry Formulas
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Problem Solving with Formulas
Shape Volume
Cube V = s3
Rectangular Prism V = lwh
Surface Area
SA = 6s2
SA = 2lw + 2wh
+2hl
2
Cylinder V = πr h SA = 2πrh + 2πr 2
Sphere A = 43 πr3 SA = 4πr 2
1. Determine which formulas fit your problem. This
formulas should contain the information given in
the problem and the quantities you are looking
for.
2. Plug the information from the problem into these
formulas.
3. Solve these equations for the unkown values. You
will need one equation per variable.
I expect you to learn all of these.
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Example 5 on page 83
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Step 2: Plug
Students are traveling in two cars to a
football game 150 miles away. The first car
leaves on time and travels at an average
speed of 48 mph. The second car starts .5
hours later and travels at an average speed of
58 mph. At these speeds, how long will it
take the second car to catch up to the first
car?
Step 1: What formula? We will use d = rt. We want
to find a time when the two cars have gone the same
distance. We will start time when the first car leaves.
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For the first car d1 = 48t. If we put some time t into
that we get the first cars position. We need to write
the same thing for the second car. Remember that car
two leaves .5 hours late so the distance it has traveled
is d2 = 58(t − .5). We want to find the time when the
cars are in the same place so d1 = d2 . That give us
three equations and three unknowns.
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Step 3: Solve
We start from d1 = d2 . We know that d1 =
48t and d2 = 58(t − .5). We substitute this in
and get 48t = 58(t − .5). Now we solve this.
48t = 58(t − .5)
48t = 58t − .5
distributive prop
48t − 58t = 58t − .5 − 58t Golden Rule
−10t = −.5
Simplify LHS
t = −.5/ − 10
Golden Rule
t = .05
Simplify RHS
Inequalities
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Properties of Inequalities
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Multiplication By a Negative
Inequalities obey the golden rule but, the direction of the inequality changes whenever
you multiply or divide by a negative number.
Example:
2 < 3
→
3 × 2 < 3 × 3 → Golden Rule
6 < 9
Observe that 2 < 3 but, −2 > −3.
We
can get the second inequality form the first
by multiplying the first by −1 on both sides.
Example:
2 < 3
→
−3 × 2 > −3 × 3 → Golden Rule
−6 > −9
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Solving an
Equations
Inequalities and Division
Example
Positive
6
6/2
3
Negative
6
6/(−2)
−3
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Absolute
Value
Lets solve |x| = 2. If x ≥ 0, then we have x = 2. If
x < 0, then −x = 2 → x = −1
< 10
< 10/2
< 5
→
→ Golden Rule
< 10
→
> 10/(−2) → Golden Rule
> −5
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Now lets solve all problems like this one. This is called
abstraction. Let say we were given some positive real
number a. (Why does it need to be positive?) Then
we could find x so that |x| = a. This works just like
before. If x ≥ 0, then x = a. IF x < 0, then −x =
a → x = −a.
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Solving a Concrete Example
Solving the Abstract Problem
|x|
1.2
|x|
a
−1.2
1.2
−a
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Nonstandard Form (Ex. 3 pg.
108)
Solve |2x-1|+3=8. First we try to simplify as much as
possible. Using the golden rule we subtract 3 from
both sides.
|2x − 1| + 3 − 3 = 8 − 3 → |2x − 1| = 5
If 2x − 1 ≥ 0 then 2x − 1 = 5. We will add 1 to both
sides. Then we divide both sides by 2. (Use the
Golden Rule twice)
a
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Continued
If 2x − 1 < 0, then −(2x − 1) = 5. We will apply the
Golden Rule three times.
−(2x−1) = 5 → 2x−1 = −5 → 2x = −4 → x = −2
So, x = 3 and x = −2 are solutions to the equation.
2x − 1 + 1 = 5 + 1 → 2x/2 = 6/2 → x = 3
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Absolute Inequalities
Picture, Picture
blah
|2x−1|
5
−2
1/2
3
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|x|
1.2
|x|
1.2
−1.2
1.2
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−1.2
1.2
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