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MAT091 :: Lecture Note :: Week 11

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### Exercise: Find Equation of a Parallel Line

Parallel lines never intersect (i.e. they don't have any points in common) and they have equal slopes.

Given the line `y = 10x + 2` find the equation of a parallel line that passes through the point `(5, 40)`.

```   y = 10x + b                    # parallel lines have the same slope
y - 40 = 10(x - 5)             # using point-slope form
y - 40 = 10x - 50              # apply distributive property
y - 40 + 40 = 10x - 50 + 40    # add 40 to both sides
y = 10x - 10                   # y = mx + b form

[check answer] substitute 5 in for x and we should get 40

y = 10(5) - 10 = 50 - 10 = 40
```

`[question]` No or Yes: The parallel line intercepts the vertical-axis at the point `(0, -10)`.

`[question]` No or Yes: The parallel line intercepts the horizontal-axis at the point `(1, 0)`.

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### Exercise: Find Equation of a Perpendicular Line

Perpendicular lines intersect (i.e. they have a point in common) and they form right angles at their point of intersection.

The slopes of perpendicular lines are negative reciprocals of each other.

```   m * mp = -1

'm' is the slope of a line and 'mp' is the slope of a perpendicular line
```

Given the line `y = 5x + 2`, find the equation of a perpendicular line that passes through the point `(5, 10)`.

```   Let A be the given line and B the perpendicular line.

slope of line A:  m = 5
slope of line B:  m = -(1/5)     # negative reciprocal

find the slope-intercept equation for line B

substitute point (5, 10) into point-slope equation...
y - 10 = -(1/5)(x - 5)

solve for y...
y - 10 + 10 = -(1/5)(x - 5) + 10     # add 10 to both sides
y = -(1/5)(x - 5) + 10               # simplify
y = -(1/5)x - -(1/5)5 + 10           # distribute -(1/5)
y = -(1/5)x + 1 + 10                 # simplify
y = -(1/5)x + 11
```

`[question]` No or Yes: The perpendicular line intercepts the vertical-axis at the point `(0, 11)`.

`[question]` No or Yes: The perpendicular line intercepts the horizontal-axis at the point `(55, 0)`.

`[exercise]` Find the point where lines A and B intersect.

```   set line A equal to line B
5x + 2 = -(1/5)x + 11

solve for x...
5x + 2 = -0.2x + 11             # divide 1 by 5
5x + 2 - 2 = -0.2x + 11 - 2     # subtract 2 from both sides
5x = -0.2x + 9                  # simplify
5x + 0.2x = 9                   # add 0.2x to both sides
5.2x = 9                        # simplify
5.2x / 5.2 = 9 / 5.2            # divide both sides by 5.2
x = 1.73                        # simplify

substitute x in one of the equations...
y = 5(1.73) + 2 = 10.65

point of intersection:  (1.73, 10.65)

check using the other line...
-(1/5)1.73 + 11 = 10.65
```

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### Intersecting Lines

Two lines that are not parallel can intersect. If the equations for the lines are known, then the point of intersection can be determined as follows.

```   line A:  y = 2x + 2
line B:  y = 5x + 5

set the lines equal to one another...
2x + 2 = 5x + 5

solve for x...
2x + 2 + -2 = 5x + 5 + -2    # add -2 to both sides
2x = 5x + 3

2x + -5x = 5x + 3 + -5x      # add -5x to both sides
-3x = 3

-3x / -3 = 3 / -3            # divide both sides by -3
x = -1

substitute for x in one of the line equations and evaluate...
y = 2(-1) + 2 = -2 + 2 = 0

point of intersection:  (-1, 0)

check answer: plug (-1, 0) into the other line equation
and see if it works...
y = 5(-1) + 5 = -5 + 5 = 0
```

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### Example Application Using Intersecting Lines

Zelmo Zeroman had to select between two linear pay scales. One choice was to receive \$10 up-front and then earn \$1 for every 8 hours worked. The second choice was to received \$2 up-front and then earn \$1 for every 5 hours worked. The following two functions define Zelmo's pay scale choices.

```           1                                  1
p(h) = ---(h) + 10                 P(h) = ---(h) + 2
8                                  5

h:  number of hours
p(h):  USD
P(h):  USD

The amount of pay is a function of hours worked.
The amount of pay depends on the number of hours worked.
```

Zelmo realized that over time the `P(h)` pay scale would result in more pay, but he wanted to learn how many hours he needed to work before `P(h)` would be greater than `p(h)`.

It was suggested that Zelmo find where `P(h)` and `p(h)` intersect when graphed.

```   observe...  1/8 = 0.125    and    1/5 = 0.2

p(h) = 0.125(h) + 10       and    P(h) = 0.2(h) + 2

let p(h) = P(h)...

0.125(h) + 10 = 0.2(h) + 2

solve for 'h'...

...subtract 0.125(h) from both sides...
-0.125(h) + 0.125(h) + 10 = -0.125(h) + 0.2(h) + 2
10 = 0.075(h) + 2

...subtract 2 from both sides...
-2 + 10 = 0.075(h) + 2 - 2
8 = 0.075(h)

...divide both sides by 0.075...
8/0.075 = 0.075(h)/0.075
106.67 = h
```

Conclusion: Zelmo should pick the `p(h)` pay scale if he is going to work less than `106.67 hours`; otherwise, he should select the pay scale defined by `P(h)`.

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### Zelmo Zeroman Borrows Money From Herb Mumford

Zelmo Zeroman borrowed \$1,000 from his "friend" Herb Mumford. Zelmo and Herb agreed that the loan was to be paid off at a rate of \$200 per week. Zelmo created the following function to help him keep track of his weekly payoff amounts.

```   p(w) = 1000 - 200(w) USD

input 'w' is week number  and  USD stands for U.S. Dollar

domain:  w ≥ 0, where 'w' is an integer
range:  0 ≤ p(w) ≤ 1000
```

The function p(w) outputs the payoff amount after w weeks.

##### Exercises
1. Re-write the function in y = mx + b form.

2. p(2) = _________   [record unit with answer]

3. The function p(w) has a slope of `_______`.

4. Explain why p(w) is a decreasing function.

5. It will take Zelmo ________ weeks to payoff the loan.

6. In this particular case, the technique used to answer the previous exercise is also the technique used to find the _______-intercept for a function.

7. Graph p(w) for the entire duration of the loan.

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### Inequalities

The following are some inequalities graphed on the number line. An end-point marked with 'x' (or ']') implies that value is included, while an end-point marked with 'o' (or ')') is excluded.

```   x > 2

---------------------o=========>
-5 -4 -3 -2 -1  0  1  2  3  4  5

x ≥ -4

---x===========================>
-5 -4 -3 -2 -1  0  1  2  3  4  5

x ≤ 3

<=========================x------
-5 -4 -3 -2 -1  0  1  2  3  4  5

x < -2

<==========o---------------------
-5 -4 -3 -2 -1  0  1  2  3  4  5
```

Linear inequalities are solved just like linear equations with one exception: the inequality sign is "flipped" whenever there is a multiply (or divide) by a negative. The term "flipped" is used to imply less-than becomes greater-than and vice versa; and that less-than-or-equal-to becomes greater-than-or-equal-to and vice versa.

```   Example 1:  8x + 1 < 2x - 5

8x - 2x + 1 < -5
6x + 1 - 1 < -5 - 1
6x < -6
x < -1

try x = 0...
8(0) + 1 = 9; 2(0) - 5 = -5
9 is not less than -5

try x = -1...
8(-1) + 1 = -7; 2(-1) - 5 = -7
-7 is not less than -7

try x = -2...
8(-2) + 1 = -15; 2(-2) - 5 = -9
-15 is less than -9

Example 2:  -2x < 5

-2x / -2 > 5 / -2      # < flipped to > (divide-by negative)
x > -2.5

try x = 5...
-2(5) < 5
-10 < 5 is true

try x = -5...
-2(-5) < 5
10 < 10 is false
```

The following uses numbers to help understand when inequalities are flipped (switched).

```   10 > 5  is true
multiply both sides by -1
10(-1) = -10  and  5(-1) = -5
-10 > -5 is false
```

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### Systems of Linear Equations

A "system" of equations is a collection of equations that are treated as an atomic unit. In other words, a system of equations is used when dealing with more than one variable.

```   A: 3x + 3y = 0
B: 12x + 2y = -20

pick one of the equations...
pick A

solve equation A for y
3x + 3y = 0
-3x + 3x + 3y = 0 + -3x
3y = -3x
3y / 3 = -3x / 3
y = -x

substitute for y in the equation B...
12x + 2(-x) = -20

solve equation B for x...
12x + -2x = -20
10x = -20
10x / 10 = -20 / 10
x = -2

pick one of the equations...
pick A

substitute for x in the equation A...
3(-2) + 3y = 0

solve equation A for y...
3(-2) + 3y = 0
-6 + 3y = 0
6 + -6 + 3y = 0 + 6
3y = 6
3y / 3 =  6 / 3
y = 2

the solution is the point (-2, 2)
```

Another system of linear equations.

```   Two numbers have a sum of 55 and a difference of 9.
Find the numbers.

let x and y be the two unknown numbers

form two equations...
A: x + y = 55
B: x - y = 9

x + y = 55
+  x - y =  9
=============
2x = 64

solve for x...
2x = 64
2x / 2 = 64 / 2
x = 32

pick an equation, substitute for x, solve for y
pick A
32 + y = 55
-32 + 32 + y = 55 + -32
y = 23

solution:  (32, 23)

[exercise] Re-do this problem using substitution.
```

Here is another system of linear equations that is solved using the addition method.

```   A: 5x + y = 10
B:  x + y = 20

multiply both sides of B by -1...
-1(x + y) = 20(-1)
-x - y = -20

5x + y =  10
+ -1x - y = -20
===============
4x     = -10

solve for x...
4x / 4 = -10 / 4
x     = -2.5

pick an equation and solve for y...
pick B
-2.5 + y = 20
y = 22.5

solution:  (-2.5, 22.5)

subtitute the solution into A and see if it works...
5(-2.5) + 22.5 = 10
-12.5 + 22.5 = 10
it works!
```

Is a system of equations necessary to solve the following problem?

```   Ten less than half of a number is thirty. What is the number?
```

No, because there is only one variable.

```   n/2 - 10 = 30
n/2 - 10 + 10 = 30 + 10
n/2 = 40
n/2 * 2 = 40 * 2
n = 80
```

Here is a exercise that can be solved using a system of equations.

```   A new nightclub had a grand opening in Scottsdale.  First night
admission was \$4 for females and \$6 for males.  90 people attended
the grand opening and the nightclub collected \$460.  How many
females and how many males attended the grand opening?
```
```   let 'f' represent # of females and let 'm' represent # of males

A: f + m = 90
B: \$4(f) + \$6(m) = \$460

pick A and solve for 'f'...
f = 90 - m

substitute 'f' into B...
4(90 - m) + 6m = 460

distribute the 4, combine like-terms, solve for 'm'...
360 - 4m + 6m = 460
360 + 2m = 460
-360 + 360 + 2m = 460 - 360
2m = 100
(1/2)2m = 100(1/2)
m = 50

substitute m = 50 into A and solve for 'f'...
f + 50 = 90
f = 40

There were 40 females and 50 males the grand opening.
```

Education.Yahoo.com:: Solving Linear Systems

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### Solving System of Linear Equations Using TI-83

The "intersection" feature of the TI-83 calculator can be used to find the solution to a system of linear equations.

```   press Y=
Y1 press CLEAR, if necessary
enter equation:  X is entered using key labeled X,T,theta,n
-2X + 3 ENTER
cursor at Y2, press CLEAR if necessary
enter equation:
(1/3)X - 4 ENTER
press GRAPH
press 2ND TRACE 5 (intersect)
First curve? ENTER
Second curve? ENTER
Guess? ENTER
Intersection x = 3 and y = -3   (3, -3)
```

Substitute `x = 3` into both equations and test if `(3, -3)` is a solution.

```   f(x) = -2x + 3
f(3) = -2(3) + 3 = -6 + 3 = -3     check

f(x) = (1/3)x - 4
f(3) = (1/3)(3) - 4 = 1 - 4 = -3   check
```

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### Assessment Exam Review

Re-do 1st 11 Weeks of Assessments.

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