Merced College; Don Power

 

INTERMEDIATE ALGEBRA - CH 2, LECTURE

 

2.1    Linear Equations

 

Def:  A linear eqn in one var is one that can be written in the form ax + b = 0

First degree:  exponent of the leading term is 1 (implied)

 

Strategy for linear and quadratic equations

0.  Floating step:  collect like terms whenever possible

1.  Clear grouping symbols:  usually by applying the distributive law

            Ex:  x(x2 + 4x -5) = 20

2.  Clear fractions (optional unless you have a variable in the denominator):  multiply by the LCD.

                  Ex (fractions):  Solve  4x/5 + 5/4 = 3x/10-1/8

                  Ex (decimals):  0.3x +0.16 = 0.25x

                  Ex (variable in denominator):  5x + 1/3 = 2/x

      3.   Linear or quadratic?  (i.e. first degree or second degree?)

            If quadratic, set everything equal to 0 and factor to get two linear equations.

                  See the discussion of the zero-factor property in lesson 5.6.

                  This also works for 3rd or higher degree polynomials

4.   Move terms with the var to one side of the equation, move other terms to the other side

First, identify which terms are out of place.

Then for each misplaced term,

      Either:

a.  Put the opposite of the misplaced term on both sides of the equation;  it can go

                        below the like terms on both sides, or

                        in-line on both sides (text's technique), or

                  b.  Move the misplaced terms across the "=" sign and change signs, or

                  c.  Add/subtract mentally

5.  Collect like terms:  If variables are involved, factor out the variable so it appears only once.

6.  Divide [or multiply by the reciprocal if you didn't clear fractions]

7.  Check by substitution

 

Identities:  Equations that are true regardless of the choice of the variable(s)

      Ex:  a(b+c) = ab + ac   The point of the properties like the distrib law is that they work for all real nrs.

 

Eqns With No Solution

      Ex:  x = x+2           The fact that you can write an eqn doesn't mean it has to have a solution.

      Ex:  sqrt(x) = -5

 

 

 

2.2    Problem Solving with Linear Equations

 

Basic Percent Problems

      The point here is that basic percent probs can be translated into equations.

      For the equation technique, there is always a conversion from percent to decimal or vice versa.

      Ex:  What percent of 32 is 28?  Ans:  87.5%

 

 

 

Proportions: 

A proportion is an equality of two ratios.  Example:  x/6 = 32/24

Solution technique:  Cross-multiply, then divide.  In the example above, x*24 = 6*32 …

The procedure is valid if whenever you double one quantity, you must also double the other

      Distance and Time traveled – yes.  Proportion works

      Distance and fuel consumed – yes

      Speed and distance – yes

      Speed and Time – No!  If you double your speed for a trip, you cut your time in half.

 

2.3    Business and Scientific Applications

 

Lecture:  Look at the even numbered problems from the Review Exercises on pages 120-121

 

Percent increase (markup) and decrease (discount)

 

Markup:  Cost + Markup = Selling Price;

                        and Cost * Markup Rate = Markup

Discount:  List Price – Discount = Sale Price;

                        and List Price * Discount Rate = Discount

 

Solving Percent Increase/Decrease Problems for the Original Amount:  Let x = original amount; then...

      The logic is that:

            old amount ± some % of old amount = new amount.  Examples/Applications:

 

Old Wage

% Raise

New Wage

Setup

Solution

x

10%

$17.60

x + .10x = 17.60

1.10x = 17.60, x = 17.6 / 1.10 = $16

                                                                              Note:  1+.10 = 1.10

                       

Cost to Store

% Markup

Selling Price

Setup

Solution

x

40%

$378.00

x + .40x = 378

1.40x = 378, x = 378 / 1.4 = $270

 

Old Price

Sale %

Sale Price

Setup

Solution

x

30%

$42

x – .30x = 42

.70x = 45, x = 42 / .70 = $60

                                                                              Note:  1 – .30 = .70

 

Old Price

Sale %

Sale Price

Setup

Solution

x

20%

$33.60

x – .20x = 33.60

.80x = 33.6, x = 33.6 / .8 = $42

 

You try it!

 

Find the old wage:

 

% Raise

New Wage

Setup

Solution

 

20%

$25.50

 

                                                  $21.25

 

Find the store's cost:

 

% Markup

Selling Price

Setup

Solution

 

30%

$110.50

 

                                                   $85

 

Find the old price:

 

Sale %

Sale Price

Setup

Solution

 

40%

$35.52

 

                                                   $54.20

 

Find the old price

 

Sale %

Sale Price

Setup

Solution

 

60%

$17.20

 

                                                     $43

 

 

 

Quantity-Rate-Value Problems:

Mixtures:  quantity of fluid, purity percentages, multiply to get the "pure stuff;"  add

Interest/investment:  amount invested vs interest rates, multiply to get interest amounts; add

Coin problems: number of coins, value of each coin, multiply to get total value of coins; add

Ticket sales:  number of tickets, value of each ticket, multiply to get total ticket sales; add

 

Distance-Rate-Time problems

 

Work-Rate Problems

 

Add the reciprocals of the times to get the reciprocal of the total time.

      (You're really adding the rates:  how many jobs per hour)

 

Solving formulas for specified variables:

 

Use general procedure for linear eqns:  include step to factor if the variable appears in 2+ terms.

Ex:  1/a + 1/b = 1/c for c;  Start by mult by LCD of abc.

Ex:  A = a + (n-1)d  for d    Alternate way to clear paren:  divide by coeff (paren) after...

 

Geometry problems:  substitute into formulas

 

2.4    Linear Inequalities

 

Addition property - same as equations

Mult property - same as equations, except:  reverse the direction of the inequality if mult or div by a neg.

      Ex:  X40  -1/2 (2x+1) £ -3/8 (x+2)            Sol:   x ³ 2/5, or "[2/5, ₯)"

      Why the change in direction:  show that -8 < -4 on nr line; mult by (-1) and show results on nr line

Transitive Property:  If a<b and b<c, then a<c

Caution:  if an inequality has a solution 10<x or x<3, do not write 10<x<3:  this would imply that 10<3, which is false.  Pictured on a number line, there are 2 intervals, so write two inequalities.

Big difference from equations is in the nature of the solutions

      Equations:  solution consists of specific values (usually one, sometimes more)

      Inequalities:  solution consists of intervals:  complete ranges of values, thus an infinite solution set

Describing intervals

      Inequality notation, i.e. using inequality symbols -- the natural result of solving the inequality

Number line (open or closed dots, with bold lines to show which intervals are included in sol set)

      Now -- Number line, parentheses or brackets to show which intervals are included in sol set

            ( ) points not included

            [ ] points are included

      "Interval notation" -- parenthesis style mirrors what you write on the number line

-- include the endpoints inside the parentheses or brackets

Types of intervals:

      Closed:  includes both endpoints

            Ex:  Solve the "continued inequality" -5 £ 2x-7 £3

      Open:  excludes both endpoints

            Ex:  Change previous problem

      Half open:  includes one, excludes the other

            Ex:  Change previous problem

      [Infinite intervals may be either open or closed, depending on the finite endpoint]

Compound Inequalities

      Set builder notation includes "and" or "or"

            “and”  conjunctive inequality:  for every number in the solution set, both conditions are true.

            “or”  disjunctive inequality:  a number is in the solution set if either condition is true.

      Ex:  1.  {x|x£-2 or x>3}                 2.  {x|x£-2 and x>3}         Which are meaningful?

            3.   {x|x>-2 and x£3}              4.  {x|x>-2 or x£3}           One interval or two?

 

2.5    Absolute Value Equations and Inequalities

 

Equations with Absolute Value:

 

Basic def of abs value

      Distance definition:  |x| or |x–0| is the dist between x and 0 on the number line.

                                    |x–c| is the distance between x and c on a number line.

            for |x| = a, since the distance is a, "x" must be either a or -a:  you get 2 equations.

                  Caution:  isolate the abs value expression on one side before writing the two equations.

            for |x| = |y|, x and y must be the same or opposite in sign

      Algebraic def:  |x| = x if x³0, but |x| = -x if x<0.

            Consequence:  you get 2 equations every time you have a variable inside absolute value;

                  in one, replace |x| by (x); in the other, replace |x| by -(x)

                  If you use this procedure, the “Caution” above is unnecessary

Ex:  single absolute value expression in the equation:

      | 9-3/5 x | +6 = 12

Ex: with two abs value expressions

      | 2/3 b - 1/4 | = | 1/6 b + 1/2 |

 

Inequalities with Absolute Value

 

Best technique:

Use the definition of absolute value above

      Test each resulting interval to see whether it really is in the solution set

      Expect one interval with < or ≤, expect two intervals with > or ≥

      Ex:

            abs(2x+1) + x < 3

            abs (3x–2)  – 8 ≥ x

           

 

Distance technique:  Visualize/contrast on number line, using the distance definition of absolute value

 

|x – 4| = 7   Two discrete points, center is +4, distance from center is 7; so solution is 4–7=–3 or 4+7=11

 

|2x+1| < 7  One (open) interval, center is –1, distance from 2x to the center (–1) is less than 7.  So the quantity 2x is between –1–7 and –1+7.  Write –8<2x<6 and divide all 3 terms by 2.

 

|y+3| ³ 7    Two (closed) intervals, centered at –3; distance from y to the center (–3) is at least 7.  So y must be to the left of  –3–7, or to the right of –3 + 7.  We have two disjoint intervals, y≤–10 or y≥4,

 

      Always, for this technique, isolate the abs value expression on one side of the inequality.

      For |x–a| < (or >) b

            locate the center a on a number line

locate/visualize the "critical points" a+b and a-b on the number line

      (by shifting b units to the left and to the right of the center a)

determine (based on the inequality symbol) whether x is

 inboard of (< )

or outboard of (>) these points

            write the inequalities based on the results

                  Note how many intervals should result from the situation

 

"no solution" and "all real nr" cases:

      |A| < negative number:  never true

      |A| > negative number:  always true

 

Alt technique: 

      Solve equations (instead of inequalities) -- don't worry about direction of symbol

      Pick a test point inside each interval (caution - not at the endpoints of the intervals)

      Check each interval by subs of the test points

      The solution consists of all the intervals that give you true results

     

Extension:  Here is an example of an absolute value equation with an interval as a solution:

      abs(x+1) − abs(x−3) −2x + 2 = 0:  Sol is the interval [−1, 3]

      Why:  if x+1 ≥ 0 and x − 3 ≤ 0 we get 0=0, which is always true.

 

 

 

 

 

 

 

McKeague:

 

2.1  Paired Data and the Rectangular Coordinate System

 

Linear equations

Definition:  any equation that can be put in the form ax + by = c [fact:  graph is a straight line]

Task:  Graph by point-plotting

Ex 3:  graph [equation of form y = mx + b] by picking three x's, calculating y's

Intercepts

Definition: x-intercept, y-intercept;  How to calculate?

Ex:  find intercepts for equation [of form ax + by = c]; plot points and graph equation

Ex 5:  Special cases
      line through the origin

      vertical

      horizontal

 

 

 

 

 

 

Using function notation:  find s(10) given s(t) = 60/t  [average speed in mph over a 1-mile course]

Notations like f(g(2))          Ex 8

#63 is an example of a step function

 

2.7  Algebra with Functions

 

+, -, *, /

Definitions:  (f+g)(x) = f(x) + g(x)

Illustrate with f(x) = x2+3x, g(x) = x/(x+1)

      Find ( f ± g)(x), (f g)(x), (f / g)(x)

Domain:  individual functions must be defined, result must be defined, and for division, gΉ0.

Find (f g)(6)

            Either find individual function results and then multiply, or

            Use the combined function (f g)

 

Additional operation:  substitution ("Composition  f o g")

The idea:  Output of one function is used as the input for the other function

      Define (f o g)(x) = f(g(x)), that is the output of g is used as the input of f.

      Ex:  calculate (f o g)(2) by first calculating g(2)

      Now find general expression (f o g)(x)

            Check:  calculate (f o g)(2) with the new formula

      Contrast (g o f)(x)  The point:  result is different:  in general composition is not commutative.

      Domain:  for (f o g)(x), g must be defined at x, and f must be defined at g(x)

Ex from geometry:

      Familiar: For a circle, C=pd and d=2r, so C=p(2r) = 2pr

      In function notation,

            C(d) = pd and d(r) = 2r

            So C(r) = C(d(r)) = C(2r) = p(2r) = 2pr     Notice the composition at the second step.

2nd Ex from geometry:

      Familiar:   For a circle, A = pr2, and r = d/2,  so A = pd2/4

      In function notation,

            A(d) = A(r(d)) = A(d/2) = p(d/2)2 = pd2/4

 

 

Return to:  Merced College; Don Power               Updated 09/02/08 by Don Power