REVIEW FOR TEST 2

Test 2 covers topics in Chapter 4 and Chapter 5.  Chapter 4 deals with concepts of solving systems of equations.  A system of equations consists of more than one linear equation each containing the same two variables.  The solution of a system is the ordered pair or pairs which make both of the linear equations true at the same time.  To determine if an ordered pair is a solution to a system, the ordered pair must “work” in all equations in the system.  For example, if (3, 4) satisfies both 2x + y = 10 and 3x + 2y = 17, then (3, 4) is said to be the solution to that system of equations.  Since 2(3) + 4 = 10 is true and 3(3) + 2(4) = 17 is true, then (3, 4) is the solution to the system.  Try (4, 3) in the system, x + 2y =10 and 3x + 5y = 3 … (4) + 2(3) = 10 is true, 3(4) + 5(3) = 3 is false; therefore, (4, 3) is not a solution to the system since (4, 3) does not satisfy both equations.  Solving a system of two equations involves finding the ordered pair or pairs that “work” in both equations.  In Chapter 4, you will learn three ways of solving systems of linear equations: 
1) graphing, 2) substitution, and 3) elimination (or adding).

There are three possible solutions to a system of equations:
First, if the lines have different slopes then they will have one solution.  The lines when graphed will intersect in one point.  These lines are said to be consistent.
Second, if the graphs are the same … same slope, same y-intercept … then the lines have infinite solutions since every point on one line will coincide with all points on the “other” line.  These lines are said to be dependent.
Third, if the lines have the same slope, but different y-intercepts, then the lines are parallel.  Since they are parallel, they will never intersect and will not have a solution.  These lines are said to be inconsistent.

Section 4.1:  Graphing a system of equations allows a “picture” of the solution.  If two linear equations are graphed on the same axis then their solution is the point of intersection. You may review graphing lines by taking the following link:  Review Test 1

If you graph two lines onto one coordinate axis and note the intersection, if any, then you will have solved that system.  Graphing systems of equations to determine a solution is a very inaccurate method, especially when there are fractional answers.  The remaining two methods are algebraic and very accurate.

To solve a linear system of equations by the method of substitution:

1) first solve for one of the variables in one of the equations … it does not matter if you solve for x or y … check both equations and solve for the variable that has the smallest coefficient,
2)  substitute the value of the variable that you solved for in step one into the other equation … the two equations must interact with one another,
3)  step 2 will result in an equation in one variable, solve this equation for the variable,
4)  use the value found in step 3 to find the value of the remaining variable, you may use either equation to find the second value,
5)  check the solution in both equations to be sure that the solution works in both equations.

Solve the system of equations … 3x – 2y  = 19 and x + y = 8 … by substitution.

First, exam both of the equations and decide for which variable to solve.  Since the second equation has both an x term and a y term with coefficients of 1, I can choose to solve for either x or y in the second equation.  I choose to solve for x in the second equation.  Solving for x, I find that x = 8 – y.  Now, I use the value for x, which is 8 – y, and substitute that value into the first equation for the value of x.

     3(8 – y) – 2y = 19 … now solve for y

     24 – 3y – 2y = 19

     – 5y = 19 – 24

     – 5y =  – 5

      y  =  1  … now use this value and substitute it into either of the original equation to find the value of x.

     3x – 2y = 19

     3x – 2(1) = 19

     3x = 19 + 2

     3x = 21

      x = 7 … the solution set is the point ( 7, 1 )

Check:  the solution set should work in both of the equations 3x – 2y  = 19 and x + y = 8.
Since 3(7) – 2(1)  = 19 and (7) + (1) = 8, then the point ( 7, 1) is the solution to this system of equations.

To solve a linear system of equations by the method of elimination:

1)  write both of the equations in standard form, x term and y term on the left-hand side of the equation and constant term on the right-hand side of the equation,
2)  place the equations in column form (one equation under the other) and “pre-add” (in your head) the x-terms and the y-terms from both equation to see if one or both of the “like” terms will be eliminated …
if neither the x terms, nor the y terms will be eliminated, then you will have to multiply one or both of the equations by some integer so that the coefficients of either the x terms or the y terms will be additive inverses … once you have made appropriate changes to one or more of the lines by multiplying then actually add the two equations, eliminating either the x terms or the y terms,
3)  step 2 will result in an equation in one variable, solve this equation for the variable,
4)  use the value found in step 3 to find the value of the remaining variable, you may use either equation to find the second value,
5)  check the solution in both equations to be sure that the solution works in both equations.

Solve the system of equations … 3x – 2y  = 19 and x + y = 8 … by elimination.
(Notice that this is the same system that I solved previously by substitution.  I want to show you the differences in the two methods.  Actually, only the first two steps are different.  Steps 3-5 are the same for both methods.)

     3x – 2y  = 19
        x + y    = 8  … “pre-adding” I would get 4x – y = 27, so I must multiply one of the lines by some integer so that the coefficients of one of the variable terms are additive inverses.  I choose to eliminate the y terms since multiplying the second line by 2 would make – 2y and + 2y additive inverses.

     (2)x +(2)y    = (2)8 … 2x + 2y = 16 … now place the equations once again in column form and add

     3x – 2y = 19
     2x + 2y = 16
     5x         =  35
       x  =  7  … now use this value and substitute it into either of the original equation to find the value of y.

     3(7) – 2y = 19

      21 – 2y  = 19

           – 2y  = 19 – 21

           – 2y  =  –2

             y = 1 … the solution set is the point ( 7, 1 )

Check:  the solution set should work in both of the equations 3x – 2y  = 19 and x + y = 8.
Since 3(7) – 2(1)  = 19 and (7) + (1) = 8, then the point ( 7, 1) is the solution to this system of equations.

Caution ... errors are often made in describing the solution set to two special cases of systems of equations.  When solving by algebraic methods, either substitution or elimination, and you find that you have “knocked out” both the x and the y variable and your resulting equation which contains only constant numbers is false like 18 = 7, then the answer to the system is “no solution” or the empty set.  The lines are parallel and therefore inconsistent.  On the other hand, if you are solving by algebraic methods and you find that you have “knocked out” both the x and the y variable and your resulting equation its true like 6 = 6 or 0 = 0, then the answer to the system is infinite solutions.  The lines are the same and therefore dependent.  [See note above explaining the three possible solutions to a system of equations.]

Solve the system, 2x + 10y = 3 and x = 1 – 5y, by substitution.  Since the second equation is solved for x, I use that value and substitute it into the first equation:

      2(1 – 5y) + 10y = 3

      2 – 10y + 10y = 3

                          2 = 3 … which is false and therefore the system has no solution.

Solve the system, 6x + 9y = 6 and 2x + 3y = 2, by elimination.  I choose to multiply the second line by – 3 in order to eliminate the x variable.

      (– 3) 2x + 3y(– 3)  = (– 3) 2  … the second line now becomes – 6x – 9y =  – 6

      Now place the equations in column form add the two equations:

      6x + 9y = 6
   – 6x – 9y = –6
              0  =  0  … which is true and therefore the system has infinite solutions.

Section 4.3:  Solving application problems using systems of equations involves setting up two equations with two variables and solving the newly formed system by either substitution or elimination.  You will have two unknown quantities.  Each unknown quantity will be identified using a unique variable.  In other words, assign one unknown using the variable “x” and the second unknown using the variable “y”.  Then you must form two equations using those variables.  Using the two equations and either the method of substitution or elimination, you will solve finding the value of each of the unknown variables.  Here are a few examples.

1)  During the 1996-1997 National Basketball Association season, the Boston Celtics played 82 games.  They lost 52 more games than they won.  What was their win-loss record that year?

First identify the unknowns using the variables, x and y … let x = wins and let y = losses

Next, set up two equations relating the two variables.  The problem states that the Boston Celtics played a total of 82 games, obviously they “won” some and they “loss” the rest.  This information gives us the first equation, x + y = 82.  Reading the second sentence, losses (y) were (=) (+) 52 more than wins (x).  This information gives us the second equation, y = x + 52.  Now, using these two equations solve using either substitution or addition.  I choose substitution since the second equation is already solved for y.

     x = wins
     y = losses

     x + y = 82
     y = x + 52

     x + x + 52 = 82
     2x + 52 = 82
     2x = 30
     x = 15 games won
     y = x + 52 = 15 + 52 = 67 games lost

     The win-loss record was 15 – 67.

2)  Josh and Langston found that the width of their basketball court was 44 feet less than the length.  If the perimeter was 288 feet, what were the length and the width of their court?

     x = width
     y = length

     x = y – 44                                     …   since the width (x) was (=) length (y) less (–) 44 feet (44)
     2x + 2y = 288                               …   use the formula 2 width + 2 length = Perimeter
     2(y – 44) + 2y = 288          … I would use substitution since the first equation is already solved for x
     2y – 88 + 2y = 288
     4y – 88 = 288
     4y = 376
     y = 94 feet long
     x = y – 44 = 94 – 44 = 50 feet wide

3)  Julie wanted to frame several family photos, including some of her recent wedding.  She went to a discount store and purchased two 11 x 14 frames and three 8 x 10 frames costing $22 (before taxes.)  Later she returned to the store and purchased one 11 x 14 frame and two 8 x 10 frames costing $13 (before taxes.)  How much did Julie pay for each of the different sized frames?

     x = cost per each 11 x 14 frame
     y = cost per each 8 x 10 frame

     2x + 3y = 22 ... purchase of two(2)11 x 14 frames(x) and(+) three(3) 8 x 10 frames(y) at(=)$22 (22)

     1x + 2y = 13     … purchase of one(1)11 x 14 frame(x) and(+) two(2) 8 x 10 frames(y) at(=)$13(13)

I choose to use the method of elimination.  I choose to multiply the second line by –2 and then add the two equations.

     –2[1x + 2y = 13] = – 2x – 4y = – 26

         2x + 3y = 22
      – 2x – 4y = – 26
              – y   = – 4
                 y   =  $4 for each 8 x 10 frame
                 x + 2y = 13 … x + 2(4) = 13 … x = $5 for each 11 x 14 frame

4)  How many gallons each of 25% alcohol and 35% alcohol should be mixed to get 20 gallons of 32% alcohol?

     x = amount of 25% alcohol
     y = amount of 35% alcohol

     x + y = 20   … since the mixture by volume will be a total of 20 gallons
    25x + 35y = 32(20)   …%(amount) + %(amount) = %(total amount)

Choose your method and solve … x = 6 gallons of 25% alcohol
                                                     y = 14 gallons of 35% alcohol

5)  A freight train and an express train leave towns 390 kilometers apart, traveling toward one another.  The freight train travels 30 kilometers per hour slower than the express train.  They pass one another 3 hours later.  What are their speeds?

     Let x = speed of the freight train
     Let y = speed of the express train

     x = y – 30   …   freight train (x) travels (=) the express train’s speed (y) but 30 kilometers per hour  
                               slower (–30)

     3x + 3y = 390   … since rate times time equals distance (r Ÿ t = d) and the total distance between the
                                   trains is 390 kilometers

     x = 50 kilometers per hour for the speed of the freight train
     y = 80 kilometers per hour for the speed of the express train

Test 2 also covers topics found in Chapter 5.  This includes simplifying algebraic expressions using the rules of exponents.  You will also add, subtract, multiply and divide polynomials.

Section 5.1 explains exponential rules:

Product Rule:

  example: 

Quotient Rule:

  example: 

Zero Exponent Rule:

  (any base number with an exponent of zero will equal 1)

Other examples:    and  

Negative Exponent Rule:

  example: 

Power Rule:

  example:  

Power Rule (Expanded):

  example: 

An example of combination of rules:

Section 5.2 defines the following mathematical vocabulary:

polynomial - "poly" means many, "nomial" refers to numbers or terms ... thus many terms
monomial - "mo" refers to one term, example:
binomial - "bi" refers to two terms, example:
trinomial - "tri" refers to three terms, example:
coefficient - the numerical part of the term (including the sign) which sits in front of the term, example the coefficients to the terms of the trinomial are in order from left to right,

To determine the degree of the term add the exponents on the variables of the term, example: would be a (2 +1 +5) = 8th degree term. (Notice that the y factor has an "understood" exponent of "1")

To determine the degree of the polynomial find the highest degreed term of the polynomial and declare this to be the degree of the polynomial, example: since the highest degreed term is 3, this is a third degree polynomial.

To write the polynomial in descending order means to line up the polynomial beginning with the highest degreed term and descending by power to the lowest degreed term, example: should be written in descending order as
 
To add polynomials, remove the parentheses and add like terms listing the terms in descending order.                                                                                                    
                          
 

To subtract polynomials, change the signs of all of the terms in the polynomial that follows the subtraction sign and then add/subtract like terms listing the terms in descending order.                                                                                        
                          

Section 5.3  MULTIPLICATION OF POLYNOMIALS

Every term in the first polynomial must multiply every term in the second polynomial using the product rule.

MONOMIAL BY POLYNOMIAL ... use the distributive property

BINOMIAL BY BINOMIAL ... FOIL [Firsts, Outers, Inners, Lasts] ... the product will be a trinomial

SPECIAL BINOMIALS ... Multiply Conjugates [Firsts, Lasts only] ... the product will be the difference of square numbers

SPECIAL BINOMIALS ... Square a Binomial ... the product will be a perfect square trinomial

POLYNOMIAL BY POLYNOMIAL ... (larger polynomial multiplication)

Section 5.4  DIVISION OF POLYNOMIALS

DIVISION BY A MONOMIAL … Divide each term in the numerator by the term in the denominator

Section 5.5  DIVISION BY A BINOMIAL...USING SYNTHETIC DIVISION

To divide a polynomial by a binomial, make sure that both the divisor and the dividend are in descending order.  [Note:  After the degree of the dividend is established, if there are any powers missing from the dividend you must insert a "place holder" in for the missing degree.]

The binomial is in the form x - c.  Identify the "c" in the binomial.  Next identify the coefficients of the terms of the dividend.  See the form below.

     5   +12    -36    -16

 To start the synthetic division, write the first coefficient below the line.           

     5   +12    -36    -16

            _______________

            5

Next, multiply the "c" by the number that is below the line and put the product below the second coefficient listed in the dividend and then add the two numbers in the second column.

     5   +12    -36    -16

                 +10                    

            5   +22

Continue multiplying and adding in the same manner until all coefficients are divided by "c."

     5   +12    -36    -16

                 +10    +44           

            5   +22    +8

     5   +12    -36    -16

                 +10    +44   +16    

            5   +22    +8        0

To interpret the answer you must understand that the quotient will always be one degree less than the dividend.  Since the example started with a 3rd degree polynomial, the quotient (answer) will be a 2nd degree polynomial.  The numbers below the line will be the coefficients of the quotient.  Thus, the quotient for the example is
5x²    +22x    + 8.  If the last number is a number other than zero, this will be the remainder.  The example above has no remainder.