REVIEW FOR TEST 3

Test 3 will cover all assignments from chapters 4 and 8.  Chapter 4 shows how to plot a point and read a point on the Cartesian coordinate system.  The x-axis runs horizontally (left and right) and the
y-axis runs vertically (up and down).  A point is read first by its’ x-location and then by its’ y-location on the graph.  The point is identified as (x,y).  The point where the two axes meet is called the origin and is designated (0,0).  Please make sure that you understand the four quadrants of this system.  They are numbered by Roman Numeral beginning with I being the upper right quadrant and (counterclockwise)
II the upper left quadrant, III the lower left quadrant, and IV the lower right quadrant.

A linear equation in two variables, x and y, has solutions for coordinate points (x,y).  To determine if a given coordinate point (x,y) is a solution to a linear equation, substitute the x and y values of the point into the linear equation.  If a true statement results, then the (x,y) point is a solution.

Example:  Verify that (3,2) is a solution to the linear equation 2x + 5y = 16.  Substitute (3,2) into the equation [where x = 3 and y = 2] to obtain 2(3) + 5(2) = 16.  Since this is a true statement, then (3,2) is a solution to the linear equation.

THIS IS VERY, VERY IMPORTANT … YOU MUST BE ABLE TO SOLVE THE EQUATION FOR Y AND IDENTIFY THE SLOPE AND Y-INTERCEPT TO ANSWER SEVERAL QUESTIONS ON THE TEST REGARDING SLOPE AND Y-INTERCEPT!

The following example shows how to solve a linear equation for the variable y:

(1)  Solve the equation for y.                       

         …subtract 2x from both sides, notice that the x term is 1^st

                      …divide all terms by

                              …the slope of the line is  and the y-intercept is

                                                …which means m = , and b = (0,-4)

Graphing lines may be accomplished using one of several methods.  You must find three points to determine a given line.  The easiest method (in my opinion) is the slope-intercept method.  If you have a linear equation (the equation which would produce the graph of a line), solving the equation for y would produce the slope-intercept equation of the line.  The form of this line would then be , where m is the slope and b is the y-intercept. Once you have the line in this form, you simply recognize the coefficient of the x (the number in front of the x) as the slope and the constant number (the number that is not in front of a either variable) as the y-intercept.

The line could then be easily graphed by first plotting the y-intercept which is “b”.  In the classroom,
I refer to “b” as the beginning point to plot for the line; it is on the y-axis.  The y-intercept (or “b”) is the point (0,b) and for the example above that point would be (0, -4).  Next using the slope, plot two additional points.  Remember that m = .  In the classroom, I say that “m” tells us how to move to the next point from the y-intercept.  The numerator (top number) of the slope tells you how many units to move vertically (up or down depending on whether the number is positive or negative) and the denominator (bottom number) of the slope tells you how many units to move horizontally (right or left depending on whether the number is positive or negative).  Since the slope in example (1) above is , then from (0,-4)  we would move up two units and to the right 3 units, putting us at the point (3,-2) where we would plot the second point.  Moving another 2 units up and 3 units to the right, we would plot the third point (0,6).  Now that we have three points, using a straight-edge we could draw a line through the points with arrows at the ends of the line indicating that there are many points (solutions) on that particular line.

To determine the x and y intercepts of a line, first let x = 0 and calculate the y value, then let y = 0 and calculate the x value.

(2)  Find the x and y intercept of the line  (above).

First let x = 0, then  and   ...  [ y-intercept ]

Next, let y = 0, then  and    …  [ x-intercept ]

Notice, that here we calculated the same points (solutions) for the line that were found by the slope-intercept method mentioned in the preceding paragraph.  To insure that you have the correct line it is always advised that you find a third point on the line.  This point could be found by letting x (or y) take on any value and calculating the corresponding value.  For example, let x = 6 then , which solving for y would produce y = -2;  a third point on this line could be (6,-2).  Plot these three points on a coordinate graph, as we did above, and you would have graphed the same line only using a different method!

Please understand and review vertical and horizontal lines and their slopes.  A vertical line (running up and down) is a line such as x = 2.  The slope of a vertical line is undefined.  A horizontal line (running left to right) is a line such as y = 2.  The slope of a horizontal line is zero.  You must be able to graph these lines and understand their slopes for the test.

The slope formula is .  This formula is used to determine the slope of the line between two points.  For example, given two points,  and , you would find the slope by substituting the appropriate values into the formula.

  ... .  The slope of the line between these points would be .

 

Parallel lines are lines whose slopes are the same; Perpendicular lines are lines whose slopes are negative reciprocals (in other words, the product of their slopes is negative 1.)
(3)  For example, given the line  (above), we have already determined that  and .  If given the line , solving for y, we would get  and find that  and .  Since the slopes are the same, these lines are parallel.

Suppose we were comparing these lines to the line -6x -4y = 16 .  Again solving for y, we would get  and find that  and .  This line would be perpendicular to the other lines since the product of their slopes equals negative 1 … .

You need to be able to find the equation of a line given a point on the line and the slope.

You will find the equation using the point-slope form: .
(4)  Find the equation of a line through (4,1) whose slope is .

                … substituting (4,1) for and

                           … solving for y

Section 4.6 deals with graphing linear inequalities.  Graphing inequalities is very similar to graphing lines.
To graph a linear inequality rewrite the inequality in the y = mx + b form and graph 3 points using the slope (m) and the y-intercept (b.)  Before you draw the line connecting the points observe the inequality.  If the inequality is £ or ³, use a solid line to connect the points on the line.  If the inequality is > or <, use a slashed or dotted line to connect the points on the line.  Next, you will have to choose whether the solution lies “above” the line or to the right (if the inequality is > or ³) or whether the solution lies “below” the line or to the left (if the inequality is < or £).  You may also decide to test a point.  Pick a point that is not on the line and substitute the test point into the inequality (it is best if you substitute the point into the original, unchanged inequality.)  If the inequality is true using the test point, then shade the region where the test point lies.  If the inequality is false using the test point, then shade the region on the opposite side of where the test point lies.

Section 4.7 is an introduction to functions.  You need to know how to determine the Domain and Range of a function, how to determine whether or not a set of points or a graph is a function, and how to evaluate a function at a given value for x.

First, the Domain of the function is the x-values of the function, the Range of the function is the

y-values of the function.

(6)  Determine the Domain and Range of the following function.

{(0,1),(2,4),(4,7),(6,10),(8,13)}

Domain = {0,2,4,6,8}

Range = {1,4,7,10,13}

A set of points is a function, if and only if, for every value of x there is a “unique” value for y.  In other words, when you look at a set of points, if an x value is repeated then it had better have the same value for y in both places!

(7)  Determine whether or not the following sets of points are or are not functions.

  (A)     {(1,2),(3,5),(4,5),(5,8),(9,1)}               … D = {1,3,4,5,9}  and  R = {1,2,5,8}

                                                                        This is a function since no value of x is repeated.

  (B)     {(1,2),(3,5),(4,5),(5,8),(1,9)}               … D = {1,3,4,5}  and  R = {2,5,8,9}

This is not a function, since x = 1, is paired with both y values of 2 and 9.

The vertical line test is another way to determine whether or not a graph is a function.  If any vertical line drawn through the graph (picture) of any equation should touch the graph in more than one point, then that equation is not a function.

Function notation is a different way to express that a particular equation is a function.  If y is replaced by the new notation, then the result is y = f(x).  To evaluate a function at a particular value for x, substitute the value given and calculate.

(8)  Find f(-1) for the function f(x) = 2x -4.
(Note: this would have formally been asked by stating y = 2x -4, find the value of y when x = -1)

f(x)   = 2x –4

f(-1) = 2(-1) -4

f(-1) =  -2 –4

f(-1)  = -6

(9)  Find f(-2) for the function .

           

           

           

           

           

Chapter 8 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 5, 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 8.1:  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.

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.

Section 8.2:  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.

Section 8.3:  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 8.4:  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

I do not like charts and you will not be given a chart on the test.  I prefer to identify variables and proceed with the problem in the same fashion as the other examples above.

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

Section 8.5:  To graph a linear inequality rewrite the inequality in the y = mx + b form and graph 3 points using the slope (m) and the y-intercept (b.)  Before you draw the line connecting the points observe the inequality.  If the inequality is £ or ³, use a solid line to connect the points on the line.  If the inequality is > or <, use a slashed or dotted line to connect the points on the line.  Next, you will have to choose whether the solution lies “above” the line or to the right (if the inequality is > or ³) or whether the solution lies “below” the line or to the left (if the inequality is < or £).  You may also decide to test a point.  Pick a point that is not on the line and substitute the test point into the inequality (it is best if you substitute the point into the original, unchanged inequality.)  If the inequality is true using the test point, then shade the region where the test point lies.  If the inequality is false using the test point, then shade the region on the opposite side of where the test point lies.

When solving a system of inequalities, graph both inequalities onto one coordinate axis and shade each inequality appropriately.  Once you have graphed both inequalities, note the intersection (overlapping) of the two graphs.  The intersection will be the solution.