Summer Assignment Pre-Calculus Honors June 2016 Dear Student: This assignment is a mandatory part of the Pre-Calculus Honors course. Students who do not complete the assignment will be placed in the regular level Pre-Calculus course at the beginning of the 2016-2017 school year. Work should be done neatly and completely with all work shown and all answers clearly indicated. A mixture of odd and even problems has been assigned in order to give you the chance to check your work on several of the problems. The assignment will be graded and weighted as a test grade (100 points) for the first marking period. You will have a test on the summer assignment material within the two weeks of school. You may work with another student, or students, and you may e-mail me for help. However, any evidence of copying will result in failure of the assignment (a test grade) and may constitute grounds for removal from the class. The important thing is that YOU understand the work. You will need a graphing calculator for part of this assignment. I recommend the TI-84 or TI-84+. The TI-nSpire is ok, but I will be using the 84 in class, so if you do not own a graphing calculator, you should buy a TI-84 this summer. If you need help, pleas do not hesitate to email me. My email address is gsalmon@springfieldschools.com. I do not necessarily check email every day, and I will be going on vacation for part of the summer. Please do not wait until the last minute to email me if you are having trouble Due Date: The assignment is due in my mailbox by Thursday, August 18 th at noon. The main office is open Monday-Thursday from 8am-12 noon and 1pm-4pm, and is closed on Fridays. If you are handing it in on the last day, make sure you are here before 4pm, otherwise you will not be able to get into the main office. Do not be late--the maximum possible grade for a late assignment is a C! If you cannot drop your assignment at school, you may mail it to the school at: Jonathan Dayton High School 139 Mountain Ave. Springfield, NJ 07081 Attn: Mr. White (Pre-Calc Honors summer assignment) If you want, you may scan your assignment and convert it to a.pdf file and e-mail it to me. The assignment is broken up into several topics which appear on the following pages. At the end of each topic are several questions that you are to answer. Please use graph paper where appropriate. Some of the questions will refer you to the textbook. You must show all work to receive credit.
Topic #1: Linear Functions (See sections 1.3 1.5 in the textbook). Linear functions are the most basic types of functions you have been working with them for years. There are three basic formats that are used to write linear functions. These formats are: Point-Slope Form: y y 1 = m x x 1 ( ) through the point x 1, y 1 ( ) where the linear function has slope m and passes Slope-Intercept Form: y = mx + b where the linear function has slope m and y-intercept 0, b ( ) Standard Form: ax + by = c. When writing a function in standard form, a, b and c must all be integers and a must be positive. No fractions or decimals should appear anywhere in a linear equation in standard form. Calculating slope: The slope of a line can be found if two points on the line are known. Since the slope of a line shows the rate of change of the y-variable in relation to the rate of change of the x-variable, the formula for slope is m = y 2 y 1 x 2 x 1. Example: Find the slope of the line passing through the points (1, 4) and (-2, 8). m = 8 4 2 1 = 4 3. The slope of the line passing through these points is 4 3 Of the three different forms, point-slope form is probably the easiest to use, although you are probably more familiar with slope-intercept form. The reason point-slope form is easier is that you can use any point on the line, not just the y-intercept which is require for slope-intercept form. Example: Find the equation of the line with slope 1 3 that passes through the point (1, -2). Use point slope form: y y 1 = m( x x 1 ) and plug the slope in for m and the coordinates of the point in for x 1 and y 1. The equation of the line is y 2 ( ) = 1 ( 3 x 1 ). Since subtracting a negative is the same as adding, the best way to express the equation is y + 2 = 1 ( 3 x 1 ). To sketch a graph of this line, begin with a point at (1, -2) and from there, count up one unit and right 3 units. Plot a second point and connect the two points using a ruler. Extend the line beyond both points and indicate that the line is infinite by putting arrows at both ends.
Converting from point-slope form to slope-intercept form. To convert from point-slope form to slope-intercept form, there are two quick steps as shown in the example below. y + 2 = 1 ( 3 x 1) Distribute. y + 2 = 1 3 x 1 3-2 - 2 Isolate y. y = 1 3 x 7 3 Note that to subtract 2 from - 1 you would need to find a common denominator 3-1 3 6 3 = 7 3 The equation of the line in slope-intercept form is y = 1 3 x 7 3 Converting from point-slope form to standard form. To convert from point-slope form to standard form, follow the steps below. y + 2 = 1 ( 3 x 1) Distribute. y + 2 = 1 3 x 1 3 Multiply through by the denominator to eliminate fractions. 3 y + 2 = 1 3 x 1 3 = 3y + 6 = x 1 Move the variables to the left and the constants to the right x -6 -x -6 x + 3y = 7 Since a, the coefficient on x must be positive, multiply through by -1 x 3y = 7 The equation of the line in standard form is x 3y = 7. Locating the zero (x-intercept) of a linear function. To find the zero of a linear function, simply plug in 0 for y and solve for x. Example: Find the zero of the function x 3y = 7. x 3( 0) = 7 ( ) x = 7 So the x intercept is 7, 0
Parallel and Perpendicular Lines Parallel lines have the same slope. To determine whether two lines are parallel, convert both to slope-intercept form and compare the slopes. Example: Determine whether y = 2x 5 and 6x 3y = 12 are parallel. The first line is already in slope-intercept form. Converting the second line yields 6x 3y = 12 3y = 6x +12 y = 2x 4 Since both lines have a slope of 2, they are parallel Example 2: Find the equation of the line parallel to y = 1 x + 6 that passes through the 4 point (-2, 9). The first line has a slope of 1 4, so using point-slope form, the answer is y 9 = 1 ( 4 x + 2). Perpendicular lines have opposite reciprocal slopes. For example, if one line has a slope of 1 2, a line perpendicular to it would have a slope of 2, or simply 2. 1 Example: Find the equation of the line perpendicular to 2x + 3y = 9 that passes through the point (1, -3). First, convert the line to slope-intercept form: 2x + 3y = 9 3y = 2x + 9 y = 2 3 x + 3 Since the original line has a slope of 2 3 the perpendicular line will have a slope of 3. Now, just use point slope form to write the equation of the new line. The 2 answer is y + 3 = 3 ( 2 x 1) Vertical and Horizontal Lines. Vertical lines are in the format x = a number. They pass through the x-axis at the number in the equation. For example, the equation x = 3 will be a vertical line passing through the x-axis at 3. Horizontal lines are in the format y = a number. They pass through the y-axis at the number in the equation. For example, the equation y = -2 will be a horizontal line passing through the y-axis at -2.
Practice Problems for Linear Functions. 1. Consider the line with a slope of 1 2 that passes through the point (-2, 4). a. Write the equation of the line in point-slope form. b. Convert the equation into slope-intercept form. c. Convert the equation into standard form. d. Locate the zero of the function e. Find the line that is parallel to the given line that passes through (8, -2). f. Find the line that is perpendicular to the given line that passes through (0, 6). g. Graph three lines the original line and the lines in parts e and f on graph paper. 2. Consider the line that passes through the points (-3, 5) and (2, -4). a. Find the slope of the line. b. Write the equation of the line in point-slope form. c. Convert from point-slope form into slope-intercept form. d. Convert from point-slope form into standard form. e. Locate the zero of the function. f. Find the line parallel to the original line that passes through the point (-1, 8). g. Find the line perpendicular to the original line that goes through (8, -3). h. Graph three lines the original line and the lines in parts f and g on graph paper. 3. The three vertices of a triangle are (-1, 4), (3, 9) and (5, -2). a. Sketch the triangle on graph paper. b. Find the equation of the lines that represent the sides of the triangle. c. Are any of the sides perpendicular? How can you tell? 4. For the following pairs of lines, determine whether they are parallel, perpendicular or neither. a. Line 1: 6x + 4y = 12 and Line 2: 2x 3y = 21 b. Line 1: x + 2y = 10 and Line 2: y 4 = 1 ( 2 x + 3) c. Line 1: 9x 3y = 6 and Line 2: The line passing through (8, 3) and (5, 12). d. Line 1: y = 5x + 2 and Line 2: 2x +10y = 80 5. Graph the following lines and find their zeros. a. y = 3 4 x 1 b. y 2 = 2 ( x +1 ) c. y + 3 = 1 ( 3 x + 2 ) d. 7x 2y = 4 e. x = 5 f. y = -1
Topic #2: Domain and Range The domain of a function refers to all of the x-values that are possible for a function to use. Many functions, including common functions such as linear and quadratic functions, can take on any x-value. When this is the case, we say that the domain of a function is all real numbers. Two special cases for domain: There are two important cases where the domain of a function is not all real numbers. The first involves the square root function y = x. Since it is impossible (at least when using only real numbers) to take the square root of a negative number, the smallest value that can possibly be used for x in this function is 0. Any number greater than or equal to zero can be used, so we say that the domain of the function is x 0. When writing the domain in interval notation, the best way to think of it is that it is expressed as farthest left x-value, farthest right x-value ( ). The second special case for the domain is rational functions (fractions). Since it is impossible to divide by zero, any number that would result in division by zero cannot be included in the domain of a function. For example, in the function f ( x) = 3 Using Interval Notation: Another way to express the domain (and range) of a function is using what is called interval notation. Interval notation states the lowest and highest values of the domain and is written almost as if you are writing the coordinates of a point. For the square root function shown at left, the lowest possible x- value is 0 and there is no highest x-value (so we say the highest value is. The domain, expressed in interval notation is. The square bracket at the left indicates that 0 is included in the domain and the parenthesis at the left indicates that is not (since it is not really a number, is never actually included., using an x-value of 1 would x 1 result in division by zero. Therefore, 1 is not included in the domain. You can say that the domain of the function is all real numbers such that x 1. Using interval notation, the domain would be written in two pieces it is express as (,1) and ( 1, ). Note the use of the parentheses, indicating that, 1, and are not included in the domain. Range: Range is expressed similar to domain, except that it refers to y-values. In the square root example above, the lowest y-value is 0 and then the graph extends upwards indefinitely, meaning that there is no highest y-value. The domain is said to be y 0, or in interval notation [0, ). When writing range in interval notation, it is expressed as lowest y-value, highest y-value ( ).
Examples: 1) y = 1 2 x + 2 2) y = ( x 2)2 + 3 ( ) Range: (, ) or Domain: (, ) Range: [3, ) or Domain:, Domain: All real numbers, Range: all real #s Domain: All real #s Range: y 3 3) ( x 1) 2 4 + ( y + 2)2 9 = 1 4) y = 2 x [ ] Range: [ 5,1] or Domain: (, ) Range: ( 0, ) or Domain: 1, 3 Domain: 1 x 3 Range: 5 y 1 Domain: All real #s Range: y > 0. Note: This is not a function it does not Note that the graph never actually hits the pass the vertical line test. x-axis, so the y-value never really reaches 0, so 0 is not included in the range.
Practice Problems for Domain and Range. Find the domain and range of each of the graphs below. 1. 2. 3. Note: an open circle indicates that a point is not included in the graph and a closed circle indicates that a point is included in the graph. 4. Also examine the following graphs in the book and state the domain and range of each graph. a. Page 10 #40 b. Page 177 #13 c. Page 177 #14 5. Generate graphs of functions that match the following domains and ranges a. Domain:, b. Domain: 3, ( ) Range: (, 2] ( ) Range: [ 2, 2]
Topic #3: Adding, Subtracting, Multiplying, Dividing and Composing Functions. (Sect. 1.2) Adding Functions: Adding functions is very easy if you are simply adding two polynomial functions as shown in Example 1 below. All you need to do is simply combine like terms. Example 1: If f ( x) = 3x 2 x +1 and g( x) = x 3 2x 2 + 8x + 4, find ( f + g) ( x) ( 3x 2 x +1) + ( x 3 2x 2 + 8x + 4) = x 3 + x 2 + 7x + 5 Adding functions that involve fractions is more difficult it requires a common denominator as shown in Example 2 below. Functions involving fractions also require what are called domain restrictions, which indicate values of x that cannot be used because they would result in a zero in the denominator. 3 Example 2: If f ( x) = 2x 1 and g( x) = x x 1, find ( f + g) ( x) 3 2x 1 + x x 1 requires a common denominator. Multiply the left fraction by x 1 2x 1 and the right one by x 1 2x 1 x 1 3 x 2x 1 x 1 2x 1 + x 1 2x 1. Distribute in the numerator, not the denominator. 3x 3 ( x 1) ( 2x 1) + 2x 2 x ( x 1) ( 2x 1) 2x 2 + 2x 3 x 1 ( )( 2x 1) Then combine like terms in the numerator Then find the values that would make the denominator = 0. The values are 1 and 1 : this is your domain restriction. 2 2x 2 + 2x 3 x 1 ( )( 2x 1) ; x 1, 1 2 This is your final answer.
Subtracting functions: Subtracting functions is very similar to adding with one important step that many students often miss. If one polynomial function is being subtracted from another, as in Example 1 below, you must be sure to distribute the negative sign to the second polynomial. Then simply combine like terms. Example 1: If f ( x) = x 2 + 3x 9 and g( x) = 3x 3 + x 2 5x 1, find ( f g) ( x). x 2 + 3x 9 ( 3x 3 + x 2 5x 1) x 2 + 3x 9 3x 3 x 2 + 5x +1 3x 3 + 8x 8 Just like adding fractions, subtracting fractions requires a common denominator and domain restrictions. Like in the polynomial example, you must remember to distribute the negative sign to the numerator of the second fraction. 2x 1 Example 2: If f ( x) = 3x + 2 and g( x) = 3x2 5 x 3, find ( f g) ( x) 2x 1 3x + 2 3x2 5 requires a common denominator. x 3 " x 3% " 2x 1% " $ ' $ ' 3x2 5% " $ ' 3x + 2 % $ ' Multiply the left by x 3 # x 3& # 3x + 2 & # x 3 &# 3x + 2 & x 3 2x 2 x 6x + 3 3x + 2 ( )( x 3) 9x3 + 6x 2 15x 10 ( 3x + 2) ( x 3) ( )( x 3) and the right by 3x + 2 3x + 2 You have to distribute (FOIL) in the numerator 2x 2 x 6x + 3 9x 3 6x 2 +15x +10 Distribute the negative in the numerator 3x + 2 9x 3 4x 2 +8x +13 Combine like terms in the numerator (Leave the denominator as is) 3x + 2 ( )( x 3) 9x 3 4x 2 +8x +13 3x + 2 ( )( x 3) ; x 2,3 Don't forget your domain restrictions. 3
Multiplying functions: Multiplying functions is actually easier in some respects than adding or subtracting. When dealing with polynomials, some people find it easier to use the box method to keep track of multiplying. Example 1: If f ( x) = x 2 3x + 2 and g( x) = 2x 2 + 7x 4, find ( f g) ( x) 2x 2 +7x 4 Multiply each thing from the top of the box with each thing on the side of the box, similar to a Punnett square. Then combine like terms from within the boxes to get the final answer of: When multiplying fractions it is not necessary to find a common denominator. Simply multiply the numerators and multiply the denominators straight across. When multiplying fractions, you do not cross multiply. x Example 2: If f ( x) = 2x 1 and g( x) = x +1 x 5, find ( f g) ( x) x 2x 1 x +1 x 5 = x x +1 2x 1 ( ) ( )( x 5) or x 2 + x ( 2x 1) ( x 5) Don t forget your domain restrictions! The final answer is x( x +1) ( 2x 1) ( x 5) ; x 1 2, 5 or x 2 + x ( 2x 1) ( x 5) ; x 1 2, 5 Dividing Functions: Dividing functions is probably the easiest of the four operations. If you are dividing polynomials, just write it as a fraction, being sure to include your domain restrictions. Example 1: If f ( x) = 2x 1 and g( x) = x 2 + 5x + 6, find f g x ( ) = 2x 1 x 2 + 5x + 6 = 2x 1 x + 2 ( )( x + 3) ; x 2, 3 f g ( x).
Note: in the example, the denominator needed to be factored in order to find the domain restriction. If you don t remember how to do this, see the section on factoring. When dividing fractions, remember these three simple words: Keep. Change. Flip. This means you keep the numerator fraction as is, change division to multiplication, and flip the denominator fraction. Then treat the problem as a multiplication problem. Example 2: If f ( x) = x 2 x x 2 x x +1 2x + 3 = x 2 2x + 3 x and g( x) = x +1 2x + 3, find ( )( 2x + 3) ( ) x +1 = x 2 x x +1 f g ( x). The domain restriction is a little tricky. Since x, 2x + 3 and x +1 were all in the denominator at some point during the problem, they all must carry with them a domain restriction. The final answer is: ( x 2) 2x + 3 ; x 0, 1, 3 x x +1 2 ( ) ( ) Composition of Functions: Students often confuse composition of functions with multiplication, when in fact they have nothing to do with one another. Composition of functions involves plugging a function into another function. There are two separate notations for composition of functions that mean exactly the same thing. Your book (for whatever reason) shows it as ( f g) x will use is f ( g( x) ). ( ). A far more common notation (and the one that I Example 1: The easiest example basically involves plugging a constant into a function. As I am sure you know, if I asked you to find f 5 simply plug in 5 everywhere you see an x. f 5 ( ) for f ( x) = x 2 2x +1 you would ( ) = 5 2 2( 5) +1 = 16. Example 2: If instead of a constant you have to plug another function in, it works the same way. Start with the outside function (in the following example, f) and everywhere you see an x, plug in the entire inside function (in the example, g). Then simplify. ( ). If f ( x) = x 2 2x +1 and g( x) = x + 3, find f g( x) In the example at left, be very carful of the expression. The common mistake is to say that this is by just squaring the x and squaring the 3. Notice that I wrote out the two (x + 3) terms and then multiplied (FOILed) them out.
Practice Problems on Adding, Subtracting, Multiplying, Dividing and Composing Functions. 1. If f x ( ) = 3x 2 + 2x 4 and g( x) = 2x 3, find ( )( x) b. ( f g) ( x) c. ( f g) ( x) a. f + g d. f g x ( ) e. f g( x) ( ) f. g( f ( x) ) 2. If f x ( ) = x 2 9x + 3 and g( x) = 2x 2 + 3x 8, find a. ( f + g) ( x) b. ( f g) ( x) c. ( f g) ( x) d. f g x ( ) e. f g( x) ( ) f. g( f ( x) ) 3. If f ( x) = 3x +1 and g( x) = x2 +1 x 3, find a. ( f + g) ( x) b. ( f g) ( x) c. ( f g) ( x) d. f g x ( ) e. f g( x) ( ) f. g( f ( x) ) 4. If f ( x) = 2x + 7 x 6 and g x a. f + g d. ( ) = x + 3 x 5, find ( )( x) b. ( f g) ( x) c. ( f g) ( x) f g x ( ) ( ) e. f g( x)
Topic #4: Factoring. Factoring is a topic that you should have mastered at this point, but you may need a little additional practice and a reminder about some of the special types of factoring. Here are several examples. Example 1: Factoring ax 2 + bx + c when a = 1. Factor x 2 7x + 6. All you need to do is find two numbers that both add up to the middle number (in this case -7) and multiply to the last number (in this case +6). These numbers are -6 and -1. Therefore the factorization of x 2 7x + 6 is ( x ( x 1). Example 2: Factoring ax 2 + bx + c when a 1. Factor 2x 2 + 5x 12. This is more complicated. There are several methods you could use to approach a problem like this, but here is the one that I prefer. 2x 2 + 5x 12 Multiply the first number and the last number. Record this as "M" The middle number is your "A" number. Here M = 24 and A = 5. As in the last example, find 2 numbers that multiply to M and add to A. The numbers are 8 and 3. 2x 2 + 8x 3x 12 Bring down the first and last term, but split the middle term into the numbers found in the previous step, keeping an x with both. 2x( x + 4) 3( x + 4) Factor the greatest common factor from the 1st two terms and from the last two terms. The expressions in parentheses must match. ( 2x 3) ( x + 4) The final factorization is the GCFs times the expression in parentheses. Example 3: Factoring a 2 b 2 (Difference of Squares). This is the easiest type of factoring. There is a basic pattern that can be followed: a 2 b 2 = ( a + b) ( a b) where a is the square root of what comes before the subtraction sign and b is the square root of what comes after the subtraction sign. Factor x 2 36 and 4x 2 9y 2 Following the pattern x 2 36 = x + 6 ( )( x 6) and 4x 2 9y 2 = ( 2x + 3y) ( 2x 3y). Example 4: Factoring a 3 b 3 and a 3 + b 3 (Difference of cubes and sum of cubes). As with the previous example, there are patterns that can be followed. a 3 b 3 = a b ( ) and a 3 + b 3 = a + b ( ) a 2 + ab + b 2 ( ) where ( ) a 2 ab + b 2 a is the cube root of what comes before the + or sign and b is the cube root of what comes after the + or sign. Notice that the pattern is the same for the two problems, its just the + and signs that are different when you factor. The three
words to remember here are Same, Opposite, Plus. The first sign (the one between a and b is the same as the one in the original problem. The second sign (between a 2 and ab) is the opposite of the one in the original question. The final sign (between ab and b 2 ) is always +. Factor x 3 27 and 8x 3 +125 In the first problem, a = x and b = 3, so the factorization is x 3 ( ) ( ) ( ) x 2 + 3x + 9 In the 2 nd one, a = 2x and b = 5, so the factorization is ( 2x + 5) 4x 2 10x + 25 Example 5: Factoring by grouping for ax 3 + bx 2 + cx + d This is essentially the method used in Example 2 on the previous page. Factor x 3 7x 2 4x + 28 x 3 7x 2 4x + 28 Find the GCF of the first two terms and the last two terms x 2 ( x 7) 4( x 7) Keep in mind the expressions in parentheses must match ( x 2 4) ( x 7) The first factor is a difference of squares that can be factored ( x + 2) ( x 2) ( x 7) This is your final answer. Practice Problems for Factoring. Factor each of the following expressions: 1) x 2 15x + 54 2) x 2 +13x 48 3) 2x 2 +13x + 20 4) 6x 2 13x 8 5) 12x 2 + 29x +15 6) x 2 225 7) 64x 4 49 8) x 3 + 343 9) 8x 3 729 10) 27x 6 64 11) x 3 2x 2 + 7x 14 12) x 3 8x 2 x +8
Topic #5: Systems of Equations (Sections 2.1 and 2.2) Solving a system of equations involves solving for two (or more) variables. Generally speaking there are 3 methods for solving systems. The first, graphing and finding the point of intersection, only works easily for systems with two variables and is not especially accurate. The other two methods, substitution and elimination, are demonstrated below. Example 1: Solving a system of equations by substitution. Solve the system of equations y = 2 3 x + 4. y = 4x 1 Since both equations are equal to y, you can replace y in the second equation with what it is equal to in the first equation essentially setting the two equations equal to one another. 2 3 x + 4 = 4x 1 Add 4x to both sides (you will need to make it 12 x since you need 3 14 x + 4 = 1 3 a common denominator). Then subtract 4 from each side. 14 3 x = 5 Divide both sides by 14, remembering to keep, change, flip 3 x = 5 1 3 14 x = 15 14 y = 4 15 14 1 y = 46 14 = 23 7 Multiply across and you have found x. Plug this into either of the original equations to solve for y. The solution is 15 14, 23 7.
Example 2: Solving a system of equations by elimination (2 variables). Solve the system of equations 4x 3y = 10 5x + 2y = 8 In a situation like this, where no one variable is easy to isolate, elimination is the most efficient method for solving the system. The initial goal is to get the same number (coefficient) but with different signs in front of one of the two variables. This is accomplished by multiplying one or both of the equations by a constant. By multiplying the top equation by 5 and the bottom equation by -4, we can achieve coefficients of 20 and -20 on x. ( ) 20x 15y = 50 ( ) 20x 8y = 32 Then add the two equations together 5 4x 3y = 10 4 5x + 2y = 8 4x 3 82 23 = 10 4x + 246 23 = 10 4x = 16 23 x = 4 23 23y = 82 y = 82 23 Plug this into either of the original equations to solve for x. The final answer is 4 23, 82 23 Example 3: Solving a system of equations by elimination (3 variables). Solve the system of equations 2x + 3y z = 11 x y + 3z = 16 5x + 2y 4z = 17 1) Choose 2 of the equations and eliminate one of the variables as in Example 2. Here I have chosen the first two equations and eliminate x. ( ) 2x + 3y z = 11 ( ) 2x + 2y 6z = 32 1 2x + 3y z = 11 2 x y + 3z = 16 5y 7z = 43
2) Do the same thing with a different pair of equations. I have chosen the 2 nd and 3 rd equations. Eliminate the same variable. ( ) 5x + 5y 15z = 80 ( ) 5x + 2y 4z = 17 5 x y + 3z = 16 1 3x + 2y + 2z = 17 7y 19z = 97 3) Use the results from the last two steps and eliminate one of the two remaining variables in the same way. I will eliminate y. ( ) 35y 49z = 301 ( ) 35y + 95z = 485 7 5y 7z = 43 5 7y 19z = 97 46z =184 z = 4 4) Substitute the solution for z back into the equation at the end of step 2 to solve for y. 7y 19( 4) = 97 7y 76 = 97 7y = 21 y = 3 5) Substitute the solutions for y and z into one of the original equations to solve for x. 2x + 3( 3) 4 = 11 2x 9 4 = 11 2x 13 = 11 2x = 2 x =1 ( ) The solution is 1, 3, 4 Be sure to check your work. This solution should work for all 3 of the original equations.
Example 4: Word problems involving systems of equations. Systems of equation can be used to solve a variety of word problems. Here is an example. In one week, a music store sold 9 guitars for a total of $3611. Electric guitars sold for $479 each and acoustic guitars sold for $339 each. How many of each type of guitar were sold? The first thing you must do is define your variables what will x stand for? What will y stand for? This can best be answered by looking at the actual question (as opposed to all the lead up information). Here, the question ultimately wants to know how many of each type of guitar were sold, so the guitar types will be your variables. Let x = the number of electric guitars and y = the number of acoustic guitars. Next, you need to set up the equations. You are given two totals one for number of guitars and one for money. Therefore you will need to have one equation that represents the number of guitars sold and one equation for the money. Here are those equations. x + y = 9 479x + 339y = 3611 Now just solve as in Example 2. ( ) 479x 479y = 4311 ( ) 479x + 339y = 3611 479 x + y = 9 1 479x + 339y = 3611 140y = 700 y = 5 They sold 5 acoustic guitars. If they sold 5 acoustic guitars, and 9 guitars in total, they must have sold 4 electric guitars.
Practice Problems for Systems of Equations. Solve each system of equations below. 1) y = 3x 10 y = 1 2 x + 4 2) y = 2x 3 3x + 2y = 10 3) 4x 7y = 71 3x 4y = 30 4) 8x 5y = 12 3x + 4y = 6 5) 4x + 2y + z = 7 2x + 2y 4z = 4 x + 3y 2z = 8 6) x + 2y + 3z = 5 3x + 2y 2z = 13 5x + 3y z = 11 Solve the following word problems. 7) An adult pass for a county fair costs $2 more than a children s pass. When 378 adult and 214 children s passes were sold, the total revenue was $2384. Find the cost of an adult pass and the cost of a child s pass. 8) Matt and Ming are selling fruit for a school fundraiser. Customers can buy small boxes of oranges and large boxes of oranges. Matt sold 3 small boxes of oranges and 14 large boxes of oranges for a total of $203. Ming sold 11 small boxes of oranges and 11 large boxes of oranges for a total of $220. Find the cost each of one small box of oranges and one large box of oranges. 9) A collection of nickels and dimes is worth $3.30. There are 42 coins in all. How many of each kind of coin are there? 10) An employee at a major corporation has $2,000 to invest in her 401K. She wants to invest part of it in a mutual fund, part in company stock and part in bonds. She has decided that she wants to invest twice as much in the mutual fund as she invests in company stock. During the first three months of her investment, the mutual fund increases in value by 4.5%, the bonds increase in value by 2.6% and the company stock falls in value by 0.2%. As a result of these changes, the value of her investment increased by a total of $58. Based on this information, how much money did she initially invest in the mutual fund, bonds and company stock? Hint: you will need to convert the percentages to decimals.
Topic #6: Inequalities (Sections 1.8 and 2.6) Example 1: Solving a basic inequality in one variable. There is no difference between solving equations and inequalities except that if you multiply or divide by a negative number, you must reverse (flip) the inequality sign. 3 2x > 7 Subtract 3 from both sides 2x > 4 Divide by -2 and reverse the inequality x < 2 Single variable inequalities are graphed on a number line with an open circle used to indicate < or > and a closed circle used to indicate < or >. -2 Example 2: Solving a compound inequality in one variable. A compound inequality involves two inequalities in the same problem. To solve, it is best to split the problem into two parts, then graph the results on the same number line. 2 < 4x + 6 16 Split this into two separate inequalities 2 < 4x + 6 and 4x + 6 16 Solve each separately 4x < 4 4x 10 Don't forget to reverse the 1st inequality x > 1 x 5 2 Graph each separately. You will be looking for where the graphs overlap. You will use one open and one solid circle -1 Example 3: Solving and graphing a linear inequality in two variables. This is no different from solving and graphing a linear equation with two exceptions. You must consider the type of line to use a solid line is used for < and > and a dotted line is used for < and >. Then you must shade the region that corresponds to the solutions to the inequality. Typically, < and < inequalities are shaded below the line and > and > are shaded above the line. Solve and graph the inequality 2x 3y > 6
2x 3y > 6 First, isolate y to make graphing the line easier 3y > 2x + 6 Don't forget to reverse the inequality because you are dividing by -3 y < 2 3 x 2 Graph the line y = 2 x 2 using a dotted line and shade below. 3 Example 4. Systems of linear inequalities. For a system of linear inequalities, you will graph multiple lines on the same graph. Your shading will create one region of the graph where the shading for all of the graphs overlaps. You will be asked to find the vertices of the shaded region. This is accomplished by setting the various lines equal to each other as in Example 1 from the section on systems of equations. Solve and graph the following system of inequalities. Then find the vertices of the shaded region. y 1 3 x + 2 y 2x 5 y 1 2 x 3 Since all of the equations are already solved for y, all you need to do is graph all three lines. Since all three inequalities involve or equal to, all three lines will be solid.
The overlapping shaded regions forms the triangle shown above. To find the vertices, set each of the equations equal to each other and solve as shown below. First equation and second equation First and third equations Second and third equations 1 3 x + 2 = 2x 5 5 3 x = 7 x = 21 5 y = 2 21 5 5 y = 17 5 The vertex where the first 2 lines intersect is at 21 5,17 5 1 3 x + 2 = 1 2 x 3 5 = 5 6 x 6 = x y = 1 ( 2 6) 3 y = 0 These lines intersect at (6, 0). 2x 5 = 1 2 x 3 2 = 5 2 x 4 5 = x y = 1 2 4 5 3 y = 17 5 The final vertex is 4 5, 17 5
Practice Problems for Inequalities. Solve and graph the following on a number line. 1. 2x + 7 < 9 2. 1 3 x 5 7 4 3. 3 2 x 6 4. 2 5x 1 9 3 5. 7 3x + 4 < 20 Graph the following inequalities on graph paper. 6. y > 1 x 3 7. 5x + 3y 15 4 8. 1 2 x 2 3 y > 1 Graph the following systems of inequalities and find the vertices of the shaded region. y 1 5 x +1 y 1 4 x + 5 9. y x 2 10. y 3 2 x +1 y 4 3 x 3 y 2x 2