ACCUPLACER MATH 0311 OR MATH 0120

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1 The University of Teas at El Paso Tutoring and Learning Center ACCUPLACER MATH 0 OR MATH 00

2 MATH 0 OR MATH 00 Page Factoring Factoring Eercises 8 Factoring Answer to Eercises 9 Rational Epressions Rational Epressions Eercises 5 Rational Epressions Answers to Eercises 6 Linear Functions 8 Linear Functions Eercises Linear Functions Answer to Eercises 5 Graphing Linear Equations 6 Systems of Linear Equations 9 Systems of Linear Equations Eercises Systems of Linear Equations Answers to Eercises Linear Inequalities 5 Equations Involving Absolute Value 7 Inequalities Involving Absolute Value 8 Linear Inequalities and Absolute Value Eercises 0 Linear Inequalities and Absolute Value Answers to Eercises Graphing Linear Inequalities in Two Variables

3 Graphing Linear Inequalities in Two Variables Eercises 5 Graphing Linear Inequalities in Two Variables Answers to Eercises 6 Radicals 8 Radicals Eercises 5 Radicals Answers to Eercises 5 Quadratic Equations 5 Quadratic Equations Eercises 60 Quadratic Equations Answers to Eercises 6

4 FACTORING Factoring is similar to breaking up a number into its multiples. For eample, 0 5. The multiples are 5 and. In a polynomial the procedure is more complicated because variables are involved in addition to numbers. There are different ways of factoring an equation depending on the compleity of the polynomial. Factoring Out the Greatest Common Factor The first thing to do when factoring is to look at all terms and break up each term into its multiples: Eamples: Factor: ( ) In this polynomial the only variable common to all is. ( 5) + + The two is also common to all terms. Therefore, this is as far as the polynomial can be factored. Factoring by Grouping Factor: The common variable is, and the smallest eponent is. The common multiple is Therefore, the greatest common factor is. ( ) You can also look at it this way: ( )( ) ( )( ) ( )( ) There will be occasions when there are no common factors for all terms. Some terms will have a variable in common, while other terms will have a different variable in common. In these cases, we can factor by grouping together the common terms. Eamples: Factor: a + ay + b + by a + ay + b + by Notice how there are two variables, and y. Group a + b + ay + by together the s and the y s and then a + b + y a + b factor out the common factor in each group. ( ) ( )

5 Factor: + y y + There are two variables present, and y. y y Start by grouping two terms at a time. ( ) + y ( ) Notice how the two polynomials inside the parentheses are similar. To make them the same change the sign in the second polynomial. ( ) y ( ) Now that they are the same, factor out the common term. y Now it is completely factored out. ( )( ) Factoring a Difference of Squares For a difference of squares, Eamples: y ( y)( + y) a b, the factors will be ( a b)( a + b). ( ) ( ) ( )( ) ( )( )( ) Factoring the Sum and Difference of Two Cubes For the addition or subtraction of two cubes, the following formulas apply: ( )( ) ( )( ) a + b a + b a ab + b a b a b a + a + bb Eamples: + 8 ( ) ( ) ( ) ( )( ) + Manipulate to be in a + b form. ( ) ( )( ) + + Follow formula y 7 Manipulate to be in a b y Follow formula. ( ) ( ) ( y ) ( y) ( y)( ) ( ) + + ( y )( y 6y 9) + + form.

6 Factoring Trinomials Factoring trinomials is based on finding the two integers whose sum and product meet the given requirements. Eample: Find two integers whose sum is and whose product is 0. Step The product is positive. The sum is negative. Therefore, both integers must be + and + negative. ( )( ) ( ) ( ) Step Possible pairs of factors that will give a positive answer:. ( )( 0) 0. ( )( ). ( )( 5) 0. ( )( ) Step The pair whose sum is is 5 and 6. Therefore, the critical integers are and 6, ( ) Factoring Trinomials of the Form + b + c To factor trinomials of the form + b + c follow the same principle described above. Find two integers whose sum equals the middle term and whose product equals the last term. Eamples: Factor: 6 Step Find two integers whose sum is equal to and whose product is equal to 6. The product is negative; therefore, we need integers that have different signs. The sum is negative, so the bigger integer is negative. Step Possible pairs of factors that will give us 6:. ( )( ) 6. ( )( ) 6. ( )( 6) 6. ( )( 6) 6 Step The pair whose sum is is and, ( + ) The critical integers are and. Step The factors for 6 are ( )( + ).. 5

7 Factor: Step Find two integers whose sum is 6 and whose product is 60. Both integers must be positive since their sum and product are positive. Step Possible pairs of factors:. ( )( 60) 60. ( )( 0) ( )( ). ( )( 0) 60. ( )( 5) ( )( ) Step The pair whose sum is 6 is (6, 0). Therefore, the critical integers are 6, 0. Step The factors for are ( + 6)( + 0 ). Factoring Trinomials of the Form a + b + c In this type of trinomial the coefficient of is not, therefore we must consider the product of Eamples: Factor: 7 Step Find two integers whose sum is 7 and whose product is ( )( ) a c. 8. The integers must be of different signs since their product is a negative number. The larger integer is negative since the sum is negative. Step Possible pairs of factors:. ( )( 8) 8. ( )( ). ( )( 8) 8. ( )( ) 8 8 Step The pair whose sum is 7 is and 8. The critical integers are and 8. Step Use the critical integers to break the first degree term into two parts: Step 5 Factor by grouping the first two and last two terms: 8 + ( ) + ( ) There are now two terms in the epression. The binomial factor in each of these two terms should be the same; otherwise, there is an error. In this case, the common binomial factor is( ). Step 6 Factor out the common binomial factor: ( ) + ( ) ( )( + ) 6

8 Factor: Step Find two integers whose sum is and whose product is 6(0) 0. Since the product is positive and the sum is negative, the two integers are negative. Step Possible pairs of factors:. ( )( 0) 0. ( )( 0) 0 7. ( )( ). ( )( 60) 0 5. ( 5)( ) 0 8. ( )( ). ( )( 0) 0 6. ( 6)( 0) Step The sum of 8 and 5 is. Therefore, the critical integers are 8 and 5. Step Use the critical integers to break the first degree term into two parts: Step 5 Factor, separately, the first two and last two terms: ( 5) ( 5) Step 6 Factor out the common binomial factor. ( 5) ( 5) ( 5)( ) 7

9 FACTORING EXERCISES ( ) 5( ) ( ) y y a ( + 5) + ( + 5) t + t y + y. ( + ) + ( + ) + 5( + ) y + y a

10 FACTORING ANSWER TO EXERCISES. ( ) 5( ) ( ) ( )( 5 ) ( ) ( ) ( )( ) + ( )( 6 + ) + ( )(( ) + ( )) ( )( + )( ) ( ) ( + ) ( + ) 9( + ) ( 9)( + ) y ( 0 + 5) y ( 5) y 9 ( + )( ) Difference of squares: ( a b ) ( a b)( a + b) ( )( + )( + ) where ( ) ( 5 y)( 5 + y) a 5 and b y y 6. ( ) ( ) a y y Difference of cubes: Difference of squares: ( a b ) ( a b)( a + b) where ( )( ) y y y 6 6 a b ( a b)( a + a + bb ) a b y and a a and b + + a a a a a ( a) Sum & Difference of Cubes: ( )( ) ( )( ) a + b a + b a ab + b a b a b a + a + bb ( + y)( y + y )( y)( + y + y ) ( + y)( y)( + y + y )( y + y ) 7. ( + 5) + ( + 5) + ( + 7)( + 7) ( + 7) 9

11 8. 6 ( )( ) ( ) ( ) t + t t t t t + t+ 9. ( ) ( ) ( )( + ). 8 5 ( + 5)( ). + y + y ( + y)( + y). ( + ) + ( + ) + 5( + ) ( + )( + )( + 5). + ( + )( ) 5. y + y+ + 6 ( + )( y+ ) 6. 8a 50 ( a+ 5)( a 5) ( + 5)( + ) ( 5)( + 5) ( + )( + 5) 0. + ( )( + ) 0

12 RATIONAL EXPRESSIONS Rational epressions can be described as a polynomial fraction, or as the ratio of two polynomials. Eamples: Working with rational epressions is similar to working with fractions. Simplify the fraction as much as possible to work with simpler terms. In the case of rational epressions we simplify by factoring each polynomial and canceling terms. Reducing Rational Epressions to the Lowest Term To reduce a rational epression to its lowest term, factor each polynomial and cancel like terms that are both in the numerator and the denominator. Eamples: Simplify ( + )( ) + ( ) Factor each polynomial, if possible. Cancel like terms. 9 Reduce to the lowest terms ( + )( ) ( + )( + ) + Multiplication of Rational Epressions Multiplying two rational epressions is done in the same way as fraction multiplication. Multiply the numerators together and the denominators together. However, it is easier if you factor each polynomial before multiplying them together. If you multiply two polynomials you ll end up with a really big polynomial, therefore it is easier to first factor and then combine.

13 Eample: Multiply + and 9 ( + )( ) ( )( ) Factor each polynomial completely. ( + )( ) ( )( + )( )( + ) ( + )( ) Division of Polynomials As with multiplication, division of polynomials is similar to fractions. The same rules apply. Multiply the first rational epression by the reciprocal of the second epression. Remember to factor each polynomial before multiplying to simplify the polynomials. Eample: Take the reciprocal of the second fraction and multiply the epressions Factor each polynomial completely. ( )( ) ( )( ) ( )( ) ( )( ) Cancel common factors. Addition and Subtraction of Rational Epressions To add or subtract rational epressions it is necessary to first find the LCD, or least common denominator, as with fractions. The LCD for a set of rational epressions is the smallest quantity divisible by each of the denominators. + Eamples: Add: Step Factor both denominators. ( )( + ) ( )( ) Step The LCD will consist of each different term obtained from the factors of both denominators.

14 The LCD is: ( )( + )( ) Step Change each rational epression to an equivalent epression having the LCD as its denominator. Multiply the missing term from the LCD to the numerator. ( )( ) ( )( + )( ) ( + )( + ) ( )( + )( ) The LCD is( )( )( ) + ; the missing factor here is ( ). The LCD is( )( )( ) + ; the missing factor here is ( + ) Step Now that we have a common denominator, an LCD, we can add the two rational epressions together. ( )( ) ( )( + )( ) + ( + )( + ) ( )( + )( ) ( )( ) + ( + )( + ) ( )( + )( ) Eliminate parentheses by distribution ( )( + )( ) Add common terms + 7 ( )( + )( ) Final answer. Subtract: Step Factor both denominators. ( )( ) ( )( ) Step The LCD will consist of each different term obtained from the factors of both denominators. The LCD is ( + )( + )( )

15 Step Change each rational epression to an equivalent epression having the LCD as its denominator. Multiply the missing term from the LCD to the numerator. ( )( ) ( ) ( ) ( )( ) ( )( )( ) ( )( ) ( ) ( ) ( ) ( )( )( ) Step Now that we have a common denominator, an LCD, we can add the two rational epressions together. ( )( ) ( )( )( ) ( ) ( )( )( ) ( )( ) ( ) ( )( )( ) ( + )( + )( ) ( + )( + )( ) 6 0 ( 8 5) ( + )( + )( )

16 RATIONAL EXPRESSIONS EXERCISES Reduce to lowest terms: Perform the indicated operations: y + y

17 RATIONAL EXPRESSIONS ANSWERS TO EXERCISES.. ( )( + ) ( + 5) 5 5 ( 5)( + 5) 5. + ( )( ) ( )( ) 5. + ( )( + ) + ( )( ) ( + )( ) 9 ( )( + ) ( )( + ) 6 9 ( )( + )( )( ) ( )( ) 8. + y y + ( + )( )( ) y ( )( + y)( y) y ( + )( )( + ) ( )( + )( )( + ) ( + ) ( )( + ) ( + )( + )( 5)( ) ( 5)( + 6)( + ) ( + )( )

18 . ( + ) + ( ) ( + ) ( )( + + 9) ( + + 9) ( )( + + 9) ( )( + + 9) ( )( + + 9) ( )( ) ( + )( ) ( + ) + ( ) ( )( )( + ) ( )( )( + ) ( ) ( + ) ( )( + ) ( + )( + ) ( ) ( ) + ( )( )( ) ( )( )( ) ( ) ( )( ) ( + 5)( ) ( + 5)( + ) ( + ) ( ) + ( + 5)( )( + ) ( + 5)( )( + ) 7

19 LINEAR FUNCTIONS As previously described, a linear equation can be defined as an equation in which the highest eponent of the equation variable is one. A linear function is a function of the form f ( ) a + b. The graph of a linear equation is a graphic view of the set of all points that make the equation true. The graph of any linear function is a straight line. A linear function can be represented in two ways, standard form and slope-intercept form. Standard form is a formal way of writing a linear equation, while slope-intercept form makes the equation easier to graph. Form Equation Note Standard A + By C A and B are not 0. A > 0 Slope-intercept y m + b m is the slope of the line and b is the y-intercept. To graph a linear function we must first identify the and y intercepts. The -intercept is the point where the graph crosses the -ais and the y-intercept is the point where the graph crosses the y-ais. To find the -intercept:. Set y 0 in the equation.. Solve for. The value obtained is the -coordinate of the -intercept.. The -intercept is the point (, 0), with the value found in step. 8

20 To find the y-intercept:. Set 0 in the equation.. Solve for y. The value obtained is the y-coordinate of the y-intercept.. The y-intercept is the point (0, y), with y the value found in step. Vertical Lines Equations of the form a are vertical lines. The -coordinate of every point on the vertical line a has the value "a," or any given number. Horizontal Lines Equations of the form y a are horizontal lines. The y-coordinate of every point on the horizontal line y a has the value "a," or any given number. 9

21 Slope The slope of a line refers to the slant or inclination of the line. The slope is the ratio of the vertical change to the horizontal change between two points on the line. The slope can also be called the rise over run ratio because it tells how many spaces to move up or down and how many spaces to move to the right. A positive sign will move the line up and a negative sign will move the line down. One important thing to remember is that the run will always be to the right, regardless of the sign. rise change in y m run change in y The formula to find the slope of a line passing through the points ( ) and ( ),y,y is: y y m Note: A horizontal line has slope of 0, while a vertical line has an undefined slope. Eample: Find the slope of the line: Use any two points on a line to calculate its slope. Your answer will be the same no matter which points you choose. y Choosing the points (, ) and (, 0): y m y

22 Parallel Lines In the y-intercept form equation, y m + b, m is the slope and b is the y-intercept. Two lines are parallel if their slopes are the same ( m m ) and their y-intercepts are different ( b b ). Eample: ) y + 9 m b 9 ) y m b In the previous eample the slope of both equations is the same and their y-intercepts are different. Therefore, the lines are parallel. 8 6 y Perpendicular Lines Two lines are perpendicular if the slopes are the negative reciprocal of each other: m Eample: ) y + 7 m b 7 ) y + m b. m The slope of equation is the negative reciprocal of the slope of equation. Therefore, the lines are perpendicular. 8 6 y

23 Finding the Equation of a Line To find the equation of a line when only a set of points or a slope is given, use the point-slope form of a linear equation formula: y y m where m is the slope of the line and ( ) ( ), y is a point on the line. Eamples: Find the equation of the line which passes through the point (, ) and whose slope is 5. Using the point-slope form equation we obtain: m( ) ( ( ) ) y y y 5 y y 5 + Resultant Equation Find the equation of the line through the points (, 5) and ( 6, ). First find the slope between these two points using the slope equation: m y y 5 6 ( ) With the slope obtained and one of the two points given, use the point-slope form equation to find the equation of the line. y y m ( ) y 5 ( ( ) ) y 5 + y + Resultant Equation Eventually, you may be asked to find the equation of a line that is parallel or perpendicular to a given line. Remember that the slopes of two parallel lines are eactly the same and that the slopes of two perpendicular lines are the negative reciprocals of each other.

24 Eamples: Find the equation of a line that is parallel to y + and that passes through the point (, 7). y + Since the equation is in slope-intercept form, the coefficient of is m the slope of the line. Remember ym+b (m is the slope!). Since the lines must be parallel, use the same slope and the given point to find the equation of the parallel line: m m ( ) ( ) y y m y 7 y 7 8 y Parallel line to y +. Find the equation of a line perpendicular to + y 9and that passes through the point (, ). + y 9 This equation is in standard form. We need to convert it to slope-intercept form: ym+b. y y 9 y + The slope is m. Since the second equation must be perpendicular to the first equation, find the negative reciprocal. m m m Using the slope m and the point (, ): y y m ( ) y ( ) ( ( ) ) y+ + y Perpendicular line to + 9.

25 LINEAR FUNCTIONS EXERCISES. Identify the false statement below a) The standard form of a linear equation is y a + b b) The point (, 0) is on the graph of y c) The graph of y passes the vertical line test for functions d) The y-intercept of y is (0, ). Identify the false statement below a) To find the -intercept of a graph, set y equal to zero and solve the equation for b) The graph of is a horizontal line three units above the -ais c) Generally speaking, it is a good idea to plot three points when constructing the graph of a linear function d) To find the y-intercept of a graph, set equal to zero and solve the equation for y. The standard form of the equation 5y 6 0 is 6 a) 6+ 5y 0 b) y c) 5y 6 0 d) 6 5y 0. Identify the true statement below a) All linear graphs are functions b) The graph of y is a function c) To find the y-intercept of a function, let y equal to zero and solve for d) The graph of is a function 5. Determine the slope and y-intercept of each equation: a) 6+ 7y b) c) y Use the point slope form to find the equation of the line which passes through (, ) and whose slope is Write in standard form the equation of the line passing through the point (, ) with slope equal to m. 8. Which of the following is the equation of a line in standard form passing through the point (, 0) and perpendicular to the line y? a) + y 6 b) + y c) y 6 d) y 9. Find the equation of a line in standard form that passes through the points (, ) and (, ). 0. Determine whether the two lines are parallel, perpendicular or either: 5y 6 and 6 0y 7

26 LINEAR FUNCTIONS ANSWERS TO EXERCISES. A is the false statement. The standard form of a linear equation is A + By C.. B is the false statement. The graph of is a vertical line units to the right of the y-ais.. D: 6 5y 0,A> 0. B: y is an eample of a linear function a) m and b 7 b) m undefinded and b does not eist c) m 0 and b 6. y y 8. B: + y is a perpendicular line to y 9. + y 7 0. The lines are parallel. 5

27 GRAPHING LINEAR EQUATIONS A linear equation can be defined as an equation in which the highest eponent of the equation variable is one. When graphed, the equation is shown as a single line. A linear equation has only one solution. The solution of a linear equation is equal to the value of the unknown variable that makes the linear equation true. A linear equation in the form of a + by c is said to be in standard form. A linear equation in the form of y m + b is said to be in slope intercept form. The solution set for linear equations in either form is not just a single number but a set of ordered pairs of numbers (,y ). The solution set can be represented graphically on a Cartesian coordinate system, and its graph is a straight line. Graphing a Linear Equation in Standard Form The easiest way to graph a linear equation in standard form is by plotting its points. Different set of points can be found by giving one of the variables a value and solving for the other variable. At the end all the points can be plotted in the graph and joined together by a straight line. Eample: Draw a graph of + y Step Assign values to one unknown and calculate the corresponding values for the other unknown. y Step Plot these points on a graph. Connect the points with a straight line. If the points do not lie in a straight line there is probably an error in one of the points used y 5 6 6

28 Graphing Linear Equations in Slope Intercept Form Linear equations in slope intercept form can also be graphed using the points system. However, the form in which the equation is written clearly tells us the slope and the y-intercept and thus makes it easier to plot with just two points. Remember that in the form y m + b, m is the slope and b is the y- intercept. This is the slope intercept form. Eample: Graph + y Step Since the equation is in standard form, solve for y to get the equation in slope-intercept form y m + b y Step Remember that in the slope-intercept form equation, y m + b, m is the slope and b is the y-intercept. By looking at the equation we can know our slope and y-intercept: m Slope b 6 y-intercept Step Using the slope and the y-intercept, graph the equation of the line. Start by finding the y-intercept on the graph. From there find the slope. Remember that the slope is rise over run, in this case the slope is m, which means that from the y-intercept point we need to move three spaces down since the slope is negative and then two places to the right. Remember that you always move to the right. The negative sign does not indicate moving to the left; it is only an indicator for going up or down. Always move to the right. y-intercept (0, 6) y slope starting from your y-intercept, move three units down (because of negative sign), and then two units to the right m Step After you have found the two points, draw a straight line connecting the two points together. This is the graph of your equation. 7

29 y draw a connecting line between the two points obtained 8

30 SYSTEMS OF LINEAR EQUATIONS Systems of linear equations refer to a set of two or more linear equations used to find the value of the unknown variables. If the set of linear equations consist of two equations there will be two unknown variables. If the set consists of three equations there will be three variables and so on. If there are more variables present than linear equations, the system cannot be solved. There are three general methods for solving systems of linear equations: graphically, by method of elimination and by substitution. Remember that in order to be able to solve the system, it is necessary that you have the same number of equations and unknown variables. Solving a System of Linear Equations Graphically If you are given a system of two linear equations with two unknowns the system can be solved and will have two answers, one for each of the variables. The graph of each linear equation will be a straight line, and the point of intersection of the two straight lines represents the solution to the system of equations. Thus, the solution to a system of two linear equations with two unknowns is an ordered pair of numbers (,y ). It is called a consistent system. Eample: + y 5 5+ y 7 Thus, the solution is and y, or (, ). 9

31 Method of Elimination This method consists on eliminating one variable by addition or subtraction of the linear equations. To be able to cancel a variable, the variable needs to have the same coefficient in both equations, however, most of the time this will not be the case. If the variables have a different coefficient, multiply the coefficients of each equation with the opposite equation. Eample: y + y Step Start by canceling one of the variables. You can choose whichever variable you want to start with. In this case we are going to start by cancelling the variable. To cancel the multiply the coefficient of in the first equation with the entire second equation, and vice versa. For this system we do not need to worry about changing signs since they are different already. ( y ) ( + y ) y 6 + 6y 6 Step Add both equations together to cancel the and solve for y. y 6 + 6y 6 y y y Step Use the variable that we just found, y, and plug it into any of the equations to obtain. y The solution is 5 and y, or (5, ). 0

32 Method of Substitution The method of substitution consists of solving one of the equations for any variable and then substituting the resultant equation into the other equation, thus leaving a one-equation, one-unknown system. Eample: + y y Step Pick one equation and solve for one variable. For this system we are going to use the first equation and solve for y. y y y y Step Use the new equation to substitute the value of y into the second equation. ( ) y Step Plug in the value of the variable into the equation obtained in step. ( ) 5 5 y y y y The solution is 5 and y, or (5, ).

33 Parallel Lines If two lines are parallel they have the same slope and thus no point of intersection since they run in the same direction and they never meet. When a system of linear equations consist of two parallel lines the system is said to have no solution since the lines have no point of intersection. The system is said to be inconsistent. Eample: ( ) + y + y 6y Parallel lines + 6y 7 + 6y y No solution Coinciding Lines When a system of linear equations consist of two lines that have the same straight line when graphed, then the two equations are equivalent and the lines are said to be coinciding lines. Since they are basically the same line, any point that satisfies one equation will satisfy the second equation. Therefore the system has an infinite number of solutions. The system is said to be dependent. Eample: y 8 y 8 y 8 y ( y ) + y 8 Coinciding lines 0 0 Infinite number of solutions

34 SYSTEM OF LINEAR EQUATIONS EXERCISES. + y 0 6+ y y 5 + 5y 6. + y + 5y 6. + y 5 + y 5. 7 y y y 8 7+ y 7. y 9 6y 8. + y y 9. y 8y y 6+ y

35 SYSTEM OF LINEAR EQUATIONS ANSWERS TO EXERCISES. + y y y 5 + 8y 0 6+ y 9 6+ y 9 y y + 5y y 8 y 7 y + ( ) 0 7+ ( ) 5 Solution (, ) Solution (, ). + y 6+ y 6. + y 5 y 0 + 5y 6 6 0y + y 6 + y ( ) y 0 9 y + No Solution Solution (, ) 5. 7 y 7 y y 8 + 0y 56 y 7 + y y 5 0y Lines Coincide 6( ) + 5y 8 y 8 Infinitely Many Solutions Solution (, 8) 7. y 9 + 6y 8. + y 9 + 6y 9 6y 9 6y y 8 6y Lines Coincide 7 7 Infinitely Many Solutions ( 7 ) + y y 7 7 Solution (, 7 7) 9. y + 8y y + y 8y 6 8y y y No Solution ( ) + y y Solution (, )

36 LINEAR INEQUALITIES When we use the equal sign in an equation we are stating that both sides of the equation are equal to each other. In an inequality, we are stating that both sides of the equation are not equal to each other. It can also be seen as an order relation; that is, it tells us which one of the two epressions is smaller, or larger, than the other one. A linear inequality is an equation in which the highest variable eponent is one. Eamples: < less than, < + 5 less than or equal to, > greater than, > 6 greater than or equal to, The solution to an inequality is the value of the variable which makes the statement, or the inequality, true. Eamples: < This inequality is telling us that is less than. Therefore, any number less than three is a possible solution. Remember that on the number line, any number to the left is less than a given number, and any number to the right of that given number is greater. Notice how is marked with an empty dot, this means that the answer does not include the number This inequality is telling us that the equation + 5 is less than or equal to the equation. Solve for to find which number, or numbers, make this equation true. + 5 Solve for by leaving the variables on one side and the numbers on the other side, just as you 8 would solve a linear equation. 8 8 Notice how 8 is marked with a black dot, this means that the answer does include the number

37 > This inequality is telling us that is greater than. Therefore, any number greater than negative two is a possible solution. 0 6 Solve for by leaving the variables on one side and the numbers on the other side. 0 As you can see, solving linear inequalities is very similar to solving linear equations. There is only one thing you need to keep in mind when multiplying or dividing a negative number, the direction or sense of the inequality is reversed. For eample, if the sign is >, after multiplying or dividing by a negative number the sign changes to <. Eample: + Solve for as in the previous eamples Multiply times the negative reciprocal to leave the alone and get an answer. 6 However, since we are multiplying by a negative the inequality sign must change from to. 6

38 EQUATIONS INVOLVING ABSOLUTE VALUE The absolute value of a number is always the positive value of that number. Since the absolute value is the distance of a number from the origin, and since distances are always positive, the absolute value is always a positive value. The same definition applies to equations involving absolute value. Any equation inside the absolute value bars must always be equal to a positive number. Eample: 8 This absolute value equation does not have a real answer. Any equation inside the absolute value bars must always be equal to a positive number. In this case, for to be equal to 8, must be : 8 However, since the equation is inside the absolute value bars it will always yield a positive answer 8 8 and never a negative answer. When the equation involving absolute value is equal to a positive number, the value of the variable inside the absolute value bars can be negative or positive; therefore, the equation must be made equal to a positive and a negative answer. The answer will have two different possible values of, and both will make the statement valid. Eamples: + 6 Make the equation equal to and and solve for in each case. + 6 and The answer to + 6 is and Leave the absolute value epression by itself on the left side of the equation. Then solve for as in the previous eample. 5 6 and The answer is and. 5 7

39 INEQUALITIES INVOLVING ABSOLUTE VALUE Inequalities involving absolute value are solved in a similar manner to equations involving absolute value. Since the variable inside the absolute value sign can be either positive or negative, both positive and negative values are used to solve for the variable. However, in the case of linear inequalities involving absolute value, the inequality is placed between the two values rather than in two different equations. Eamples: < To solve this linear inequality use both positive and negative values of and place the inequality in the middle without the absolute value sign. The inequality sign will be the same on both sides. < < This notation is read: is greater than but less than Write both positive and negative values of on each side of the inequality and leave it in the middle. + 0 Solve for by moving each term to both sides of the inequality Remember that when dividing by a negative number the direction of the inequality sign changes. This notation is read: is greater than but less than

40 We eplained previously that equations that contain an absolute value cannot be equal to a negative number since an absolute value will always give you a positive number. In the case of linear inequalities, if an inequality is greater than or equal to a negative number, then there are infinite solutions. If the inequality is less than or equal to a negative number, then there is no real solution to the inequality. Eamples: + The absolute value will give you a positive number which will always be greater than any negative number. The inequality is true and has infinite solutions. + The absolute value will give you a positive number and no positive number will ever be less than or equal to a negative number. The inequality is false and has no real solution. 9

41 LINEAR INEQUALITIES AND ABSOLUTE VALUE EQUATIONS EXERCISES Solve for : ( + ) < 8 7. ( + ) >

42 LINEAR INEQUALITIES AND ABSOLUTE VALUE EQUATIONS ANSWER TO EXERCISES ( + ) < < < 5 < ( + ) > > > < or 8 0 or

43 GRAPHING LINEAR EQUATIONS IN TWO VARIABLES To graph a linear inequality with one variable we use the number line. To graph a linear inequality with two variables we use a two-dimensional graph. As with linear equations, linear inequalities can be written in standard form or slope intercept form. Standard Form: Slope-Intercept Form: A + By < C or A + By > C A + By C or A + By C y < m + b or y > m + b y m + b or y m + b To graph a linear inequality with two variables it is probably easier and recommended to change it to slope intercept form. There are two important rules to remember when graphing a linear inequality. If the inequality sign is > or <, the line is dashed. If the inequality is or, then the line is solid. Also, for inequalities that are greater than > or greater or equal to the graph needs to be shaded above the line. For inequalities that are less than < or less or equal to the graph needs to be shaded below the line. Eamples: Graph the solutions of + y 6: Step Since the equation is in standard form, change it to slope intercept form. + y 6 y y y + Slope-Intercept Form Step Graph the slope-intercept form equation. Since the inequality has the sign the line is solid.

44 Step Since the equation has the sign, the graph is shaded above the line. The shaded area in the graph is the graphic answer to the inequality. It means that any point above the line, including the line, satisfies the inequality and can be a possible answer. Graph the solutions of y< Step Since the equation is in standard form, change it to slope intercept form. y< y< + + y > Remember to switch the inequality sign when dividing by a negative. y > Slope-Intercept Form Step Graph the slope-intercept form equation. Since the inequality has the sign > the line is dashed. Step Since the equation has the sign >, the graph is shaded above the line.

45 Solving a System of Linear Inequalities To solve a system of linear inequalities we need to do it graphically since the solution to a system of linear inequalities is the set of points whose coordinates satisfy all the inequalities in the system, or where both shaded areas overlap. Graph the lines as previously shown. y< + Eample: Determine the solution to the following system of inequalities y Step Change the second inequality to slope-intercept form. y y y + Remember to switch the inequality sign since we are dividing by a negative number. y Step Graph both inequalities in slope-intercept form and shade the solution for each inequality. Since the first inequality has the sign < the line is dashed. The second inequality has the sign so the line is solid. y Solution to the system of inequalities y< +

46 GRAPHING LINEAR INEQUALITIES IN TWO VARIABLES EXERCISES Graph each inequality:. y. y < y Determine the solution to each system of inequalities:. y < y y 5 y y < 6 y y y y 0 + y 6 + y 9. 0 y y 6 + 6y 8 0 y y y 800 5

47 GRAPHING LINEAR INEQUALITIES IN TWO VARIABLES ANSWERS TO EXERCISES

48

49 RADICALS For every operation there is an opposite that can be used to cancel the operation or make it equal to zero. Eample: + inverse inverse inverse inverse The opposite of eponents are called radicals. If the eponent is a square, or, the opposite would be a square root; if the eponent is, the opposite would be a cubic root, and so on. Eamples: inverse 8 inverse 8 5 inverse 5 Radicals can also be written in eponent notation; however, in this case the eponent would be a fraction. Eamples: The epression 8 is called a radical epression, where is called the inde, is the radical sign, and 8 is called the radicand. The inde of a radical epression must always be a positive integer greater than. When no inde is written it is assumed to be, or a square root. Negative Radicals The only restriction that eists for negative signs and radicals is that there cannot be a negative sign under an even root since there is no real solution to this problem. However, a negative sign can eist in front of a radical or under odd roots and still be able to obtain a real number. Eamples: 6 Negative sign under even root, no real answer. or ( )( ) 6 ( )( ) 6 Negative sign under odd root, real answer. 8 ( )( )( ) 8 8 Negative sign is outside the radical, the sign does not affect the Calculations. It is only carried out to the final answer ( )( ) 8 8

50 Radicals Containing Large Numbers When working with large numbers under a radical sign, it is always easier to break down the number and work in smaller parts rather than trying to find an eact root. Eample: 55 Break down the number into smaller numbers. 5 Find the root of each of the smaller numbers. 5 Write final answer with single digits first and radicals at the end. 5 Sometimes it is not easy to find two small numbers that when multiplied give you the large numbers. In these cases, you may break down the large number little by little. Eample: 55 Follow the same procedure as described above. It takes a little longer, but at the end the same result is 05 5 obtained Multiplication and Division of Radicals Radical multiplication is similar to polynomial multiplication, the same distributive property applies. For instance, if you had to multiply ( + ), you would take the through the parentheses to get + 6. With radicals it works the same way. Eamples: ( + 5) ( ) + ( 5) Distribute the outside radical to each term inside the parentheses ( + 5)( 6 )

51 Rationalizing the Denominator When dealing with fractions, a final answer cannot contain radicals in the denominator; therefore, it is necessary to eliminate any radical from the denominator. The process of removing the radical from the denominator is called rationalizing the denominator. Eamples: Simplify the following epression: To simplify the epression we need to eliminate the radical in the denominator by rationalizing the denominator. It is necessary to remove the radical sign from and leave the by itself. To do this, multiply by another and you'll get. Remember that, and when you multiply the eponents add together. Therefore, or. However, don t forget to multiply the numerator as well to compensate the change. Multiply the denominator by to eliminate the radical. + 6 When the denominator is a little more comple, multiply the denominator by its conjugate and the radicals will disappear. The conjugate is the same denominator but with a different sign in between: in this case ( )

52 RADICALS - EXERCISES Simplify: ( + )( ) y y ( 6 ) 5

53 RADICALS - ANSWERS TO EXERCISES y + y y

54 QUADRATIC EQUATIONS A quadratic equation is always written in the form of: a + b + c 0 where a 0 The form a + b + c 0 is called the standard form of a quadratic equation. Eamples: This is a quadratic equation written in standard form. + This is a quadratic equation that is not written in standard form but can be once we set the equation to: This too can be a quadratic equation once it is set to 0. 0 (standard form with c0). Solving Quadratic Equations by Square Root Property When a, where a is a real number, then your ± a Eamples: 9 0 y y y 5 ± 9 y ± 5 ± y ± 5 Solving Quadratic Equations by Factoring It can also be solved by factoring the equation. Remember to always check your solutions. You can use direct substitution of the solutions in the equation to see if the solutions satisfy the equation. Eamples: Factoring ( )( ) Set it equal to 0 and solve for 5

55 Now check if, and are the solutions of Check: 5( ) ( ) ( )( + ) Another method of checking the solutions is by using one of the following statements: b The sum of the solutions or The product of the solutions c a a where a, b, and c are the coefficients in a + b + c 0. Now we check if and are the solutions of Check: Using the sum of the solutions 7 + ( ) Based on the original equation b a Now by using the product of the solutions ( ) 7 Based on the original equation c a 8 0 Rewrite in standard form by multiplying each side of the equation by ( + )( )

56 Check: ( ) Solutions must be checked in the original equation to avoid any errors. Solution Using the Quadratic Formula Factoring is useful only for those quadratic equations which have whole numbers. When you encounter quadratic equations that can not be easily factored out, use the quadratic formula to find the value of : b b ac ± a Eamples: Rewrite in standard form, where a,b, and c 8 ± ( )( 8 ) ( ) Plug in numbers into the equation ± 6 ( ) ± 6, The two rational solutions + 0 ( ) ± ( ) ( )( ) ( ) ± 6 ± 6,, The two rational solutions 6 6 In some cases you encounter repeated rational solutions. And to prove you have the right values you use the discriminant which gives you information about the nature of the solutions to the equation. 55

57 Based on the epression b ac, which is under the radical in the quadratic formula it can be found in the equation a + b + c 0. I. When the discriminant is equal to 0, the equation has repeated rational solutions. Eample: + 0 By using the discriminant b ( ) ± ( ) ( )( ) ( ) ± 0 ac ( ) ( )( ) 0, Repeated rational solutions II. When the discriminant is positive and a perfect square, the equation has two distinct rational solutions. Eample: + 0 By discriminant b ac ( ) ( )( ) ( ) ± ( ) ( )( ) ( ) ±, Two distinct rational solutions III. When the discriminant is positive but not a perfect square, the equation has two irrational solutions. Eample: The discriminant b ac ( ) ( )( 6) 0 ± ( ) ( )( 6 ) ( ) ± 0 ± 0 Two irrational solutions IV. When the discriminant is negative, the equation has two comple number solutions. Eample:

58 The discriminant b ac ( ) ( )( 6) 8 ± ( ) ( )( 6 ) ( ) ± 8 Two comple number solutions Solution by Completing the Square One more method of solving quadratic equations is by completing the square. Eample: Solve by completing the square. ) If the leading coefficient is not, use the multiplication (or division) property of equality to make it : In this case the leading coefficient is already ) Rewrite the equation by sending the constant to the right side of the equation: ) Divide the numerical coefficient the middle term by, then square it, and add it to both sides of the equation, but leave the square form on the left side of the equation: ( ) ( ) ( ) ( ) Middle term coefficient 6 6 ( ) ) Once you found the squared number rewrite the equation as follows: ( ) + 6+ Bring down the variable and put it inside the parentheses Use the sign of the middle term. In this case it is +. ( + ) Write the squared number. In this case it is. The resultant binomial is ( + ) 57

59 5) Using the square root property clear the term. ( + ) ± The square root of a squared term is the term by itself. + ± 6) Solve for the variable. + ± The ± notation is used because the square root can have both positive and negative answers ( + ) ( ) And ( ) ( ) 5+ 7) Check your solution ( ) ( ) Both solutions are true: ( ) ( ) Let s keep practicing with one more. Eample: 5 0 One way to make the leading coefficient is ( 5) ( 0) by multiplying both sides of the equation by 5 0 Move the constant to the right side of the equation Divide the middle term coefficient by, square it, and add it to both sides of the equation: 58

60 Write the squared number in binomial form. ± 6 Find the square root of both sides and don t forget the ± sign. ± Send the other number to the right side of the 6 equation. ± Try to solve it using the square root. If not possible leave it in radical form. ± + Solve for. ± Final answer. 59

61 QUADRATIC EQUATIONS EXERCISES Solve each of the following equations by the method of your choice and check your solutions y y 0. y + y y + 6y y y y ( 5)( + 8) 0. ( + 6)( )

62 QUADRATIC EQUATIONS ANSWER TO EXERCISES ( )( ) 0 ( + 5)( + ) ( + )( ) 0 ( + ) 0 ( )( + ) ( + 7)( ) ( 7)( + ) ( + ) 0 ( ) 0 ( )( + ) y y ( y )( y + ) y 0 y + 0 ( 5)( + ) 0 y y y + y

63 ( y + ) 0 y + 0 y, ( )( + ) ± 0 ( )( ) ± ± ± ( )( ) ( ) 5. 9y + 6y y ± 6 ( 9 )( 8 ) ( y 5)( y + 5) 0 ( 9) 6 ± y y 5 6 ± 8 8 y y y y ( y )( y ) y 0 y 0 ( + )( ) 0 y y

64 ( 5 + )( 5) 0 ( 6 )( ) ± ( 8) 0 ( + )( ) ( 5)( + 8) 0. ( + 6)( ) ( )( 0 ) ( + 7)( ) 0 ± ( 5) ( + 5) ± 9 ( )( 5) ± 9 6

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