Download presentation

Presentation is loading. Please wait.

Published byManuel Standing Modified over 3 years ago

1
**Solving Linear Inequalities Solving Linear Inequalities**

We now extend our work on linear equations to linear inequalities. These are very similar to linear equations except that the expression on the left side is generally not equal to the one on the right. Linear Inequalities 7/2/2013

2
**Solving Linear Inequalities**

Meanings and Solutions What does f(x) < g(x) mean? f(x) < g(x) is an inequality It says f(x) is less than g(x) (algebraically smaller than g(x)) … for certain values of x This may be TRUE for some values of x and FALSE for others Inequalities: Meanings and Solutions What does it mean for an expression to be less than another expression? Much like an equation, an inequality is a statement that can be either TRUE or FALSE conditionally, always or never. For inequalities containing variables, we generally want to solve the inequality by finding which values of the variable make it TRUE. Strict inequalities are indicated by the less-than symbol, <, or the greater-than symbol, >. Non-strict inequalities, defined by ≤ (less-than-or-equal) or ≥ (greater-than-or-equal) are those that allow for the possibility of equality. In solving a conditional inequality we are looking for all solutions of the inequality, that is, all values of the variable that make the inequality true. 7/2/2013 Linear Inequalities 2 Linear Inequalities 7/2/2013

3
**Solving Linear Inequalities**

Meanings and Solutions Other inequality forms include ≤ , > , and ≥ Now what do we do with inequalities ? We solve them, i.e. find their solutions Convert to equivalent inequality, i.e. one with same solutions Inequalities: Meanings and Solutions What does it mean for an expression to be less than another expression? Much like an equation, an inequality is a statement that can be either TRUE or FALSE conditionally, always or never. For inequalities containing variables, we generally want to solve the inequality by finding which values of the variable make it TRUE. Strict inequalities are indicated by the less-than symbol, <, or the greater-than symbol, >. Non-strict inequalities, defined by ≤ (less-than-or-equal) or ≥ (greater-than-or-equal) are those that allow for the possibility of equality. In solving a conditional inequality we are looking for all solutions of the inequality, that is, all values of the variable that make the inequality true. 7/2/2013 Linear Inequalities 3 Linear Inequalities 7/2/2013

4
**Solving Linear Inequalities**

Finding Solutions What is a solution for an inequality? A solution for f(x) < g(x) is a value of x that makes the inequality TRUE Any x that makes the inequality false is not a solution Inequalities: Finding Solutions A solution for and inequality is simply a value of the variable for which the inequality is a true statement. Thus, for f(x) < g(x), any value of x which makes the inequality true is a solution. Any value of x which makes the inequality false is not a solution. In solving a conditional inequality (one which is true only under certain conditions) we are looking for all solutions of the inequality, that is, all values of the variable that make the inequality true. In general we are seeking the solution set for the inequality. Unlike solution sets for equations, solution sets for inequalities are generally infinite sets. This means we cannot list all the members and must use intervals or set-builder notation to describe them. Some inequalities, especially those dealing with more complex functions than linear functions, may have solution sets composed of two or more discrete sets of solutions; that is, the solution set might described as union of several intervals. It is possible that some inequalities have an empty solution set, that is, have no solution. Thus there may be no solution, there is always a solution set. 7/2/2013 Linear Inequalities 4 Linear Inequalities 7/2/2013

5
**Solving Linear Inequalities**

Finding Solutions What is the solution set for an inequality? The solution set for the inequality is the set of all solutions The solution set might be in several discrete pieces The solution set might be empty Inequalities: Finding Solutions A solution for and inequality is simply a value of the variable for which the inequality is a true statement. Thus, for f(x) < g(x), any value of x which makes the inequality true is a solution. Any value of x which makes the inequality false is not a solution. In solving a conditional inequality (one which is true only under certain conditions) we are looking for all solutions of the inequality, that is, all values of the variable that make the inequality true. In general we are seeking the solution set for the inequality. Unlike solution sets for equations, solution sets for inequalities are generally infinite sets. This means we cannot list all the members and must use intervals or set-builder notation to describe them. Some inequalities, especially those dealing with more complex functions than linear functions, may have solution sets composed of two or more discrete sets of solutions; that is, the solution set might described as union of several intervals. It is possible that some inequalities have an empty solution set, that is, have no solution. Thus there may be no solution, there is always a solution set. 7/2/2013 Linear Inequalities 5 Linear Inequalities 7/2/2013

6
**Solving Linear Inequalities**

Examples 1. 3(2x + 3) < 4x + 1 2. – x(x – 4) ≥ x – 4 TRUE for all x < –4 , FALSE otherwise TRUE for –1 ≤ x ≤ 4 , FALSE otherwise Inequalities: Examples What makes each of these inequalities true (or false) ? The first three examples are conditional inequalities, i.e. true for some values of x and false for other values. The first inequality is a true statement only if x is less than -4, so the solution set contains all such values of x. The second inequality is true only in case x is between -1 and 4, inclusive. This solution set can be described as the closed interval [-1,4]. The third inequality is true only for x greater than or equal to 2y – 4. This introduces a second variable, y. The solutions are now ordered pairs (x,y), not just simple numbers. This means the solution set is a subset of the real plane, not a subset of the real numbers. The fourth inequality is logically false, that is, false no matter what value of x is chosen. This is because x2 is always greater-than-or-equal-to 0, making the right hand side always greater-than-or-equal-to 4. There is no choice of x that makes the right hand side less than 3. The fifth inequality is logically true, that is, true no matter what value of x is chosen, since there is no choice of x that can make it false. The equality part of less-than-or-equal-to is always true. Question: What are the solution sets for the above ? Linear Inequalities 6 7/2/2013 Linear Inequalities 7/2/2013

7
**Solving Linear Inequalities**

Examples 3. 2y – 4 ≤ x TRUE for all (x,y) where y ≤ x + 2 , 1 2 FALSE otherwise 4. 3 > 4 + 2x, for x > 0 Logically FALSE WHY ? Inequalities: Examples What makes each of these inequalities true (or false) ? The first three examples are conditional inequalities, i.e. true for some values of x and false for other values. The first inequality is a true statement only if x is less than -4, so the solution set contains all such values of x. The second inequality is true only in case x is between -1 and 4, inclusive. This solution set can be described as the closed interval [-1,4]. The third inequality is true only for x greater than or equal to 2y – 4. This introduces a second variable, y. The solutions are now ordered pairs (x,y), not just simple numbers. This means the solution set is a subset of the real plane, not a subset of the real numbers. The fourth inequality is logically false, that is, false no matter what value of x is chosen. This is because x2 is always greater-than-or-equal-to 0, making the right hand side always greater-than-or-equal-to 4. There is no choice of x that makes the right hand side less than 3. The fifth inequality is logically true, that is, true no matter what value of x is chosen, since there is no choice of x that can make it false. The equality part of less-than-or-equal-to is always true. Question: What are the solution sets for the above ? 7/2/2013 Linear Inequalities 7 Linear Inequalities 7/2/2013

8
**Solving Linear Inequalities**

Examples ≤ 7 Logically TRUE WHY ? 6. (x – 2)2 + 3 ≥ 1, for x ≥ 1 Logically TRUE WHY ? Inequalities: Examples What makes each of these inequalities true (or false) ? The first three examples are conditional inequalities, i.e. true for some values of x and false for other values. The first inequality is a true statement only if x is less than -4, so the solution set contains all such values of x. The second inequality is true only in case x is between -1 and 4, inclusive. This solution set can be described as the closed interval [-1,4]. The third inequality is true only for x greater than or equal to 2y – 4. This introduces a second variable, y. The solutions are now ordered pairs (x,y), not just simple numbers. This means the solution set is a subset of the real plane, not a subset of the real numbers. The fourth inequality is logically false, that is, false no matter what value of x is chosen. This is because x2 is always greater-than-or-equal-to 0, making the right hand side always greater-than-or-equal-to 4. There is no choice of x that makes the right hand side less than 3. The fifth inequality is logically true, that is, true no matter what value of x is chosen, since there is no choice of x that can make it false. The equality part of less-than-or-equal-to is always true. Question: What are the solution sets for the above ? Linear Inequalities 8 7/2/2013 Linear Inequalities 7/2/2013

9
**Inequalities: Rules of the Road**

Solving Linear Inequalities Linear Inequalities Addition Rule: If a > b then a + c > b + c for any real c Examples If 7 > 3 then > 3 + 4 If 7 > 3 then 7 – 9 > 3 – 9 If 2x + 5 > 3 then (2x + 5) – 5 > 3 – 5 Rules of the Road Shown here are the basic rules for handling inequalities. These are just the addition and multiplication rules for equations modified slightly for inequalities. The first rule, additive cancellation, is identical to that for equations. We may add (or subtract) any expression from both sides of the inequality and preserve the truth of the statement and the direction of the inequality. The second rule, multiplicative cancellation, like that for equations allows multiplication only by non-zero multipliers. However, unlike equations positive and negative multipliers have different effects: the truth of the inequality statement is still preserved, BUT for negative multipliers the direction of the inequality is reversed The examples illustrate these two points. OR 2x > -2 … an equivalent inequality 7/2/2013 Linear Inequalities 9 Linear Inequalities 7/2/2013

10
**Inequalities: Rules of the Road**

Solving Linear Inequalities Linear Inequalities Multiplication Rule 1: If a > b and c > 0 then ac > bc Examples If 7 > 3 and 5 > 0 then 7(5) > 3(5) If 2x + 6 > 8 then ½(2x + 6) > ½(8) Rules of the Road Shown here are the basic rules for handling inequalities. These are just the addition and multiplication rules for equations modified slightly for inequalities. The first rule, additive cancellation, is identical to that for equations. We may add (or subtract) any expression from both sides of the inequality and preserve the truth of the statement and the direction of the inequality. The second rule, multiplicative cancellation, like that for equations allows multiplication only by non-zero multipliers. However, unlike equations positive and negative multipliers have different effects: the truth of the inequality statement is still preserved, BUT for negative multipliers the direction of the inequality is reversed The examples illustrate these two points. OR x + 3 > 4 … an equivalent inequality 10 7/2/2013 Linear Inequalities Linear Inequalities 7/2/2013

11
**Inequalities: Rules of the Road**

Solving Linear Inequalities Linear Inequalities Multiplication Rule 2: If a > b and c < 0 then ac < bc Examples If 7 > 3 and -5 < 0 then 7(-5) < 3(-5) If 2x + 6 > 8 then -½(2x + 6) < -½(8) Rules of the Road Shown here are the basic rules for handling inequalities. These are just the addition and multiplication rules for equations modified slightly for inequalities. The first rule, additive cancellation, is identical to that for equations. We may add (or subtract) any expression from both sides of the inequality and preserve the truth of the statement and the direction of the inequality. The second rule, multiplicative cancellation, like that for equations allows multiplication only by non-zero multipliers. However, unlike equations positive and negative multipliers have different effects: the truth of the inequality statement is still preserved, BUT for negative multipliers the direction of the inequality is reversed The examples illustrate these two points. OR -x – 3 < -4 … an equivalent inequality Question: What if c = 0 ? 11 7/2/2013 Linear Inequalities Linear Inequalities 7/2/2013

12
**Solving Linear Inequalities**

Definition A linear inequality in one variable is one that can be written ax + b > 0 in standard form, with a ≠ 0 … also includes forms with ≥ , < , ≤ Linear Inequalities: Definition We define a linear inequality using the same expression found in the definition of linear equation: ax + b. The difference is that the equality is replaced by any of the inequality symbols: <, ≤, >, ≥ . The examples illustrate some of the possible forms of linear inequalities. Each of these can be rewritten in standard form using the rules for multiplication and addition of inequalities. The goal is reduce the right hand side to 0. In the first example, we can subtract 1 from (or add -1 to) both sides, giving 2x – 5 – 1 < 1 – 1 or 2x – 6 < 0 In the second example, we can rewrite, first by subtracting 5 from (or adding -5 to) both sides: 3 – 4x – 5 ≥ 5 – 5 or -4x – 2 ≥ 0 or multiplying by -1 (and thus reversing the inequality), -1(-4x – 2) ≤ -1(0) or 4x + 2 ≤ 0 This can be further simplified by multiplying by ½ (i.e. dividing by 2) ½ (4x + 2) ≤ ½ (0) or 2x + 1 ≤ 0 In the third example we can subtract (2x + 1) from both sides (2x + 1) – (2x + 1) ≤ (3x + 7) – (2x + 1) or 0 ≤ x + 6 and in standard form (with 0 on the right) this is: x + 6 ≥ 0 . 12 7/2/2013 Linear Inequalities Linear Inequalities 7/2/2013

13
**Solving Linear Inequalities**

Examples 1. 2x – 5 < 1 2. 3 – 4x ≥ 5 3. 2x + 1 ≤ 3x + 7 2x – 6 < 0 -4x – 2 ≥ 0 4x + 2 ≤ 0 -x – 6 ≤ 0 Linear Inequalities: Definition We define a linear inequality using the same expression found in the definition of linear equation: ax + b. The difference is that the equality is replaced by any of the inequality symbols: <, ≤, >, ≥ . The examples illustrate some of the possible forms of linear inequalities. Each of these can be rewritten in standard form using the rules for multiplication and addition of inequalities. The goal is reduce the right hand side to 0. In the first example, we can subtract 1 from (or add -1 to) both sides, giving 2x – 5 – 1 < 1 – 1 or 2x – 6 < 0 In the second example, we can rewrite, first by subtracting 5 from (or adding -5 to) both sides: 3 – 4x – 5 ≥ 5 – 5 or -4x – 2 ≥ 0 or multiplying by -1 (and thus reversing the inequality), -1(-4x – 2) ≤ -1(0) or 4x + 2 ≤ 0 This can be further simplified by multiplying by ½ (i.e. dividing by 2) ½ (4x + 2) ≤ ½ (0) or 2x + 1 ≤ 0 In the third example we can subtract (2x + 1) from both sides (2x + 1) – (2x + 1) ≤ (3x + 7) – (2x + 1) or 0 ≤ x + 6 and in standard form (with 0 on the right) this is: x + 6 ≥ 0 . x + 6 ≥ 0 Question: Can each of these be written in standard form ? 7/2/2013 Linear Inequalities 13 Section v5.0 Linear Inequalities 13 2/6/2013 7/2/2013

14
**Solutions of Linear Inequalities**

Solving Linear Inequalities Definition A solution for a linear inequality in one variable is a value of the variable that makes the inequality TRUE The set of all solutions for an inequality is the solution set for the inequality Solutions for Linear Inequalities A solution is a number. The solution set is the set of all solutions. One must distinguish between numbers and sets of numbers. In general, inequalities have many solutions which is why we focus on solution sets. In the example, we wish to find all values of x that make the given inequality a true statement. As in solving equations, we wish to isolate x using the rules for inequalities: x – 5 < 1 2x < or 2x < 6 ½(2x) < ½(6) or x < 3 There are infinitely many values of x that are less than 3 and satisfy the original equality. Two common ways to express this are: Set builder notation: { x | x < 3 } Note the set symbols “{“ and “}” ; these are important Interval notation: (– , 3) Note the open interval symbols “(“ and “)” which indicate the endpoints are not included Clearly 3 is not included in the solution set. This can be verified by putting 3 into the original inequality: 2(3) – 5 < 1 ? Since 1 is not less than 1, we see that 3 is not a solution. 14 7/2/2013 Linear Inequalities ∞ Linear Inequalities 7/2/2013

15
**Solutions of Linear Inequalities**

Solving Linear Inequalities Example 2x – 5 < 1 Some solutions are: 2, 2.5, 1, 0, -5.4, … Solution Set Notation Set notation: { x x < 3 } Interval notation: ( – , 3) How many solutions? Solutions for Linear Inequalities A solution is a number. The solution set is the set of all solutions. One must distinguish between numbers and sets of numbers. In general, inequalities have many solutions which is why we focus on solution sets. In the example, we wish to find all values of x that make the given inequality a true statement. As in solving equations, we wish to isolate x using the rules for inequalities: x – 5 < 1 2x < or 2x < 6 ½(2x) < ½(6) or x < 3 There are infinitely many values of x that are less than 3 and satisfy the original equality. Two common ways to express this are: Set builder notation: { x | x < 3 } Note the set symbols “{“ and “}” ; these are important Interval notation: (– , 3) Note the open interval symbols “(“ and “)” which indicate the endpoints are not included Clearly 3 is not included in the solution set. This can be verified by putting 3 into the original inequality: 2(3) – 5 < 1 ? Since 1 is not less than 1, we see that 3 is not a solution. Question: Does this interval include 3 ? 7/2/2013 Linear Inequalities 15 ∞ Linear Inequalities 7/2/2013

16
**Solving Linear Inequalities**

Example 1: Analytical Method Solve: Solving: Solution Set is: 3 – 2x ≤ 5 – 2x ≤ 5 – 3 = 2 x ≥ – 1 { x | x ≥ – 1 } or [ – 1, ) Example 1 We show the solution set of the inequality 3 – 2x ≤ 5, algebraically and graphically. By identifying each side of the inequality as a function of x, say y1 on the left and y2 on the right, we can see where the graph of y1 is less than (i.e. below) the graph of y2. We sketch the graphs of y1 and y2 and find their point of intersection at x = -1. Note that the graph of y1 is below (that is, y1 < y2) the graph of y2 for all x > -1. We note that, since the inequality is a non-strict inequality, we include the point where x = -1. Projecting all these points on the graph of y2 onto the x-axis we find the solution set to be the interval [-1, ). We note that the interval is open on the right since is not a number but just an indicator showing there is no upper bound on the solution set. We can also express the solution set as the set of numbers greater than or equal to -1, that is { x | x ≥ -1 }. We can “graph” the solution set, as shown, on the number line in order to visualize where the solutions are located relative to all the real numbers. Linear Inequalities 16 7/2/2013 Linear Inequalities 7/2/2013

17
**Solving Linear Inequalities**

Example 1: Graphical Method Solve: Solutions: Solution Set is: y x 3 – 2x ≤ 5 y1 = 3 – 2x y2 = 5 y1 = 3 – 2x y2 = 5 (-1, 5) y1 = y2 1 3 5 -1 -3 -5 – [ Example 1 We show the solution set of the inequality 3 – 2x ≤ 5, algebraically and graphically. By identifying each side of the inequality as a function of x, say y1 on the left and y2 on the right, we can see where the graph of y1 is less than (i.e. below) the graph of y2. We sketch the graphs of y1 and y2 and find their point of intersection at x = -1. Note that the graph of y1 is below (that is, y1 < y2) the graph of y2 for all x > -1. We note that, since the inequality is a non-strict inequality, we include the point where x = -1. Projecting all these points on the graph of y2 onto the x-axis we find the solution set to be the interval [-1, ). We note that the interval is open on the right since is not a number but just an indicator showing there is no upper bound on the solution set. We can also express the solution set as the set of numbers greater than or equal to -1, that is { x | x ≥ -1 }. We can “graph” the solution set, as shown, on the number line in order to visualize where the solutions are located relative to all the real numbers. -1 { x | x ≥ – 1 } or [ – 1, ) Linear Inequalities 17 7/2/2013 Linear Inequalities 7/2/2013

18
**Solving Linear Inequalities**

Example 2: Analytical Method Solve: Solving: Solution Set is: 5x – 1 < 2x + 11 5x – 2x < 3x < 12 x < 4 Example 2 We again identify one side of the inequality as a function y1 and the other side as a function y2. Finding intersection of the two graphs at point (4, 19), we see that the graph of y1 is below the graph of y2 for all x < 4. Projecting this portion of the graph onto the x-axis gives us the solution set graphically. Since the inequality is a strict inequality, the point of intersection (4, 19) is excluded and hence 4 is excluded from the solution set. This gives us { x | x < 4 } or (–, 4) as the solution set. The solution set is shown graphically on the real-number line. { x | x < 4 } or ( – , 4 ) Linear Inequalities 18 7/2/2013 Linear Inequalities 7/2/2013

19
**Solving Linear Inequalities**

Example 2: Graphical Method Solve: Graphically: Solution Set is: y x 5x – 1 < 2x + 11 y2 = 2x + 11 y1 = 5x – 1 y1 = 5x – 1 y2 = 2x + 11 (4, 19) 2 4 -2 -4 – ) Example 2 We again identify one side of the inequality as a function y1 and the other side as a function y2. Finding intersection of the two graphs at point (4, 19), we see that the graph of y1 is below the graph of y2 for all x < 4. Projecting this portion of the graph onto the x-axis gives us the solution set graphically. Since the inequality is a strict inequality, the point of intersection (4, 19) is excluded and hence 4 is excluded from the solution set. This gives us { x | x < 4 } or (–, 4) as the solution set. The solution set is shown graphically on the real-number line. y1 < y2 4 { x | x < 4 } or ( – , 4 ) 19 7/2/2013 Linear Inequalities Linear Inequalities 7/2/2013

20
**Solving Linear Inequalities**

Notes on Notation Solving Linear Inequalities Linear Inequalities Symbols Sometimes the same symbols are used to mean different things – just as words in English can have different meanings Points and Intervals The point (2, 3) and the open interval (2, 3) use the same notation The difference is determined by context Notes on Notation Various concepts in mathematics have different origins in place, time and culture. This often leads to similar or identical notations with quite different meanings – just as in the English language (as with most languages) certain words or phrases can have vastly different meanings. In the examples shown we note that the notation (m, n) is used both to represent an ordered pair, possibly representing the coordinates of a point in the plane, and also to represent the interval from m to n, excluding the endpoints. We interpret these symbols based on the context in which it is used. For example, if we say “the point (m, n)” then clearly m and n are coordinates of point in the plane; but, if we say “the interval (m, n)” then it is obvious that m and n are the endpoints of an open interval (excludes the endpoints) and that m < n. Sometimes students confuse symbols to say nonsensical things, such as (3, 7) or [3, 7] (open and closed intervals from 3 to 7) when they mean the set of solutions 3 and 7, written {3, 7}. Misusing symbols is analogous to using bad grammar in English (or any other natural language). It creates confusion and leaves an impression of ignorance. Note that is not a number so that it can never be the endpoint of a closed interval. In other words, we never write next to a square bracket “[“ or “]” to indicate the interval begins or ends at . 20 7/2/2013 Linear Inequalities Linear Inequalities 7/2/2013

21
**Solving Linear Inequalities**

Notes on Notation Solving Linear Inequalities Linear Inequalities Points of Confusion Do not write (3, 7) or [3, 7] when you mean { 3, 7 } Never write [–, 7 ] , [ , 7 ) , (3, ] , or [3, ] WHY ? Notes on Notation Various concepts in mathematics have different origins in place, time and culture. This often leads to similar or identical notations with quite different meanings – just as in the English language (as with most languages) certain words or phrases can have vastly different meanings. In the examples shown we note that the notation (m, n) is used both to represent an ordered pair, possibly representing the coordinates of a point in the plane, and also to represent the interval from m to n, excluding the endpoints. We interpret these symbols based on the context in which it is used. For example, if we say “the point (m, n)” then clearly m and n are coordinates of point in the plane; but, if we say “the interval (m, n)” then it is obvious that m and n are the endpoints of an open interval (excludes the endpoints) and that m < n. Sometimes students confuse symbols to say nonsensical things, such as (3, 7) or [3, 7] (open and closed intervals from 3 to 7) when they mean the set of solutions 3 and 7, written {3, 7}. Misusing symbols is analogous to using bad grammar in English (or any other natural language). It creates confusion and leaves an impression of ignorance. Note that is not a number so that it can never be the endpoint of a closed interval. In other words, we never write next to a square bracket “[“ or “]” to indicate the interval begins or ends at . 21 7/2/2013 Linear Inequalities Linear Inequalities 7/2/2013

22
**Horizontal Intercept Method**

Solving Linear Inequalities Linear Inequalities Rewrite inequality in form y < 0 (or ≤ , ≥ , > ) Example 5x – 2x – 1 – 11 < 0 3x – 12 < 0 Let y = 3x – 12 y = 3x – 12 = 0 for x = 4 y x 5x – 1 < 2x + 11 y = 3x – 12 < 0 (4, 0) Horizontal Intercept Method This is just a variation on the graphical method of solution discussed earlier. We first set up the inequality by rewriting it in such a way as to have 0 on one side of the inequality (either side will do). In the example we transform the inequality 5x – 1 < 2x + 11 to the inequality 3x – 12 < 0 or equivalently x – 4 < 0 We identify the expression on the left as the function y = 3x – 12 = x – 4 and the right side as the function y = 0. Finding the intersection point of the graphs at (4,0) we see that the graph of y = 3x – 12 lies below the graph of y = 0 for all x less than 4. Since the inequality is a strict inequality, we omit 4 from the solution set. We can write the solution set as either { x | x < 4 } or as the interval (–, 4). y = 3x – 12 < 0 Linear Inequalities 22 7/2/2013 Linear Inequalities 7/2/2013

23
**Horizontal Intercept Method**

Linear Inequalities Solving Linear Inequalities Rewrite inequality in form y < 0 (or ≤ , ≥ , > ) Example y = 3x – 12 = 0 for x = 4 For y < 0 , we have x < 4 Solution set: { x x < 4 } or (– , 4) (4, 0) y x y = 3x – 12 5x – 1 < 2x + 11 Horizontal Intercept Method This is just a variation on the graphical method of solution discussed earlier. We first set up the inequality by rewriting it in such a way as to have 0 on one side of the inequality (either side will do). In the example we transform the inequality 5x – 1 < 2x + 11 to the inequality 3x – 12 < 0 or equivalently x – 4 < 0 We identify the expression on the left as the function y = 3x – 12 = x – 4 and the right side as the function y = 0. Finding the intersection point of the graphs at (4,0) we see that the graph of y = 3x – 12 lies below the graph of y = 0 for all x less than 4. Since the inequality is a strict inequality, we omit 4 from the solution set. We can write the solution set as either { x | x < 4 } or as the interval (–, 4). y = 3x – 12 < 0 – 2 4 -2 -4 ) 7/2/2013 Linear Inequalities 23 Linear Inequalities 7/2/2013

24
**Absolute Values in Inequalities**

Solving Linear Inequalities Linear Inequalities Recall: Basic Absolute Value Facts 1. x ≥ 0 for all real x 2. x = –x for all real x a = –a , for a < 0 a , for a ≥ 0 Absolute Values in Inequalities We recall the definition of absolute value to remind ourselves that when we talk about any real number a, we leave open the possibility that a might be either positive or negative. The first two facts are then quite obvious: |x| is never negative, regardless of the value of x itself, x and –x have exactly the same absolute value. The next two facts are a bit less obvious, which is why the animations show what happens if x is positive as well as what happens if x is negative. Assuming x is not zero, the trivial case, we start by placing |x| on the positive side of 0 and less than the given number b. We see that b must be positive so that –b is surely negative and is shown in its position relative to the other numbers. If we remove the absolute value, then x might be positive and –x negative, as shown in the animation. Or, as the animation shows, x might be negative and –x positive, in which case |–x| is located exactly where |x| is located. This follows from the above properties since |x| and |–x| are equal. Clearly, then whether x is positive or negative it lies between –b and b. The last animated fact shows that if |x| is larger than a positive number b, then either x > b (if x is positive) or x < –b (if x is negative). Here again, the animation leads us through the possibility that x is positive and hence x > b, and then through the possibility that x is negative, so that –x > b, making x < –b. Again we note that |–x| and |x| are equal, placing them at the same location on the number line. 24 7/2/2013 Linear Inequalities Linear Inequalities 7/2/2013

25
**Absolute Values in Inequalities**

Solving Linear Inequalities Linear Inequalities Basic Absolute Value Facts 3. If x < b then –b < x < b 4. If x > b > 0 WHY ? –b –x x –x x –x x b Absolute Values in Inequalities We recall the definition of absolute value to remind ourselves that when we talk about any real number a, we leave open the possibility that a might be either positive or negative. The first two facts are then quite obvious: |x| is never negative, regardless of the value of x itself, x and –x have exactly the same absolute value. The next two facts are a bit less obvious, which is why the animations show what happens if x is positive as well as what happens if x is negative. Assuming x is not zero, the trivial case, we start by placing |x| on the positive side of 0 and less than the given number b. We see that b must be positive so that –b is surely negative and is shown in its position relative to the other numbers. If we remove the absolute value, then x might be positive and –x negative, as shown in the animation. Or, as the animation shows, x might be negative and –x positive, in which case |–x| is located exactly where |x| is located. This follows from the above properties since |x| and |–x| are equal. Clearly, then whether x is positive or negative it lies between –b and b. The last animated fact shows that if |x| is larger than a positive number b, then either x > b (if x is positive) or x < –b (if x is negative). Here again, the animation leads us through the possibility that x is positive and hence x > b, and then through the possibility that x is negative, so that –x > b, making x < –b. Again we note that |–x| and |x| are equal, placing them at the same location on the number line. then either x > b or x < -b WHY ? b –b x –x –x x x 25 7/2/2013 Linear Inequalities Linear Inequalities 7/2/2013

26
**Absolute Values in Inequalities**

Solving Linear Inequalities Linear Inequalities Examples 1. │ 3 │ < 7 then –7 < 3 < 7 2. │–3│ < 7 then –7 < –3 < 7 –7 │3│ 3 7 Absolute Values: Examples Examples 1 and 2 show that 3 and -3 between 0 and 7 and between –7 and 0, respectively. In either case, the absolute values of –3 and 3 are less than 7. Example 3 shows that the absolute value of –7 is greater than 3. Removal of the absolute value symbols leaves –7 less than 3. Note that the only other possibility is that –7 is greater than 3, which is of course false. So, as promised, either –7 > 3 or –7 < –3. In Example 4 we have a variable expression whose absolute value is less than 9. Removing the absolute value leaves x+2 trapped between –9 and 9. By subtracting 2 from all three expressions we see that x itself is trapped between –11 and 7. Example 5 shows the absolute value of x+2 greater than 7, which is of course greater than 0. For x positive, this makes x+2 greater than 7 and leaves x greater than 5. For x negative, the difference between x and 2, in absolute value, is less than –7. We could rewrite this as │ x – 2 │> 7. So, for x negative, we find x < –9. –7 –3 │–3│ 7 Linear Inequalities 26 7/2/2013 Linear Inequalities 7/2/2013

27
**Absolute Values in Inequalities**

Solving Linear Inequalities Linear Inequalities Examples 3. │ –7 │ > 3 Rule: Replace absolute value with each of the two forms indicated in the definition then either –7 > 3 or –7 < –3 –7 –3 3 │–7│ 7 Absolute Values: Examples Examples 1 and 2 show that 3 and -3 between 0 and 7 and between –7 and 0, respectively. In either case, the absolute values of –3 and 3 are less than 7. Example 3 shows that the absolute value of –7 is greater than 3. Removal of the absolute value symbols leaves –7 less than 3. Note that the only other possibility is that –7 is greater than 3, which is of course false. So, as promised, either –7 > 3 or –7 < –3. In Example 4 we have a variable expression whose absolute value is less than 9. Removing the absolute value leaves x+2 trapped between –9 and 9. By subtracting 2 from all three expressions we see that x itself is trapped between –11 and 7. Example 5 shows the absolute value of x+2 greater than 7, which is of course greater than 0. For x positive, this makes x+2 greater than 7 and leaves x greater than 5. For x negative, the difference between x and 2, in absolute value, is less than –7. We could rewrite this as │ x – 2 │> 7. So, for x negative, we find x < –9. Linear Inequalities 27 7/2/2013 Linear Inequalities 7/2/2013

28
**Absolute Values in Inequalities**

Solving Linear Inequalities Linear Inequalities Examples 4. If │x + 2 │< 9 then –9 < x + 2 < 9 and –11 < x < 7 –11 –9 x x + 2 │x + 2 │ x x + 2 7 9 Absolute Values: Examples Examples 1 and 2 show that 3 and -3 between 0 and 7 and between –7 and 0, respectively. In either case, the absolute values of –3 and 3 are less than 7. Example 3 shows that the absolute value of –7 is greater than 3. Removal of the absolute value symbols leaves –7 less than 3. Note that the only other possibility is that –7 is greater than 3, which is of course false. So, as promised, either –7 > 3 or –7 < –3. In Example 4 we have a variable expression whose absolute value is less than 9. Removing the absolute value leaves x+2 trapped between –9 and 9. By subtracting 2 from all three expressions we see that x itself is trapped between –11 and 7. Example 5 shows the absolute value of x+2 greater than 7, which is of course greater than 0. For x positive, this makes x+2 greater than 7 and leaves x greater than 5. For x negative, the difference between x and 2, in absolute value, is less than –7. We could rewrite this as │ x – 2 │> 7. So, for x negative, we find x < –9. 28 7/2/2013 Linear Inequalities Linear Inequalities 7/2/2013

29
**Absolute Values in Inequalities**

Solving Linear Inequalities Linear Inequalities Examples 5. If x + 2 > 7 > 0 then either x > 5 or x < –9 x –9 –│x + 2│ x + 2 –7 5 x 7 │x + 2 │ x + 2 Absolute Values: Examples Examples 1 and 2 show that 3 and -3 between 0 and 7 and between –7 and 0, respectively. In either case, the absolute values of –3 and 3 are less than 7. Example 3 shows that the absolute value of –7 is greater than 3. Removal of the absolute value symbols leaves –7 less than 3. Note that the only other possibility is that –7 is greater than 3, which is of course false. So, as promised, either –7 > 3 or –7 < –3. In Example 4 we have a variable expression whose absolute value is less than 9. Removing the absolute value leaves x+2 trapped between –9 and 9. By subtracting 2 from all three expressions we see that x itself is trapped between –11 and 7. Example 5 shows the absolute value of x+2 greater than 7, which is of course greater than 0. For x positive, this makes x+2 greater than 7 and leaves x greater than 5. For x negative, the difference between x and 2, in absolute value, is less than –7. We could rewrite this as │ x – 2 │> 7. So, for x negative, we find x < –9. Linear Inequalities 29 7/2/2013 Linear Inequalities 7/2/2013

30
**Compound Inequalities**

Solving Linear Inequalities Linear Inequalities More Than One Inequality 1. Find the solution set for x ≥ –3 and x < 7 Rewriting: –3 ≤ x and x < 7 Solution set is { x –3 ≤ x < 7 } OR compounding: –3 ≤ x < 7 Compound Inequalities These are simply a concatenation of multiple inequalities. So each solution x must satisfy one or more inequalities, depending on how they are structured. When the inequalities are logically joined by “and” then each solution must satisfy all the inequalities. If they are joined by “or” then each solution must satisfy at least one of the inequalities. In the first example the two inequalities are joined by “and” so that each x in the solution must satisfy both inequalities. By rewriting the inequalities with the sense of the inequality pointing in the same direction, we can write the two inequalities as a single three-part inequality, as shown. Thus x must be greater-than-or-equal-to -3 and x must be less than 7. So the solution set can be written either as { x | -3 ≤ x < 7 } or [ -3, 7). In the second example the inequalities can’t be rewritten as one three-part inequality since x cannot be both less than -3 and greater than or equal to 7. As the example shows, if we attempted this rewrite, the implication would be that: 7 is less than -3 – which we all know is false! So we solve each inequality separately and join the solution sets. These sets are { x | x < -3 } = (–, -3) and { x | x ≥ 7 } = [7, ) which we join with the set union operator { x | x < -3 } { x | x ≥ 7 } = (–, -3) [7, ) . OR [ –3, 7 ) – 1 3 5 -1 -3 -5 7 9 11 -7 -9 [ ) 30 7/2/2013 Linear Inequalities Linear Inequalities 7/2/2013

31
**Compound Inequalities**

Solving Linear Inequalities Linear Inequalities More Than One Inequality 2. Find the solution set for x < –3 OR x ≥ 7 Can’t rewrite as 7 ≤ x < –3 Solution set is { x x < –3 } { x x ≥ 7 } WHY ? OR ( – , –3) [ 7, ) Compound Inequalities These are simply a concatenation of multiple inequalities. So each solution x must satisfy one or more inequalities, depending on how they are structured. When the inequalities are logically joined by “and” then each solution must satisfy all the inequalities. If they are joined by “or” then each solution must satisfy at least one of the inequalities. In the first example the two inequalities are joined by “and” so that each x in the solution must satisfy both inequalities. By rewriting the inequalities with the sense of the inequality pointing in the same direction, we can write the two inequalities as a single three-part inequality, as shown. Thus x must be greater-than-or-equal-to -3 and x must be less than 7. So the solution set can be written either as { x | -3 ≤ x < 7 } or [ -3, 7). In the second example the inequalities can’t be rewritten as one three-part inequality since x cannot be both less than -3 and greater than or equal to 7. As the example shows, if we attempted this rewrite, the implication would be that: 7 is less than -3 – which we all know is false! So we solve each inequality separately and join the solution sets. These sets are { x | x < -3 } = (–, -3) and { x | x ≥ 7 } = [7, ) which we join with the set union operator { x | x < -3 } { x | x ≥ 7 } = (–, -3) [7, ) . – 1 3 5 -1 -3 -5 7 9 11 -7 -9 ) [ Linear Inequalities 31 7/2/2013 Linear Inequalities 7/2/2013

32
**Compound Inequalities**

Solving Linear Inequalities Linear Inequalities Example Find the solution set for – 2 – t 5 3 4 < Clear the fractions: – 3 4 20 ( ) < 2 – t 5 < 3 4 20 ( ) Simplify: 8 – 4t < – 15 Compound Inequalities Here we are given a three-part inequality. We solve in a way similar to solving equations. First clear the fractions by multiplying all three parts by a common multiple, the least common multiple being the most efficient. We choose 20, the product of 4 and 5, as the common multiple. The example shows a new three-part inequality with no fractions. The variable term is isolated by subtracting 8 from all parts and then multiplying by –¼ , which reverses the direction of the inequalities. Rewriting the resulting inequality with the smallest term on the left and the less-than symbol separating the terms, we get -7/4 < t < 23/4 The solution set is then straight forward: { t | -7/4 < t < 23/4 } = (-7/4, 23/4) which is “graphed” on the real-number line as shown. –23 < –4t < 7 Note: – 7 4 23 > t Inequalities reversed 32 7/2/2013 Linear Inequalities Linear Inequalities 7/2/2013

33
**Compound Inequalities**

Solving Linear Inequalities Linear Inequalities Example Find the solution set for – 2 – t 5 3 4 < – 7 4 23 > t – 7 4 23 < t Rewriting Solution set Compound Inequalities Here we are given a three-part inequality. We solve in a way similar to solving equations. First clear the fractions by multiplying all three parts by a common multiple, the least common multiple being the most efficient. We choose 20, the product of 4 and 5, as the common multiple. The example shows a new three-part inequality with no fractions. The variable term is isolated by subtracting 8 from all parts and then multiplying by –¼ , which reverses the direction of the inequalities. Rewriting the resulting inequality with the smallest term on the left and the less-than symbol separating the terms, we get -7/4 < t < 23/4 The solution set is then straight forward: { t | -7/4 < t < 23/4 } = (-7/4, 23/4) which is “graphed” on the real-number line as shown. – 7 4 23 < t { | } 23 4 , – 7 ( ) OR – 2 4 -2 -4 6 8 10 -6 ) ( 33 7/2/2013 Linear Inequalities Linear Inequalities 7/2/2013

34
**Compound Inequalities**

Solving Linear Inequalities Linear Inequalities Example Find the solution set for 10x + 2(x – 4) < 12x – 10 Simplifying: 10x + 2x – 8 < 12x – 10 –8 < –10 Compound Inequalities: Example In this example we isolate x by using the rules for addition and multiplication. In the process we reduce the inequality to an absurdity, that is -8 < This is logically false, so there is no solution for the given inequality. But there is a solution set, namely the empty set, written either { } or ∅. Note that ∅ ≠ { ∅ }, since the set { ∅ } is a set with one element, namely another set, while ∅ is itself a set, a set with no elements. A logically FALSE statement ! There is no value of x … that will make this true ! Linear Inequalities 34 7/2/2013 Linear Inequalities 7/2/2013

35
**Compound Inequalities**

Linear Inequalities Solving Linear Inequalities Example Find the solution set for 10x + 2(x – 4) < 12x – 10 There is no value of x … that will make this true ! We say the solution set is EMPTY ! Compound Inequalities: Example In this example we isolate x by using the rules for addition and multiplication. In the process we reduce the inequality to an absurdity, that is -8 < This is logically false, so there is no solution for the given inequality. But there is a solution set, namely the empty set, written either { } or ∅. Note that ∅ ≠ { ∅ }, since the set { ∅ } is a set with one element, namely another set, while ∅ is itself a set, a set with no elements. We write: Solution set is O … OR just { } Note: O ≠ { } WHY ? 7/2/2013 Linear Inequalities 35 Linear Inequalities 7/2/2013

36
**Solving Linear Inequalities**

Think about it ! Linear Inequalities 36 7/2/2013 Linear Inequalities 7/2/2013

Similar presentations

Presentation is loading. Please wait....

OK

Time for a BREAK! You have 45 Minutes.

Time for a BREAK! You have 45 Minutes.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Scrolling message display ppt on ipad Science ppt on carbon and its compounds Ppt on eia report on natural gas Ppt on indian culture vs western culture Ppt on series and parallel combination of resistors in series Dr appt online Ppt on contributor personality development Training ppt on time management Animated ppt on magnetism quiz Ppt on blood stain pattern analysis definition