Chapter 6 Solving Linear Inequalities

6.1 Solving Inequalities by Addition and Subtraction Set builder Notation : –A way to write a solution set Ex: if the solution set is all #s less than or equal to 58 <> Less than Fewer than Greater than More than At most No more than Less than or equal to At least No less than Greater than or equal to If the variable is on the open side = shade right If the variable is on the closed side = shade left Solve by Addition Solve by Subtraction ex: t – 45 < 13ex: 8 – 2y -1y ex: s + 19 > 56ex: 5p + 7 6p t < 58 { t | t < 58} y + 2y 8 y { y | y 8} s > 37 {s | s > 37} -5p 7 p { p | p 7}

6.2 Solving Inequalities by Multiplication and Division If you multiply or divide by a negative number, you must change the inequality sign Solve by Multilplication Ex: s < Ex: p 7.5 Solve by Division { s | s < 35} { p | p 7.5} 12x 60 /12 x 5 { x | x 5} q < 136 /-8 q > -17 { q | q > -17}

6.3 Solving Multi-Step Inequalities Multi-step Inequality Distributive Property Write and Solve an InequalityEmpty Sets and All Real Numbers -32 c > 40 {c | c > 40} d – 2(8d – 9) > -2d - 4 3d – 16d + 18 > -2d d + 18 > -2d d 18 > 11d > 11d /11 2 > d {d | d < 2} Three times a number minus eighteen is at least five times the number plus twenty-one -3x / { x | x -19.5} 8(t +2)- 3(t – 4) < 5(t -7) + 8 8t t + 12 < 5t t + 28 < 5t t - 5t 28 < -27 Empty set If it is true – all real numbers If it is false – it is an empty set { x | x is a real number}

6.4 Solving Compound Inequalities Intersection = and Union = or Ex: Graph a Compound Inequality {x | -2 x < 3} Ex: Solve and Graph a Compound Inequality -5 < x – 4 < 2 -5 < x - 4x – 4 < < x + 4 x < {x | -1 < x < 6 }

6.5 Absolute Value Open Sentences Absolute Value: The distance from zero on a number line –The positive value of the number Ex: Make a frayer- Foldable (diamond in the center) |-5| = |5| = |-6| = |2| =

Solving Absolute Value Equations: a.The expression inside the absolute value bars is positive b.The expression inside the absolute value bars is negative Ex: | x + 7| = 4 a.x + 7 = 4b. x + 7 = x = -3 x = -11 Solution set: (-3, -11)

If the absolute value bars equal a negative number- it is an empty set Write an absolute value equation: | x - # | = # # half way between # spaces from each point 33 | x – 2 | = 3 Ex: | x + 16| = 2

Absolute Value Function f(x) = |x| and f(x) must be greater than or equal to zero f(x) = |x| can be written as: f(x)= { -x if x < 0 x if x 0 Graph of f(x) = |x| x |x|

6.6 Solving Absolute Value Inequalities A. the expression in absolute value bars is positive B. the expression in absolute value bars is negative (also flip the inequality sign) On the back of the 6.5 frayer |x| = n x = -n or x = n |x| -n |x| > n x > n or x < -n

Ex 1 : | t + 5 | < 9 t + 5 < 9t + 5 > t < 4t > < t < 4

Ex 2 : |2x + 8| 6 2x x x -2 2x -14 /2 x -1x x -7 or x -1

Ex 3 : | 2y – 1 | -4 Ex 4 : (your turn) | 2k + 1| > 7 the absolute value cannot be less than zero so y is all numbers *If it is a negative number- the answer is empty set

6.7 Graphing Inequalities with Two Variables The equation makes the line to define the boundary The shaded region is the half-plane 1.Get the equation into slope-intercept form 2.List the intercept as an ordered-pair and the slope as a ratio 3.Graph the intercept and use the slope to find at least 2 more points 4.Draw the line (dotted or solid) 5.Test an ordered-pair not on the line 1.If it is true shade that side of the line 2.If it is false shade the other side of the line Make a frayer- Foldable (diamond in the center)

or Dotted Line Solid Line Ex 1 : y 2x - 3 m = b = -3 = (0, -3) Use a solid line because it is Test: (0, 0) 0 2(0) – – false (shade other side)

Ex 2 : y – 2x < 4 y – 2x < 4 + 2x +2x y < 2x + 4 m = b = 4 = (0, 4) Test: (0, 0) 0 < 2(0) < < 4 true (shade this side) Use a dotted line because it is <

Ex 3 : 3y - 2 > -x + 7 3y – 2 > -x y > -x + 9 /3 /3 /3 y > - x + 3 m = - b = 3 = (0, 3) Test: (0, 0) 0 > - (0) > > 3 false (shade other side) Use a dotted line because it is >