The Subtitles

In this week’s discussion we will be solving for “and” compound inequality my problem to solve is 0 ≤ 4 + 2x ≤ 22. An inequality is formed by the intersection of two simple inequalities as shown in this example, 0 ≤ 4 + 2x and 4 + 2x ≤ 22.   This said 4+2x must be equal to or less than 22 and also grater or equal to 0.

0 < 4+2x < 22 Original problem
0-4 < 4-4+2x < 22-4 Subtract 4 from all three parts of the inequality.
-4 < 2x < 18
-4 < 2x< 18 Divide all three parts by 2.
2       2     2
-2 < x < 9 Final answer
<________-2 _________x________9__________>
Any value of x greater than or equal to -2 and less than or equal to 9 will make an inequality true.   Also an intersection would be [-2, infinity sign) n (- infinity sign, 9] also known as [-2, 9] in an interval notation.
My second problem to solve for this week is an “or” compound inequality 3x + 6 < -3 or 5 – x ≤ 1. This problem is made by the union of two sets of inequalities, example 3x + 6 < -3 and 5 – x ≤ 1. The final solution says that x is greater than any number on the left side of -3 and   4 or any number greater than 4.

3x+6<-3 Original problem
3x<-6-3 Subtract 3 from -6 to get -9
3x<-9
3x<-9     Divide each tern by 3
3       3

x<-9   Reduce expression
    3
x<-3 Final answer

<______-3____________________     4______>
5-x < 1 Original problem
-x < -4 Since 5 doesn’t have a variable we move it and add 1 to-5 and get -4
-x*-1 < -4*-1
x > 4 When multiplying a inequality by negative value, the inequality sign is flipped.