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.