Week 2: Student Response to Discussion Two

4:37pmJul 27 at 4:37pm

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I was given two compound inequalities to solve: one, an "And" inequality, and the other an "Or" inequality.

-5 < 4x + 1 < 3 The first compound inequality which is the "And" one is as such because 4x + 1 must be

-6 < 4x < 2 both greater than - 5 and less than or equal to 3. We subtract 1 from all three parts. Then,

-3/2 < x < 1/2 we divide 4 from all three sides-- simplifying -6/4 into -3/2. This leaves us with: x > -3/2 and

x < 1/2 or, algebraically, -3/2 < x < 1/2. Here is a graph plotting the solution set:

<--------------(-------|----]-------------> with brackets for "or equal" and parenthesis for no equal

-3/2 (1 1/2) 0 1/2 The intersection of these sets would be (-3/2,∞) ∩ (-∞,1/2]

or written in interval notation, (-3/2, 1/2).

3x + 2 < 2 or 3 - x > 11 The next compound inequality is the "Or'', which means there are two conditions

3x < 0 -x > 8 here and they both can't be true at the same time but with an "x" from the solution

x < 0 x < -8 one of them can be. So, we solve both of them separately, subtracting 2 from both

x < 0 or x < -8 sides of the left inequality and subtracting 3 from both sides of the right one. Then,

divide 3 in the left inequality and -1 from the right one, flipping the symbol because

the division of the negative number, leaving x<0 or x<-8. Here is the graph:

<-------------|-------------------------)|> Because x<0 overlaps the space of x<-8, it's a solid

-8 0 block after 0 and into ∞. The union of these two is

[∞,0) ∩ (-8,0).

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