Answer: Choice 3) 0, 3.9, and 5.1
Basically almost every value of that set but leaving out 10.2
--------------------------------------------------
--------------------------------------------------
Explanation:
The given set of values is {0, 3.9, 5.1, 10.2}
Let's check each number of that set one value at a time.
--------------------------------------------------
Checking x = 0
Replace every copy of x with 0, then use PEMDAS to simplify each side
We need to see if we get a true inequality or not
[tex]8.9 - 0.3x \ge 0.8x [/tex]
[tex]8.9 - 0.3*0 \ge 0.8*0 [/tex]
[tex]8.9 - 0 \ge 0 [/tex]
[tex]8.9 \ge 0 [/tex]
The last inequality is TRUE (8.9 is greater than 0).
So x = 0 is part of the solution set.
--------------------------------------------------
Checking x = 3.9
Repeat the same steps as done previously (with x = 0).
This time we replace every x with 3.9, but the remaining general steps are the same more or less.
[tex]8.9 - 0.3x \ge 0.8x [/tex]
[tex]8.9 - 0.3*3.9 \ge 0.8*3.9 [/tex]
[tex]8.9 - 1.17 \ge 3.12 [/tex]
[tex]7.73 \ge 3.12 [/tex]
The last inequality is TRUE (7.73 is greater than 3.12).
The value x = 3.9 is also part of the solution set.
--------------------------------------------------
Checking x = 5.1
Follow the same outline. This time we use 5.1 in place of x.
[tex]8.9 - 0.3x \ge 0.8x [/tex]
[tex]8.9 - 0.3*5.1 \ge 0.8*5.1 [/tex]
[tex]8.9 - 1.53 \ge 4.08 [/tex]
[tex]7.37 \ge 4.08 [/tex]
The last inequality is TRUE (7.37 is greater than 4.08).
The value x = 5.1 joins in with the other two values that are part of the solution set.
--------------------------------------------------
Checking x = 10.2
Now replace every x value with 10.2
[tex]8.9 - 0.3x \ge 0.8x [/tex]
[tex]8.9 - 0.3*10.2 \ge 0.8*10.2 [/tex]
[tex]8.9 - 3.06 \ge 8.16 [/tex]
[tex]5.84 \ge 8.16 [/tex]
The last inequality is FALSE (5.84 is not larger than 8.16, nor is it it equal; 5.84 is smaller than 8.16).
Because we have a false inequality result, we can say that x = 10.2 is NOT part of the solution set.
If you want, you can use a red pen to cross off 10.2 to visually remind you that 10.2 is not part of the solution set.
--------------------------------------------------
So in summary we found x = 0, x = 3.9 and x = 5.1 to satisfy (make true) the original inequality.
The only value, from that set, to make the original inequality false is the value x = 10.2
This is why the answer is choice 3.
--------------------------------------------------
Side Note (extra optional info)
If we solve the given inquality for x, we get the following:
[tex]8.9 - 0.3x \ge 0.8x [/tex]
[tex]8.9 - 0.3x + 0.3x \ge 0.8x + 0.3x [/tex]
[tex]8.9 \ge 1.1x [/tex]
[tex]1.1x \le 8.9[/tex]
[tex]\frac{1.1x}{1.1} \le \frac{8.9}{1.1}[/tex]
[tex]x \le 8.09[/tex]
The value 8.09 is approximate
Since the solution set is x values that satisfy [tex]x \le 8.09[/tex], this basically means any value smaller than 8.09 will work. Those values include: 0, 3.9 and 5.1. The value 10.2 is not smaller than 8.09 which is why it is excluded.
