solve: 3+|2y-1|>1 graph the solution set on a number line.
Subtract 3 from each side to get the valid inequality
|2y-1|> -2
If 2y>1, then the term inside | | is positive and
2y -1 > -2
2y > -1
Since you already required 2y>1, the allowed value of y are > 1/2 in this case.
IF the term inside || is negative, then
2y < 1 and
1 - 2y > -2
-2y > -3
2y < 3
y < 3/2 are also allowed solutions
The solution is all real numbers. No matter what value of y you select, the inequality is valid. That should have been immediately obvious, since the term in || cannot be less tnan zero.
To graph the solution set on a number line, we will plot two points representing the boundaries of the solution.
First, let's plot the point y = 1/2 since it represents the lower bound for the solution set. This point is included in the solution.
Next, let's plot the point y = 3/2 since it represents the upper bound for the solution set. This point is not included in the solution.
Now, draw an open circle at y = 1/2 and an open circle at y = 3/2 on the number line.
Lastly, draw a line connecting the two points. This line represents all the values of y that satisfy the inequality.
The solutions to the inequality 3 + |2y - 1| > 1 are all real numbers. This means that the entire number line is shaded.
To graph the solution set on a number line, we need to plot the values of y that satisfy the inequality.
First, let's mark the values of y > 1/2 on the number line. We indicate this by shading the portion of the number line to the right of 1/2.
Next, let's mark the values of y < 3/2 on the number line. We indicate this by shading the portion of the number line to the left of 3/2.
Since the solution is all real numbers, we will shade the entire number line. This indicates that any value of y satisfies the inequality.
Therefore, the solution set is represented by shading the entire number line.