( / means absolute value)
Solve: 3+/2y-1/>=1. Graph the solution set on a number line.
The number line I have is:
-5 -4 -3 -2 -1 0 1 2 3 4 5 6
I'm confused on what to do.
3+ │2y-1│ ≥ 1 or
│2y-1│ ≥ -2
this has an intuitive solution.
by the definition of absolute value, the left side can only be positive or zero.
any positive value or zero ≥ -2
so the solution would be the entire number set. So draw a solid line on your number line with arrows in either direction.
I have to show my work so is that ALL I have to do? 3+ ©¦2y-1©¦ ¡Ý 1 or
©¦2y-1©¦ ¡Ý -2
by saying,
"by definition │anything│ ≥ -2" you have shown your work.
To solve the inequality 3|2y-1| ≥ 1, you can break it down into two separate inequalities:
1) 3(2y-1) ≥ 1
2) 3(-(2y-1)) ≥ 1
For the first inequality, you can distribute the 3 across the parentheses:
6y - 3 ≥ 1
Next, add 3 to both sides of the inequality:
6y - 3 + 3 ≥ 1 + 3
6y ≥ 4
To isolate y, divide both sides of the inequality by 6:
6y/6 ≥ 4/6
y ≥ 2/3
For the second inequality, you can distribute the 3 across the parentheses and change the sign of the inequality:
-6y + 3 ≥ 1
Next, subtract 3 from both sides of the inequality:
-6y + 3 - 3 ≥ 1 - 3
-6y ≥ -2
To isolate y, divide both sides of the inequality by -6 (and remember to flip the inequality symbol):
(-6y)/(-6) ≤ (-2)/(-6)
y ≤ 1/3
Now, you have two solutions: y ≥ 2/3 and y ≤ 1/3. To graph the solution set on a number line, you can plot these values:
-5 -4 -3 -2 -1 0 1 2 3 4 5 6
|_________|___________|
The interval from 2/3 to positive infinity (including 2/3) is represented by a closed circle on 2/3, and the ray extending to the right.
On the other hand, the interval from negative infinity to 1/3 (including 1/3) is represented by a closed circle on 1/3, and the ray extending to the left.
Thus, the solution set is the combination of these two intervals on the number line.