Now we can find the boundary points. We will do that by temporarily considering the related equation instead. We find a related equation by changing the inequality sign to an equals sign.
Inequality:& |3x-5|≥2
Equality:& |3x-5|=2The absolute value of a number measures the distance from 0 to that number. In this case, the distance from 0 to 3x-5 is 2. The only numbers whose distance to 0 is 2 are 2 and - 2. Therefore, 3x-5 can be either 2 or - 2.
|3x-5|=2 ⇒ l3x-5= 2 3x-5= - 2
To find the solutions to the absolute value equation, we need to solve both of these cases for x.
lc3x-5=2 & (I) 3x-5=-2 & (II)
(I), (II): LHS+5=RHS+5
l3x=7 3x=3
(I), (II): .LHS /3.=.RHS /3.
lx_1= 73 x_2=1
The boundary points are located at x= 73 and x=1. Let's mark them on a number line. Notice that the inequality is non-strict. This means the boundary points are part of the solution set. To determine where we should shade the number line, we will test three numbers — one in each of the intervals defined by the boundary points.
Now, we will substitute the numbers for x in the given inequality. If the inequality holds true, we should shade this region. Otherwise, we do not shade it.
|c|c|c|
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x & 2|3x-5|≥ 4 & Evaluate [0.5em]
[-1em]
& 2|3( )-5|? ≥4 & 10 ≥ 4 ✓ [0.5em]
[-1em]
2 & 2|3(2)-5|? ≥4 & 2 ≱ 4 * [0.5em]
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4 & 2|3(4)-5|? ≥4 & 14 ≥ 4 ✓ [0.5em]
The inequality is true to the left of the first boundary point and to the right of the second boundary point. Therefore, we will shade these regions.
b By performing inverse operations we can solve the inequality by isolating x.
The solution set to the inequality is x<3. Therefore, we should shade everything that is lower than 3. Note that this is a strict inequality which means we will keep the endpoint open.
