Equations with absolute value:

Where k is a positive number; if k is a negative number, the equation is impossible (absolute value is always positive).
How to solve:


Then:
1. |x+7|=12
x+7=12 V -x-7=12
x=5 V -x=19
x=5 V x=-19
{-19, 5}
2. |2x+4|=8
2x+4=8 V -2x-4=8
2x=4 V -2x=12
x=2 V x=-6
3. 3|3k|=27
3×3k=27 V 3×(-3k)=27
9k=27 V -9k=27
k=3 V k=-3
{-3, 3}
4. 5|b+8|=30
5×(b+8)=30 V 5×(-b-8)=30
5b+40=30 V -5b-40=30
5b=-10 V -5b=70
b=-2 V b=-14
{-14, -2}
5. |m+9|=5
m+9=5 V -m-9=5
m+9=5 V m+9=-5
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The inequality should read:

Subtract 2 from all the terms:

Divide all the terms by -5 to get x by itself (note that you flip the inequality signs as you divide by a negative):

The inequality is 2 ≤ x < 8.
A closed dot represents ≤, and an open dot represents <. Since x can represent all values between 2 and 8, you will shade in between 2 and 8 on the number line. x is greater than or equal to 2, so there will be a closed dot on 2. x is less than 8, so there will be an open dot on 8.
The answer is 'number line with a closed dot on 2 and an open dot on 8 and shading in between'.
Because this is a second degree equation you will have 2 solutions. When you take the square root of a number you have to account for both the positive and negative roots. Since the square root of 289 is 17 then your solutions are +17 and -17.