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
This means Square root (^1/2). this is kind of hard to read. anyways. 32^(1/2) separates into 4^(1/2) and 8^(1/2), which further separates into 4^(1/2) and 2^(1/2), root 4 becomes 2, their are two root 4's, so you get 2 x 2, and you are left with 2^(1/2) now why go to all this trouble. because now you can multiply the 4 you created (2x2) times, the 7... giving you 28*2^(1/2) now subtract it from the the other one with root 2. -5*2^(1/2), giving you 23*2^(1/2)-...idk what that last bit is. if its a odd number then this is the end of the problem, if you can get it to root 2. then do that and simplify.
Answer:
Step-by-step explanation:
s(1-30/100)=420
s(100-30)/100=420
s(70/100)=420
s(7/10)=420
7s=4200
s=600