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
The solution would be like this for this specific problem:
sin(θ°) = √(2)/2
θ° = 360°n + sin⁻¹(√(2)/2) and θ° = 360°n + 180° −
sin⁻¹(√(2)/2)
θ° = 360°n + 45° and θ° = 360°n + 135° where n∈ℤ
360°*0 + 45° = 45°
360°*0 + 135° = 135°
360°*1 + 45° = 405°
<span>sin(225°) = -√(2)/2
</span>225 has an angle where sin theta= -(sqrt2)/2 therefore, the value of theta
cannot be 225 degrees.
(-3,2)))))))))))))))))))))))))))))))))))