We'll subtract 8 both
sides:
7 l 3x-4 l = 15 - 8
7 l
3x-4 l = 7
We'll divide by
7:
l 3x-4 l = 1
We'll have 2
cases:
1) 3x-4 for 3x-4
>=0
We'll add 4 both
sides:
3x>=4
We'll
divide by 3:
x>=4/3
The
interval of admissible values for x is [4/3 ,
+infinite)
We'll solve the
equation:
3x-4 = 1
We'll add 4
both sides:
3x = 5
x =
5/3
Since x = 5/3 belongs to the interval [4/3 ,
+infinite), the equation has the solution x = 5/3.
2) -3x+4
for 3x-4 <0
The solution of the equation has to
belong to the interval
(-infinite,
4/3)
No comments:
Post a Comment