Question 958406: need help just cant figure out these problems , have to solve the equations
|2x-7|=3
|x-6|=|2x+1|
|x-4|<3
|x/3 - 1|≥2
thanks
Found 3 solutions by lwsshak3, ikleyn, greenestamps:
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! need help just cant figure out these problems , have to solve the equations
|2x-7|=3
case 1:
2x-7=3
2x=10
x=5
case 2:
-(2x-7)=3
2x-7=-3
2x=4
x=2
solution: (x=5 or x=2)
..
|x-6|=|2x+1|
case 1
x-6=2x+1
x=-7
..
case 2
x-6=-(2x+1)
x-6=-2x-1
3x=5
x=5/3
solution: x=5/3 or x=-7
..
|x-4|<3
case 1
x-4<3
x<7
..
case2
-(x-4)<3
x-4<-3
x<1
solution: x<7 or x<1
..
|x/3 - 1|≥2
case 1
x/3-1≥2
x/3≥3
x≥9
case2
x/3-1≥-2
x/3≥-1
x≥-3
solution: x=9 or x=-3
note: cases 1 and 2 check out the possibility that what's between the absolute signs could be positive or negative.
Answer by ikleyn(53937)  (Show Source):
You can put this solution on YOUR website! .
need help just cant figure out these problems , have to solve the equations
(a) |2x-7|=3
(b) |x-6|=|2x+1|
(c) |x-4|<3
(d) |x/3 - 1|≥2
thanks
~~~~~~~~~~~~~~~~~~~~~~~~~~~
In the post by @lwsshar3, the solutions and the answers to problems (c) and (d) are incorrect.
I came to provide correct solutions to these problems.
.. problem (c)
|x-4| < 3
case 1
x-4 < 3
x < 7
..
case2
-(x-4) < 3
x-4 > -3
x > 1
solution: 1 < x < 7.
.. problem (d)
|x/3 - 1| ≥ 2
case 1
x/3-1 ≥ 2
x/3 ≥ 3
x ≥ 9
case2
x/3-1 <= -2
x/3 <= -1
x <= -3
solution: x <= -3 or x >= 9.
Solved correctly.
Answer by greenestamps(13367)  (Show Source):
You can put this solution on YOUR website!
The other tutors solve the problems by considering different cases; that is the standard formal algebraic method.
For problems (a), (c), and (d) -- in which there is an absolute value expression on only one side -- the solutions can be found by a different method.
The statement  can be interpreted to say "the difference between (variable) x and (fixed) a is equal to b".
For (c), this can lead directly to the answer. The equation  says that the difference between x and 4 is equal to 3, which means the values of x are 4-3=1 and 4+3=7. Then since the statement is in fact a strict "less than" inequality, 1 and 7 are the boundaries of the solution set, so the solution in interval notation is (1,7).
For (a) and (d) we need to multiply the whole inequalities by appropriate constants to make "x" appear by itself in the absolute value expression.
(a)
The solutions are the two numbers for which the difference between the number and 3.5 is 1.5 -- i.e., 3.5-1.5 = 2 and 3.5+1.5 = 5.
(d)
The boundaries of the solution set are the two numbers whose difference from 3 is 6 -- i.e., 3-6 = -3 and 3+6 = 9. Then since the statement is an inclusive "greater than" inequality, the solution set in interval notation is (-infinity,-3] U [9,infinity).
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