I x+3 I - I 2x-1 I = 2 How do you solve this?
2008-12-27 4:30 pm
How do you solve this? I solved for x and got 2,6, -4r3, and 0. When I checked, 6 and -4r3 did not equal to 2. Did I do something wrong or do I just not include those numbers into my final answer?
回答 (9)
2008-12-27 4:49 pm
✔ 最佳答案
x + 3 = 0 => x = -3 and 2x - 1 = 0 => x = 1/2Dividing number line into three parts,
(- ∞, -3), (-3, 1/2) and (1/2, ∞)
Case 1:
x ∈ (- ∞, -3)
=> l x+ 3 l = - x -3 and l 2x -1 l = -2x + 1
So I x+3 I - I 2x-1 I = 2
=> -x -3 + 2x -1 = 2
=> x = 6
This solution is ruled out as 6 does not belong to (- ∞, -3)
Case 2:
x ∈ (-3, 1/2)
=> x + 3 +2x -1 = 2
=> 3x = 0
=> x = 0
Case 3:
x ∈ (1/2, ∞)
=> x + 3 - 2x + 1 = 2
=> - x = -2
=> x = 2
Case 4: x = -3 or 1/2
None of these values of x satisfy the given equation.
Answer: x = { 0, 2 }.
2008-12-27 4:45 pm
|x + 3| - |2x - 1| = 2
|x + 3| = 2 + |2x - 1|
x + 3 = 2 + 2x - 1
x - 2x = 2 - 3 - 1
-x = -2
x = -2/-1
x = 2
-(x + 3) = 2 - (2x - 1)
-x - 3 = 2 - 2x + 1
-x + 2x = 2 + 3 + 1
x = 6
-(x + 3) = 2 + 2x - 1
-x - 3 = 2 + 2x - 1
-x - 2x = 2 + 3 - 1
-3x = 4
x = 4/-3
x = -4/3
x + 3 = 2 - (2x - 1)
x + 3 = 2 - 2x + 1
x + 2x = 2 - 3 + 1
3x = 0
x = 0/3
x = 0
(when I substitute the answer back, x is only equal to 0 and 2.)
∴ x = 0 , 2
|x + 3| = 2 + |2x - 1|
x + 3 = 2 + 2x - 1
x - 2x = 2 - 3 - 1
-x = -2
x = -2/-1
x = 2
-(x + 3) = 2 - (2x - 1)
-x - 3 = 2 - 2x + 1
-x + 2x = 2 + 3 + 1
x = 6
-(x + 3) = 2 + 2x - 1
-x - 3 = 2 + 2x - 1
-x - 2x = 2 + 3 - 1
-3x = 4
x = 4/-3
x = -4/3
x + 3 = 2 - (2x - 1)
x + 3 = 2 - 2x + 1
x + 2x = 2 - 3 + 1
3x = 0
x = 0/3
x = 0
(when I substitute the answer back, x is only equal to 0 and 2.)
∴ x = 0 , 2
2008-12-27 4:43 pm
Here , there are two points of inflexion -3 and 1/2
So, the intervals we have to check for are x< -3 ,-3<x<1/2 and x>1/2
For x< -3
I x+3 I - I 2x-1 I = 2
[-(x+3)] - [-(2x -1)] =2
-x-3+2x -1 =2
x-4=2
x=6
If we check for x = 6 ,equation is not satisfied and so, x=6is not a solution
For -3<x<1/2
I x+3 I - I 2x-1 I = 2
I x+3 I - I 2x-1 I = 2
x+3 - [-( 2x-1)] = 2
x+3+2x -1 = 2
3x+2 = 2
3x =0
x=0
If we check for x =0 ,equation is satisfied .So, x=0is a solution
For x>1/2
I x+3 I - I 2x-1 I = 2
(x+3)-(2x-1) = 2
x+3 - 2x +1 =2
-x+4 = 2
-x = -2
x = 2
If we put x=2,equatoin is satisfied
So, x = 0 and x = 2 are the solutions
So, the intervals we have to check for are x< -3 ,-3<x<1/2 and x>1/2
For x< -3
I x+3 I - I 2x-1 I = 2
[-(x+3)] - [-(2x -1)] =2
-x-3+2x -1 =2
x-4=2
x=6
If we check for x = 6 ,equation is not satisfied and so, x=6is not a solution
For -3<x<1/2
I x+3 I - I 2x-1 I = 2
I x+3 I - I 2x-1 I = 2
x+3 - [-( 2x-1)] = 2
x+3+2x -1 = 2
3x+2 = 2
3x =0
x=0
If we check for x =0 ,equation is satisfied .So, x=0is a solution
For x>1/2
I x+3 I - I 2x-1 I = 2
(x+3)-(2x-1) = 2
x+3 - 2x +1 =2
-x+4 = 2
-x = -2
x = 2
If we put x=2,equatoin is satisfied
So, x = 0 and x = 2 are the solutions
2008-12-27 4:37 pm
(x + 3) - (2x - 1) = 2
x + 3 - 2x + 1 = 2
x = 4 - 2
x = 2
Answer: x = 2
Proof:
(2 + 3) - (2[2] - 1) = 2
5 - (4 - 1) = 2
5 - (3) = 2
5 - 3 = 2
2 = 2
x + 3 - 2x + 1 = 2
x = 4 - 2
x = 2
Answer: x = 2
Proof:
(2 + 3) - (2[2] - 1) = 2
5 - (4 - 1) = 2
5 - (3) = 2
5 - 3 = 2
2 = 2
2008-12-27 4:45 pm
Case 1: x+3 > 0 & 2x-1 > 0
|x+3| = x+3 & |2x-1| = 2x-1
x+3 - (2x-1) = 2
-x +4 =2
-x=-2
x=2 (solution 1)
Case 2:
x+3 > 0 2x-1 < 0
|x+3| = x+3 & |2x-1| = -(2x-1) =-2x+1
(x+3) - (-2x+1) = 2
x+3+2x-1 =2
3x+2=2
x=0
Case 3:
x+3 < 0 2x-1 > 0
|x+3| = -(x+3) = -x-3
}2x-1| = 2x-1
-x-3 -(2x-1) = 2
-x-3-2x+1 =2
-3x=4
x=-4/3
case 4:
x+3 < 0 & 2x-1 < 0
|x+3| = -(x+3) = -x-3
|2x-1| = -(2x-1) = -2x+1
-x-3 -(-2x+1) = 2
-x-3+2x-1 =2
x-4=2
x=6
You are correct. 6 and -4/3 produce -2 rather than 2.
The correct solution is x=2,0
|x+3| = x+3 & |2x-1| = 2x-1
x+3 - (2x-1) = 2
-x +4 =2
-x=-2
x=2 (solution 1)
Case 2:
x+3 > 0 2x-1 < 0
|x+3| = x+3 & |2x-1| = -(2x-1) =-2x+1
(x+3) - (-2x+1) = 2
x+3+2x-1 =2
3x+2=2
x=0
Case 3:
x+3 < 0 2x-1 > 0
|x+3| = -(x+3) = -x-3
}2x-1| = 2x-1
-x-3 -(2x-1) = 2
-x-3-2x+1 =2
-3x=4
x=-4/3
case 4:
x+3 < 0 & 2x-1 < 0
|x+3| = -(x+3) = -x-3
|2x-1| = -(2x-1) = -2x+1
-x-3 -(-2x+1) = 2
-x-3+2x-1 =2
x-4=2
x=6
You are correct. 6 and -4/3 produce -2 rather than 2.
The correct solution is x=2,0
2008-12-27 4:39 pm
x+3-(2x-1)=2 or x+3-(-2x+1)=2
x+3-2x+1=2 or x+3+2x-1=2
-x=-2 or 3x+2=2
x=2 or x=0
x+3-2x+1=2 or x+3+2x-1=2
-x=-2 or 3x+2=2
x=2 or x=0
2008-12-27 4:38 pm
1) remove the absolute values and do the problem the same
2) replace the absolute values with parenthesis and put a negative sign infront of the paren, then solve
2) replace the absolute values with parenthesis and put a negative sign infront of the paren, then solve
參考: IM GUESSING
2008-12-27 8:22 pm
In an absolute value equation use the original one, and then change the sign on another one.
x+3-2x-1 = 2
-x= 0
x= 0
And
x+3-2x-1 = -2
-x= -4
x= 4
So x = 0, x = 4, remember that when you have a negative value like -x, you have to multiply both sides by -1 to get a positive variable. :o)
x+3-2x-1 = 2
-x= 0
x= 0
And
x+3-2x-1 = -2
-x= -4
x= 4
So x = 0, x = 4, remember that when you have a negative value like -x, you have to multiply both sides by -1 to get a positive variable. :o)
2008-12-27 4:51 pm
1. case:2x-1 >=0 ,than x+3 >=0 too, so |x+3|=x+3 and |2x-1| = 2x-1
x+3-(2x-1) = 2 <=> x = 2
2.case: x+3 >=0 and 2x-1< 0, so we have
I x+3 I - I 2x-1 I = 2 <=> x+3 +2x-1 = 2 <=> x = 0
3.) case: x+3 < 0 and 2x-1< 0, than
| x+3 I - I 2x-1 I = -(x+3) + (2x-1) = 2 <=> x=6 which is impossibile.
x+3-(2x-1) = 2 <=> x = 2
2.case: x+3 >=0 and 2x-1< 0, so we have
I x+3 I - I 2x-1 I = 2 <=> x+3 +2x-1 = 2 <=> x = 0
3.) case: x+3 < 0 and 2x-1< 0, than
| x+3 I - I 2x-1 I = -(x+3) + (2x-1) = 2 <=> x=6 which is impossibile.
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