a. (-infinity, - 6 / 7)
b. (-6 / 7, 12 / 7)
c. (-infinity, 12 / 7)
d. (-infinity, - 6 / 7) u (12 / 7, infinity)
solve the inequality: |7k - 3| - 6 < 3?
2009-02-16 1:37 pm
回答 (8)
2009-02-16 1:42 pm
✔ 最佳答案
|7k - 3| - 6 < 3|7k - 3| < 9
7k - 3 < 9
or
7k - 3 > -9
7k - 3 < 9
7k < 12
k < 12/7
7k - 3 > -9
7k > -6
k > -6/7
k ∈ (-6/7, 12/7)
B
2009-02-16 10:14 pm
|7k - 3| - 6 < 3
|7k - 3| < 3 + 6
|7k - 3| < 9
7k - 3 < 9, 7k - 3 > -9
7k - 3 < 9
7k < 9 + 3
7k < 12
k < 12/7
7k - 3 > -9
7k > -9 + 3
7k > -6
k > -6/7
â´ -6/7 < k < 12/7
(answer b)
|7k - 3| < 3 + 6
|7k - 3| < 9
7k - 3 < 9, 7k - 3 > -9
7k - 3 < 9
7k < 9 + 3
7k < 12
k < 12/7
7k - 3 > -9
7k > -9 + 3
7k > -6
k > -6/7
â´ -6/7 < k < 12/7
(answer b)
2009-02-16 9:55 pm
case 1---- 7k<3
-7k+3-6<3
-7k-3<3
-7k<6 so,k belongs to( 6/7,infinity)
7k>6
k>6/7
case2---- 7k>3
7k-3-6<3
7k-9<3 so k belongs to( -infinity,12/7)
7k<12
k<12/7
on taking union c is the ans.
-7k+3-6<3
-7k-3<3
-7k<6 so,k belongs to( 6/7,infinity)
7k>6
k>6/7
case2---- 7k>3
7k-3-6<3
7k-9<3 so k belongs to( -infinity,12/7)
7k<12
k<12/7
on taking union c is the ans.
2009-02-16 9:52 pm
|7k - 3| - 6 < 3 ; add 6 to both sides
|7k - 3| < 9
7k - 3 < 9 (1)
or
-(7k-3) < 9 ; distribute the minus
-7k+3 < 9 (2)
Solve for k in (1) and (2)
7k < 12 ; divide 7 by both sides
k < 12/7
-7k < 6 ; divide by -7, reverse the sign
k > -6/7
So your answer is b.
|7k - 3| < 9
7k - 3 < 9 (1)
or
-(7k-3) < 9 ; distribute the minus
-7k+3 < 9 (2)
Solve for k in (1) and (2)
7k < 12 ; divide 7 by both sides
k < 12/7
-7k < 6 ; divide by -7, reverse the sign
k > -6/7
So your answer is b.
2009-02-16 9:47 pm
|7k - 3| - 6 < 3
|7k - 3| < 9
7k - 3 < 9 ==> k<12/7 {.12/7 ) }
or
- 7k + 3 < 9 ==> k> -6/7 {. (-6/7 }
so b
|7k - 3| < 9
7k - 3 < 9 ==> k<12/7 {.12/7 ) }
or
- 7k + 3 < 9 ==> k> -6/7 {. (-6/7 }
so b
2009-02-16 9:47 pm
I was wrong with my first answer, but lets quickly double check. If we try -1, 0, and 2 that will allow us to pick an answer.
-1 give us | -10 | - 6 which equals 4, so we already know it has to be B
0 gives us | -3 | - 6 which equals -3 which is less than 3
2 gives us | 14 -3 | - 6 which equals 5 which is not < 3
We have imperically proven it is B in addition to the other fine proofs actually doing the work.
-1 give us | -10 | - 6 which equals 4, so we already know it has to be B
0 gives us | -3 | - 6 which equals -3 which is less than 3
2 gives us | 14 -3 | - 6 which equals 5 which is not < 3
We have imperically proven it is B in addition to the other fine proofs actually doing the work.
2009-02-16 9:44 pm
|7k - 3| - 6 < 3
|7k - 3| < 9
When k >= 3/7:
7k-3 < 9
7k < 12
k < 12/7
Therefore 3/7 <= k < 12/7 is valid.
When k <=3/7:
3-7k < 9
-7k < 6
7k > -6
k > -6/7
Therefore -6/7 < k <= 3/7 is valid.
So the answer is b. (-6 / 7, 12 / 7)
|7k - 3| < 9
When k >= 3/7:
7k-3 < 9
7k < 12
k < 12/7
Therefore 3/7 <= k < 12/7 is valid.
When k <=3/7:
3-7k < 9
-7k < 6
7k > -6
k > -6/7
Therefore -6/7 < k <= 3/7 is valid.
So the answer is b. (-6 / 7, 12 / 7)
2009-02-16 9:44 pm
|a| < b means -b<a<b
so |7k-3| - 6 < 3 => |7k-3|<9 so
-9<7k-3<9 (add 3 to all the sides)
-6<7k<12
divide by 7:
-6/7 < k < 12/7 or k = (-6/7 , 12/7) (b)
so |7k-3| - 6 < 3 => |7k-3|<9 so
-9<7k-3<9 (add 3 to all the sides)
-6<7k<12
divide by 7:
-6/7 < k < 12/7 or k = (-6/7 , 12/7) (b)
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