Ahhh....I see. So for say 6≤x<11 x is indeed positive and that case is true. Then we could test the negative case 1<x<6 in which case x is positive so for both the positive and negative cases, x is a positive #.
Thanks!
Your method is almost correct, remember to include an = sign (as I told you before).
Here we'll work on the intervals, so
Positive x\geq{6}: 2x-12<10 ==> 2x<22 ==> x<11
You have found the range of solutions x<11, correct. But we are considering values x\geq{6}, so we have to MERGE the intervals into
6\leq{x}<11 (just united them). This right here is the result of the analysis, so the equation |2x-12|<10 (in its positive case) holds true into that range of values.
This is an important concept to understand.
Negative x<6: -2x+12<10 ==> -2x<-2 ==> x>1
Same thing here, merge the intervals into 1<x<6. So the function |2x-12|<10 (in its negative case) holds true into that range of values. Same thing as above.
But we are studying the overall function |2x-12|<10, so we have to merge the intervals in which its separated cases hold true, so
1<x<6+6\leq{x}<11 = 1<x<11
And here is my result
The question asks is x positive? YES! It's in the range 1-11 so it's positive
When you study and abs value, first and foremost pay attention to the interval you are considering.
If you are asked a specific value of x=?, then just check weather x is in the interval you are considering.
If you have an inequality, MERGE the intervals you have => so you'll find the range into which the equation holds true.
If the intervals have NO COMMON VALUES example x>5 and x<2, that equation NEVER holds true.
Thanks!
Zarrolou wrote:
WholeLottaLove wrote:
Hi there.
I revisited this problem and attacked it in the way you have taught me to go about solving problems. Namely, for #1 I checked for which values would 2x-12 be negative and positive and then went about establishing the positive and negative cases as follows:
2x-12 is positive when x>6.
2x-12 is negative when x<6.
Therefore,
Positive: 2x-12<10 ==> 2x<22 ==> x<11
Negative: -2x+12<10 ==> -2x<-2 ==> x>1
Here is my problem. For example, regarding the positive case. 2x-12 is positive when x>6 and indeed for the positive case we find that x<11. Therefore, x may or may not be greater than 6. I see that for this problem you combine the statements into 1<x<11 but even then, couldn't x be less than 1? I understand the concepts used to solve this problem but I'm not sure I would be able to differentiate between two (seemingly) different problems that demand (seemingly) different approaches to solve them.
Thanks!
I revisited this problem and attacked it in the way you have taught me to go about solving problems. Namely, for #1 I checked for which values would 2x-12 be negative and positive and then went about establishing the positive and negative cases as follows:
2x-12 is positive when x>6.
2x-12 is negative when x<6.
Therefore,
Positive: 2x-12<10 ==> 2x<22 ==> x<11
Negative: -2x+12<10 ==> -2x<-2 ==> x>1
Here is my problem. For example, regarding the positive case. 2x-12 is positive when x>6 and indeed for the positive case we find that x<11. Therefore, x may or may not be greater than 6. I see that for this problem you combine the statements into 1<x<11 but even then, couldn't x be less than 1? I understand the concepts used to solve this problem but I'm not sure I would be able to differentiate between two (seemingly) different problems that demand (seemingly) different approaches to solve them.
Thanks!
Your method is almost correct, remember to include an = sign (as I told you before).
Here we'll work on the intervals, so
Positive x\geq{6}: 2x-12<10 ==> 2x<22 ==> x<11
You have found the range of solutions x<11, correct. But we are considering values x\geq{6}, so we have to MERGE the intervals into
6\leq{x}<11 (just united them). This right here is the result of the analysis, so the equation |2x-12|<10 (in its positive case) holds true into that range of values.
This is an important concept to understand.
Negative x<6: -2x+12<10 ==> -2x<-2 ==> x>1
Same thing here, merge the intervals into 1<x<6. So the function |2x-12|<10 (in its negative case) holds true into that range of values. Same thing as above.
But we are studying the overall function |2x-12|<10, so we have to merge the intervals in which its separated cases hold true, so
1<x<6+6\leq{x}<11 = 1<x<11
And here is my result
The question asks is x positive? YES! It's in the range 1-11 so it's positive
When you study and abs value, first and foremost pay attention to the interval you are considering.
If you are asked a specific value of x=?, then just check weather x is in the interval you are considering.
If you have an inequality, MERGE the intervals you have => so you'll find the range into which the equation holds true.
If the intervals have NO COMMON VALUES example x>5 and x<2, that equation NEVER holds true.