Any math teachers out there?

[Minnie*J*]

I am studying for at test and one of the questions is to solve an equation with an absolute value and I don't know how to do it. Anyone know?

-3lx+4l>15

My thinking is that any number for x +4 doesn't matter because the absolute signs are going to make that equation positive, then multiply by -3 and that number is going to be negative and that is always going to be less than 15. So is this a typo or am I missing something?

Thanks.

 

[Cinderella94]

So start by dividing both sides by -3, which would flip your inequality sign. Then your new equation is |x+4|<-5. But then...I don't know what you'd do sorry!

 

[PrincesCJM]

-3|x+4|>15
First divide both sides by -3. When dividing by a negative, the inequality symbol changes to get |x+4|<-5
To then solve from here, split the inequality into two pieces x+4<-5 and x+4>-5
Subtract 4 from both sides of both of the equations to get x<-9 and x>-9
To me, then the answer can be any number but -9.

 

[Cinderella94]

-3|x+4|>15
First divide both sides by -3. When dividing by a negative, the inequality symbol changes to get |x+4|<-5
To then solve from here, split the inequality into two pieces x+4<-5 and x+4>-5
Subtract 4 from both sides of both of the equations to get x<-9 and x>-9
To me, then the answer can be any number but -9.

Ooooh yes! This is right.

ETA: Maybe I have no idea what I'm talking about but I know the step I posted above is right!

 

[crashbb]

-3|x+4|>15
First divide both sides by -3. When dividing by a negative, the inequality symbol changes to get |x+4|<-5
To then solve from here, split the inequality into two pieces x+4<-5 and x+4>-5
Subtract 4 from both sides of both of the equations to get x<-9 and x>-9
To me, then the answer can be any number but -9.

I don't see how that works. Try subbing in some numbers.

X=1

-3 x |1-4| = -3 x |-3| = -3 x 3 = -9

-9 is NOT bigger than 15

X=-10

-3 x |1 - -10| = -3 x |1+10| = -3 x |11| = -33

Again, not bigger than 15

I think you ignored the absolute brackets.

 

[ajwomic]

Math teacher here! You guys were so close! After dividing by -5 and changing the sign, you split it like this

X+4< -5

and

-(x+4)< -5 or -x-4 < -5

The first gives you x < -9

and the second gives you x >1

So the answer includes all values greater than 1 and less the negative 9.

 

[crashbb]

Math teacher here! You guys were so close! After dividing by -5 and changing the sign, you split it like this

X+4< -5

and

-(x+4)< -5 or -x-4 < -5

The first gives you x < -9

and the second gives you x >1

So the answer includes all values greater than 1 and less the negative 9.

Again, I think you are ignoring the ABSOLUTE brackets (it looks like you just used regular brackets).

 

[ajwomic]

You actually are not ignoring them when you split it into 2 equations. When splitting it you are basically saying it could be positive or negative inside the brackets and still satisfy the inequality. So we let one side be the positive "version" x+4 and the other be the negative "version" -(x+4) or distributed -x-4. Does this make sense?

 

[princessmom29]

I am studying for at test and one of the questions is to solve an equation with an absolute value and I don't know how to do it. Anyone know?

-3lx+4l>15

My thinking is that any number for x +4 doesn't matter because the absolute signs are going to make that equation positive, then multiply by -3 and that number is going to be negative and that is always going to be less than 15. So is this a typo or am I missing something?

Thanks.

here is my answer: you are right, there is no real solution.
The absolute value of anything is always a positive number, so no matter what number you substitute in lx+4l is always positive. You cannot ever multiply a positive number by -3 and get and answer greater than 15, so there is no real number that will make the inequality true. So, the answer is "no real solution".

 

[crashbb]

You actually are not ignoring them when you split it into 2 equations. When splitting it you are basically saying it could be positive or negative inside the brackets and still satisfy the inequality. So we let one side be the positive "version" x+4 and the other be the negative "version" -(x+4) or distributed -x-4. Does this make sense?

It's been ages (I do biostatistics now).

I'm still not getting it though.

Let's plug in 2 (which, according you should work)

-3 x |2+4| = -3 x 6 = -18

This is NOT bigger than 15.

Or let's try -10 (again,according to you, this should work).

-3 x |-10 + 4| = -3 x |-6| = -3 x 6 = -18

Again, NOT bigger than 15.

What am I missing here?

BTW do you mean OR, rather than AND. When you say - "So the answer includes all values greater than 1 and less the negative 9. ". Because, of course, there are no values that are greater than 1 AND less then negative 9.

 

[ajwomic]

A clarification....your solution is all numbers greater than 1 AND less than negative 9. There are not any so no solution. I should have been more clear on that step.

 

[princessmom29]

You actually are not ignoring them when you split it into 2 equations. When splitting it you are basically saying it could be positive or negative inside the brackets and still satisfy the inequality. So we let one side be the positive "version" x+4 and the other be the negative "version" -(x+4) or distributed -x-4. Does this make sense?

you would have to isolate the absolute value to do that. by definition the absolute value MUST be psoitive so lx+4l <-5 can never be true so there is no solution.

 

[crashbb]

A clarification....your solution is all numbers greater than 1 AND less than negative 9. There are not any so no solution. I should have been more clear on that step.

Okay, *that* makes sense now.

 

[ajwomic]

you would have to isolate the absolute value to do that. by definition the absolute value MUST be psoitive so lx+4l <-5 can never be true so there is no solution.

You can always isolate the absolute value and algebraically show why there is no solution which is what I did. You can absolutely deduce it logically as well, but some are more complex and by isolating and solving, you can determine solutions or no solutions for just about any absolute value inequality.

 

[princessmom29]

You can always isolate the absolute value and algebraically show why there is no solution which is what I did. You can absolutely deduce it logically as well, but some are more complex and by isolating and solving, you can determine solutions or no solutions for just about any absolute value inequality.

I till don't think what you did is valid. When you split inequalities you are solving for 2 solution sets, one where the answer inside the inequality is positive, and one where it is negative. By setting it up the way you did, it makes it appear as if there are in fact 2 solution sets when in fact there are not.
lx+4l<-5 would break into:
x+4 <-5 and -(x+4) > 5 yielding 2 solution sets
x< -9 OR x> 1
and absolute value is always an OR solution, never an AND. it is 2 divergent soultion sets, excluding all values inbetween. this is not the correct way to solve this problem.

 

[tasha99]

I think you can just look at this and see there is no solution.

-3lx+4l>15

Okay, so divide both sides by -3 to get the absolute value on one side. That's

|x+4| < -5

But wait, the absolute value is always a positive number. There is no way the absolute value of x+4 is going to be less than -5. It's a trick question and the answer is "no solution." I think . . . doing it my "logic it out" way . . .

 

[tasha99]

I till don't think what you did is valid. When you split inequalities you are solving for 2 solution sets, one where the answer inside the inequality is positive, and one where it is negative. By setting it up the way you did, it makes it appear as if there are in fact 2 solution sets when in fact there are not.
lx+4l<-5 would break into:
x+4 <-5 and -(x+4) > 5 yielding 2 solution sets
x< -9 OR x> 1
and absolute value is always an OR solution, never an AND. it is 2 divergent soultion sets, excluding all values inbetween. this is not the correct way to solve this problem.

I agree with this. There are plenty of absolute value inequalities that have 2 solution sets. Not sure why this would be any different. I came up with the same answer doing it that way (x<-9 or x >1) but it doesn't work when plugging numbers in. I think it doesn't work because there is no solution, but I'm not sure how you would officially say that other than "wait, this makes no sense." There must be a way, though. Or maybe there is a rule that you can't have an absolute value on one side less than a negative value on the other. That's like saying +<-. It's a falsehood to begin with.

 

[RachelEllen]

-3|X+4| > 15

I agree, there's no solution as written. Whatever X is, the absolute will come out positive and turn negative when you multiply by negative 3.

However, it's kind of an interesting situation in theroy. When you have an absolute value that's > X, you end up with two points on the number line that create outward rays and every number on the ray is the solution (ie every number but the ones between them, an "or" situation)

When you have a absolute thats < X, you end up with a line segment with the answer being everything between the points and nothing on the outside (an "and" situation).

So, you can also have a situation where |X| is > than a negative number. There the rays actually point towards each other and pass, so the answer is every single number on the line.

Or, in this case, you have a solution where the answer is a line segment where the left side of the line is actually to the right of the right side of the line, i.e. no solution.

This may make sense to no one else, but I'm a really visual math person and I thought this problem brought up some cool stuff.

 

[BCDisneyFanatic]

If you have an absolute value inequality which contains a 'less than' sign, then your two solution sets are inclusive (i.e., you use an "and"). If it contains a 'greater than' sign, you use an "or".

Think of it this way...

|x| <3 ...in this case, we can divide our solution into x<3 and x>-3. Our solution set includes -2, -1, 0,1, and 2.

|x| >3 ...in this case, we divide our solution into x>3 or x<-3. Our solution set includes all numbers less than -3 and all numbers greater than 3, but does not include -2, -1, 0, 1 and 2.

For the OP's equation, since it's a 'less than' inequality:

|x+4| < -5

So, x+ 4 < -5 AND -(x+4) < -5; or, x < -9 AND x > 1 . Since this is not possible, there is no solution to the inequality.

 

[tasha99]

If you have an absolute value inequality which contains a 'less than' sign, then your two solution sets are inclusive (i.e., you use an "and"). If it contains a 'greater than' sign, you use an "or".

Think of it this way...

|x| <3 ...in this case, we can divide our solution into x<3 and x>-3. Our solution set includes -2, -1, 0,1, and 2.

|x| >3 ...in this case, we divide our solution into x>3 or x<-3. Our solution set includes all numbers less than -3 and all numbers greater than 3, but does not include -2, -1, 0, 1 and 2.

For the OP's equation, since it's a 'less than' inequality:

|x+4| < -5

So, x+ 4 < -5 AND -(x+4) < -5; or, x < -9 AND x > 1 . Since this is not possible, there is no solution to the inequality.

Thank you! That makes sense.