Inequality equation help please

[Kmaughan]

hi, I'm trying to understand how to do my homework. I'm currently working on inequalities but I dont understand how to do these equations at all??

the solution set x=2 or x=4 is consistent with an equation of the form

a. |ax+b|<c where c is greater than zero
b. |ax+b>c where c is greater than zero
c. |ax+b|= c where c is less than zero
d. |-ax+b| < where c is less than zero

Can some body please help I don't understand this at all

 

[Ishuda]

hi, I'm trying to understand how to do my homework. I'm currently working on inequalities but I dont understand how to do these equations at all??

the solution set x=2 or x=4 is consistent with an equation of the form

a. |ax+b|<c where c is greater than zero
b. |ax+b>c where c is greater than zero
c. |ax+b|= c where c is less than zero
d. |-ax+b| < where c is less than zero

Can some body please help I don't understand this at all

I'm not sure what is wanted exactly because of the lack of information on a and b. However one can say something about c. and d. just because of the definition of absolute value which can be stated as |x| is equal to x if x in non-negative and is equal to -x if x is negative.

So, for c., Can the absolute value of something be equal to a negative number? For d., can the absolute value of something be negative, that is less than some negative number c [assuming the statement was d. |-ax+b| < c where c is less than zero].

 

[Steven G]

hi, I'm trying to understand how to do my homework. I'm currently working on inequalities but I dont understand how to do these equations at all??

the solution set x=2 or x=4 is consistent with an equation of the form

a. |ax+b|<c where c is greater than zero
b. |ax+b>c where c is greater than zero
c. |ax+b|= c where c is less than zero
d. |-ax+b| < where c is less than zero

Can some body please help I don't understand this at all

1st, to answer these questions it does not matter what is between the absolute bar.

2nd, what is the absolute value of a number. The answer is that it is the same number between the absolute value bar but without the sign.

Assume for ease that c=5 for parts a and b.

| whatever | < 5 means all numbers when we ignore their signs is less than 5. For positive number we get all positive numbers less than 5. 0<5 so 0 can be between the bars. Now which negative numbers can we put between the bars, ie which negative numbers when we ignore their signs are less then 5. That answer is all numbers between -5 and 0. Putting this all together, what can go between the bars are all numbers between -5 and 5. This is not the answer since you want just two numbers to be the answer, not a range. For the record we have -5< ax+b <5

| whatever | > 5 means all numbers when we ignore their signs are greater than 5. 0 is not greater than 5. Positive number greater than 5 are greater than 5. Negative numbers that when we ignore their signs are greater than 5 are numbers less than -5. Again, we have many values that work, not just two values. For the record we would solve ax+b >5 and ax + b <-5

|whatever|= c where c is less than zero. This is asking which numbers that when you ignore its sign equals c (say -5). Well there is no such number since when you ignore the sign of any number you either get 0 or positive which is never less that c (-5)


|whatever| < c, where c is less than zero. Say c= -5. This is asking for which numbers when you ignore their sign will give you a number less then -5. Again this can't happen.


However if you had |whatever| =c , and c is positive, say c=5. Then this is asking which numbers when you ignore its sign is 5. This has two answers, -5 and 5. So we would solve ax+b=-5 and ax+b =5 and for the correct choices of a and b the solution could be x=2 and x=5

So the answer to your question is no choices work.

I hope this helps!

 

[Steven G]

Since x=2 or x=4: (x-2)(x-4) = 0 : x^2 - 6x + 8 = 0
So a=1, b=-6, c=8
Play with that...

Hmm Dennis I do not think you are correct (but usually I am wrong about these things). Let y=ax+t for ease of typing.

Then |y|<c where c is greater than zero has an interval (-c<y<c), not just two values (like x=2 and x=4)

Then |y|>c where c is greater than zero has two intervals namely y<-c and y>c, not just two values (like x=2 and x=4)

Then |y|= c where c is less than zero has no solutions

and |y| < c where c is less than zero has no solutions.

No choice has just two solutions.

Am I missing something??

 

[Steven G]

No.
I was giving him something "to look at"; not a solution...

OK, cool. Thanks!

 

[Mrspi]

hi, I'm trying to understand how to do my homework. I'm currently working on inequalities but I dont understand how to do these equations at all??

the solution set x=2 or x=4 is consistent with an equation of the form

a. |ax+b|<c where c is greater than zero
b. |ax+b>c where c is greater than zero
c. |ax+b|= c where c is less than zero
d. |-ax+b| < where c is less than zero

Can some body please help I don't understand this at all

Only ONE of those choices is an equation, and since the question is which EQUATION would have a solution consistent with a solution set of x = 2 or x = 4, we only need to look at that equation.

|ax + b| = c where c is less than 0
That's the only answer choice containing an equation.
And that equation says that the absolute value of something is less than 0.
But an absolute value can't be less than 0.....the absolute value of anything must be greater than or equal to 0. After all, an absolute value is the distance between a number and 0 on the number line, and a distance can't be negative.

So, that can't be the right answer.....

NONE of these answer choices is correct.

I hope this helps you.