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# Inequalities - Graphical form help needed!!! Tweet

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1. Inequalities - Graphical form help needed!!!
Having trouble with the equation shown, I know how to get the values of x but the problem is what way does the sign go?

I sort of understand which way it is when working the equation out algebraically. I want to be able to know which way the sign is supposed to go directly from the graph of the function. Is it even possible?
I can draw it on my calculator with no problem, but when writing the solution of x, how am I supposed to know which way the sign is supposed to be?

Thanks, will +rep
Last edited by Doctor.; 27-05-2012 at 16:53.
2. Re: Inequalities - Graphical form help needed!!!
(Original post by Sgt.Incontro)
Abs(x+2) > 2x+1
x^2+4x+4>2x+1
x^2+2x+3>0
(x+1)(x+2)>0

One method is by drawing the graph of y=(x+1)(x+2).

The solutions are x>2 and x<-1, as the curve is below the x-axis, in the negative y region between these two coordinates. That is how you can tell.

Graphing is usually the easiest method, another method is substituting points into the given equation, but this method is better IMO.

If you have any questions, feel free to ask me.
You're awesome Thank for the quick response.

I see that you've squared the modulus side, but not the other side. I thought "whatever you do to the right, you have to do to the left" would apply here?

Also I am not sure but:
x^2+2x+3>0
(x+1)(x+2)>0
That isn't correct?

I just drew the graph, I see it's in the region you said but how I know it's x>2 and x<-1?

I do find Graphing much easier, so really want to use it to the fullest Sorry for all the questions, I really need to master this. Need to get an amazing mark in my C3 exam

thank you so so so much
3. Re: Inequalities - Graphical form help needed!!!
(Original post by Doctor.)
You're awesome Thank for the quick response.

I see that you've squared the modulus side, but not the other side. I thought "whatever you do to the right, you have to do to the left" would apply here?

Also I am not sure but:
x^2+2x+3>0
(x+1)(x+2)>0
That isn't correct?

I just drew the graph, I see it's in the region you said but how I know it's x>2 and x<-1?

I do find Graphing much easier, so really want to use it to the fullest Sorry for all the questions, I really need to master this. Need to get an amazing mark in my C3 exam

thank you so so so much
Ignore my previous post, it was rubbish I admit.

I apologize; I have been studying all day, and I was completely off the plot.

YES, square BOTH sides to get rid of the modulus bracket on the left.

Simplify the expression, and then draw the graph of it.

Sorry for the mistake once again. I will post again, shortly.

P.S I am not usually rubbish, I got 92 in C3, no reason why you can't get better
Last edited by Sgt.Incontro; 27-05-2012 at 17:49.
4. Re: Inequalities - Graphical form help needed!!!
Abs(x+2)>2x+1
x^2+4x+4>4x^2+4x+1

Then simplify the above to get:
(3x^2)-3<0

On your calculator, draw the graph of y=(3x^2)-3

Since your y-function ((3x^2)-3) is less than 0 from the inequality above, the valid solutions lie where the function is NEGATIVE.

This is the region

x>-1
x<1

So sorry again.
5. Re: Inequalities - Graphical form help needed!!!
(Original post by Sgt.Incontro)
Abs(x+2)>2x+1
x^2+4x+4>4x^2+4x+1

Then simplify the above to get:
(3x^2)-3<0

On your calculator, draw the graph of y=(3x^2)-3

Since your y-function ((3x^2)-3) is less than 0 from the inequality above, the valid solutions lie where the function is NEGATIVE.

This is the region

x>-1
x<1

So sorry again.
It's fine

"Since your y-function ((3x^2)-3) is less than 0 from the inequality above, the valid solutions lie where the function is NEGATIVE."

-Just done the graph, the answer in the book says only "x<1" So how do we know that we don't include the x>-1?
6. Re: Inequalities - Graphical form help needed!!!
Anyone know?
7. Re: Inequalities - Graphical form help needed!!!
bump..
8. Re: Inequalities - Graphical form help needed!!!
(Original post by Doctor.)
It's fine

"Since your y-function ((3x^2)-3) is less than 0 from the inequality above, the valid solutions lie where the function is NEGATIVE."

-Just done the graph, the answer in the book says only "x<1" So how do we know that we don't include the x>-1?
You have been stuck on it from a long time, so now i am going to give a full solution.

Draw the graph of y=2x+1 and y=|x+2| on the same sketch.

Here are the graphs:

We can see that x+2 intersects with 2x+1, solving x+2=2x+1 gives x=1, so the both graphs interesect at x=1.

Now we need to consider the region in which |x+2|>2x+1, they both intersect at x=1, so the region must be x<1 or x>1. By seeing the graph you can deduce that the range is x<1.
9. (Original post by raheem94)
You have been stuck on it from a long time, so now i am going to give a full solution.

Draw the graph of y=2x+1 and y=|x+2| on the same sketch.

Here are the graphs:

We can see that x+2 intersects with 2x+1, solving x+2=2x+1 gives x=1, so the both graphs interesect at x=1.

Now we need to consider the region in which |x+2|>2x+1, they both intersect at x=1, so the region must be x<1 or x>1. By seeing the graph you can deduce that the range is x<1.
Really appreciate all the effort you put into that so thank ii very much.

Just one small question - How do you deduce by looking at the graph it must be x<1?
10. Re: Inequalities - Graphical form help needed!!!
(Original post by Doctor.)
Really appreciate all the effort you put into that so thank ii very much.

Just one small question - How do you deduce by looking at the graph it must be x<1?
You don't actually need the graph because you know that X+2>2X+1 and this means that X must be smaller than 1 because the lower the value of X,the less the value of timsing by 2 will be, on the other side of the greater than symbol and thus making X+2 bigger than 2x+1
11. Re: Inequalities - Graphical form help needed!!!
(Original post by Doctor.)
Really appreciate all the effort you put into that so thank ii very much.

Just one small question - How do you deduce by looking at the graph it must be x<1?
The black line is 2x+1 and the green line is |x+2|. When you look at the left side of x=1, you can see the green line is always above the black line, hence |x+2|>2x+1.

Looking at the right side of x=1, you can see that the black line is always above the green line, hence 2x+1>|x+2| in that region.

Do you get it?
12. (Original post by raheem94)
The black line is 2x+1 and the green line is |x+2|. When you look at the left side of x=1, you can see the green line is always above the black line, hence |x+2|>2x+1.

Looking at the right side of x=1, you can see that the black line is always above the green line, hence 2x+1>|x+2| in that region.

Do you get it?
I see what you mean, but I can't see why x<1

I understand how you got the equation, but how are you working out that x<1

Sorry, this is really confusing for me
13. Re: Inequalities - Graphical form help needed!!!
(Original post by Doctor.)
I see what you mean, but I can't see why x<1

I understand how you got the equation, but how are you working out that x<1

Sorry, this is really confusing for me
Look at the graphs the other posters put up.

On the part of that graph where x<1, you can see that the green function, is higher/larger than the other function. This is where the valid interval is.

The interval you look at must be this, since the green function represents the modulus expression, which is greater than the other function, as originally stated in the question.

Does that help a bit?
14. Re: Inequalities - Graphical form help needed!!!
(Original post by Doctor.)
I see what you mean, but I can't see why x<1

I understand how you got the equation, but how are you working out that x<1

Sorry, this is really confusing for me
For graphic evidence,just look at the point of intersection and if X+1 is greater than 2x+1,it must be above the line because higher up the line,gives a higher value because it is more positive-this bit is hard to explain why and then if you put the point of intersection higher,what do you notice?-the x value starts to decrease.
15. Re: Inequalities - Graphical form help needed!!!
(Original post by Doctor.)
I see what you mean, but I can't see why x<1

I understand how you got the equation, but how are you working out that x<1

Sorry, this is really confusing for me
My graph shows two equations y=|x+2| and y=2x+1.

In the region, x<1, the green line is always higher than the line 2x+1.

To check this sub in some values,
sub in x=0.5(this is less than 1) you get y=|0.5+2|=2.5 and y=2(0.5)+1=2. Hence you can see that |x+2|>2x+1 because 2.5>2.

Sub in a value greater than 1
Sub in x=1.5, you will get, y=|1.5+2|=3.5 and y=2(1.5)+1=4, hence |x+2|<2x+1 because 3.5<4.

Does this make any sense?
16. (Original post by raheem94)
My graph shows two equations y=|x+2| and y=2x+1.

In the region, x<1, the green line is always higher than the line 2x+1.

To check this sub in some values,
sub in x=0.5(this is less than 1) you get y=|0.5+2|=2.5 and y=2(0.5)+1=2. Hence you can see that |x+2|>2x+1 because 2.5>2.

Sub in a value greater than 1
Sub in x=1.5, you will get, y=|1.5+2|=3.5 and y=2(1.5)+1=4, hence |x+2|<2x+1 because 3.5<4.

Does this make any sense?
Ah I understand the testing with values you just did.
So you say x<1 because the modulus function is bigger than 2x+1 at all times when x<1??

I think that makes sense lol

Ps thank pi everyone for helping me, I'm just going to try a slightly harder quesion and see what I get with that
17. Re: Inequalities - Graphical form help needed!!!
(Original post by Doctor.)
Ah I understand the testing with values you just did.
So you say x<1 because the modulus function is bigger than 2x+1 at all times when x<1??

I think that makes sense lol

Ps thank pi everyone for helping me, I'm just going to try a slightly harder quesion and see what I get with that
Yes, now you are finally correct!

Though i only said about testing values because you were failing to understand, usually by looking at the graph you can easily deduce this.
18. Re: Inequalities - Graphical form help needed!!!
(Original post by Sgt.Incontro)
Ignore my previous post, it was rubbish I admit.

I apologize; I have been studying all day, and I was completely off the plot.

YES, square BOTH sides to get rid of the modulus bracket on the left.

Simplify the expression, and then draw the graph of it.

Sorry for the mistake once again. I will post again, shortly.

P.S I am not usually rubbish, I got 92 in C3, no reason why you can't get better
May be you aren't always rubbish but the method you were trying to use looked rubbish to me lol

I don't see the point of squaring both sides in this question, squaring can add extra solutions so i always try to avoid it unless the question isn't possible without it.
19. (Original post by raheem94)
My graph shows two equations y=|x+2| and y=2x+1.

In the region, x<1, the green line is always higher than the line 2x+1.

To check this sub in some values,
sub in x=0.5(this is less than 1) you get y=|0.5+2|=2.5 and y=2(0.5)+1=2. Hence you can see that |x+2|>2x+1 because 2.5>2.

Sub in a value greater than 1
Sub in x=1.5, you will get, y=|1.5+2|=3.5 and y=2(1.5)+1=4, hence |x+2|<2x+1 because 3.5<4.

Does this make any sense?
(Original post by Dalek1099)
For graphic evidence,just look at the point of intersection and if X+1 is greater than 2x+1,it must be above the line because higher up the line,gives a higher value because it is more positive-this bit is hard to explain why and then if you put the point of intersection higher,what do you notice?-the x value starts to decrease.
(Original post by Sgt.Incontro)
Look at the graphs the other posters put up.

On the part of that graph where x<1, you can see that the green function, is higher/larger than the other function. This is where the valid interval is.

The interval you look at must be this, since the green function represents the modulus expression, which is greater than the other function, as originally stated in the question.

Does that help a bit?
OMG YOU ARE ALL ABSOLUTE LEGENDS!!!!!! (as if I just got this happy after doing a equation correctly lool)

I just did the following equation: |x-2|>/|2x-3|

I just found where they both intersect first. Then I looked at the graph and saw |x-2| is always greater than or equal to |2x-3| when the |x-2| bit is ABOVE the other one.

And I got the signs tge correct way!!!! Thank you all sosososos much.

I will +rep you all ASAP