Further Inequations involving the pronumeral on the denominator (1 Viewer)

[academical]

I really just need someone to explain this to me if at all possible

I don't understand how to solve them.

Here are some questions:

Q1. (1)/(x-1)>4

Q2. (1)/(z+3)<-5

Q3. (8p+7)/(2p-9)>5

Q4. (5m+4)/(2m)<(1/4)

Thanks in advance for your help

 

[RivalryofTroll]

1/(x-1) > 4

Method 1: You times (x-1)^2 to both sides (TIMES DENOMINATOR SQUARED TO BOTH SIDES)
1(x-1) > 4(x-1)^2
x-1 > 4(x^2 - 2x + 1)
x-1 > 4x^2 - 8x + 4
0 > 4x^2 - 9x + 5
4x^2 - 9x + 5 < 0
(4x - 5 ) (x - 1) < 0
Draw a concave up (since x^2 is positive) parabola and labelling the intercepts as x = 5/4 and x = 1
SHADE THE REGION BELOW SINCE IT IS LESS THAN.
You see that it is in between 5/4 and 1.
So 1<x<5/4

Method 2: Making the RHS = 0
1/(x-1) - 4 > 0
[1- 4(x-1)] / (x-1) > 0 - Common demoninator
(5-4x)/(x-1) > 0
Times (x-1)^2 to both sides.
(5-4x)/(x-1) > 0
x = 5/4 and x = 1 are the parabola intercepts so shade the above region, SINCE IT IS MORE THAN, of the concave down parabola since x^2 is negative in this case.
Therefore, 1<x<5/4


NOTE: If it's 1/(x-1) >= 4, then the answer would be 1<x<= 5/4 BECAUSE x=/= 1 (since the denominator cannot be 0 for x-1)

Do the rest yourself (just apply these methods and you'll be fine... Also read how your textbook does it)

THINGS TO NOTE: If a/b > 0 then a times b > 0 (or a.b > 0)
Do it graphically when solving the inequality
If it's >= or <= then if the denominator is x-a then x cannot equal to a since the denominator =/= 0.....

THINGS TO DO NEXT TIME:
- Listen to your teacher
- Read the textbook carefully

 

[qrpw]

The other method that most textbooks and teachers use is the "gatepost" or "boundary points" or whatever. Personally I always use this one as it requires less work and is faster. Although the method above will always work as well, once you get into harder Year 11/12 inequalities, multiplying both sides by squares gets very fiddly and annoying.

So for 1/(x-1)>4
Find the "boundarypoints". One will be 1, as x cannot be 1 or you will have a zero on the denominator.

Solve for the equality.






So the boundary points are 1 and 5/4.

Now test the three regions on the number line. Less than 1, between 1 and 5/4, and more than 5/4. For example test x=0, substitute into 1/(x-1)>4. It is not true, therefore the region does not work. Test the other two, and you will get 1<x<5/4

 

[D94]

You just need to remember that since they are inequalities, we don't know whether the denominator is positive or negative for what ever values of x which satisfy the domain. If for some values, the denominator is negative, and then you just multiply it through, the inequality sign may change, thus giving you an incorrect result if you didn't change the sign. Hence why one of better methods is to multiply through by the square of the denominator because this ensures that you multiply through by a positive number, so the sign doesn't change.

 

[academical]

THINGS TO DO NEXT TIME:
- Listen to your teacher
- Read the textbook carefully

I really appreciate your help

Our teacher hasn't gone through this in class though, hence why I needed help... He told us that we have a test this week and what topics would be in it - half of which he hasn't taught us, but we were supposed to apparantly know

And we were given Maths In Focus - I didn't understand their explanation.

Once again, thank you soooo much though

 

[academical]

The other method that most textbooks and teachers use is the "gatepost" or "boundary points" or whatever. Personally I always use this one as it requires less work and is faster. Although the method above will always work as well, once you get into harder Year 11/12 inequalities, multiplying both sides by squares gets very fiddly and annoying.

So for 1/(x-1)>4
Find the "boundarypoints". One will be 1, as x cannot be 1 or you will have a zero on the denominator.

Solve for the equality.






So the boundary points are 1 and 5/4.

Now test the three regions on the number line. Less than 1, between 1 and 5/4, and more than 5/4. For example test x=0, substitute into 1/(x-1)>4. It is not true, therefore the region does not work. Test the other two, and you will get 1<x<5/4

This is the method my text book used - I just couldn't follow it haha.

Thanks heaps

 

[RivalryofTroll]

I really appreciate your help

Our teacher hasn't gone through this in class though, hence why I needed help... He told us that we have a test this week and what topics would be in it - half of which he hasn't taught us, but we were supposed to apparantly know

And we were given Maths In Focus - I didn't understand their explanation.

Once again, thank you soooo much though

Also, if the denominator is surrounded by absolute value, then you are allowed to times the denominator over to the other side as absolute value ensures that the denominator is positive. You can't times the denominator over if it can be positive or negative since you don't know whether the sign flips or not but absolute value means it will always be positive.

So if 69/(|x-2|) > 1
Then 69 > 1.|x-2|

 

[RivalryofTroll]

This is the method my text book used - I just couldn't follow it haha.

Thanks heaps

My teacher calls it the critical points method.
It's the one that requires the least amount of algebra work.
I'd recommend you use it OR making the RHS = 0.

Though, I've heard from my teacher that the critical points method (or whatever you wanna call it) is an all or nothing method.
Means, if you get it wrong, you might not get working out marks (or less marks than the 'times denominator squared to both sides method' methods).

If you want to ensure 100% that you will get the maximum amount of working out marks even if you get it wrong then the denominator squared to both sides is the safest route.

 

[academical]

Also, if the denominator is surrounded by absolute value, then you are allowed to times the denominator over to the other side as absolute value ensures that the denominator is positive. You can't times the denominator over if it can be positive or negative since you don't know whether the sign flips or not but absolute value means it will always be positive.

So if 69/(|x-2|) > 1
Then 69 > 1.|x-2|

Ohh, ok...that makes a lot of sense, thanks

I really can't thak you enough!!!

Rep'd haha!

 

[academical]

If you want to ensure 100% that you will get the maximum amount of working out marks even if you get it wrong then the denominator squared to both sides is the safest route.

Thanks for sharing that... I didn't know

 

[Magical Kebab]

1/(x-1) > 4

Method 1: You times (x-1)^2 to both sides (TIMES DENOMINATOR SQUARED TO BOTH SIDES)
1(x-1) > 4(x-1)^2
x-1 > 4(x^2 - 2x + 1)
x-1 > 4x^2 - 8x + 4
0 > 4x^2 - 9x + 5
4x^2 - 9x + 5 < 0
(4x - 5 ) (x - 1) < 0
Draw a concave up (since x^2 is positive) parabola and labelling the intercepts as x = 5/4 and x = 1
SHADE THE REGION BELOW SINCE IT IS LESS THAN.
You see that it is in between 5/4 and 1.
So 1<x<5/4

Method 2: Making the RHS = 0
1/(x-1) - 4 > 0
[1- 4(x-1)] / (x-1) > 0 - Common demoninator
(5-4x)/(x-1) > 0
Times (x-1)^2 to both sides.
(5-4x)/(x-1) > 0
x = 5/4 and x = 1 are the parabola intercepts so shade the above region, SINCE IT IS MORE THAN, of the concave down parabola since x^2 is negative in this case.
Therefore, 1<x<5/4


NOTE: If it's 1/(x-1) >= 4, then the answer would be 1<x<= 5/4 BECAUSE x=/= 1 (since the denominator cannot be 0 for x-1)

Do the rest yourself (just apply these methods and you'll be fine... Also read how your textbook does it)

THINGS TO NOTE: If a/b > 0 then a times b > 0 (or a.b > 0)
Do it graphically when solving the inequality
If it's >= or <= then if the denominator is x-a then x cannot equal to a since the denominator =/= 0.....

THINGS TO DO NEXT TIME:
- Listen to your teacher
- Read the textbook carefully

No offense but that's a waste of time. Square the top (both lhs and rhs) by the denominator than solve for x. After that, test a number between what u got for x and if it is true, draw the lines in and if it is false draw the lines out. Done.

 

[RealiseNothing]

No offense but that's a waste of time. Square the top (both lhs and rhs) by the denominator than solve for x. After that, test a number between what u got for x and if it is true, draw the lines in and if it is false draw the lines out. Done.

Boundary point method is even quicker.

 

[Magical Kebab]

Boundary point method is even quicker.

Haven't heard of it. I have only seen the method in Maths in focus and that method I discussed.

 

[RealiseNothing]

Haven't heard of it. I have only seen the method in Maths in focus and that method I discussed.

The two boundary points are:

1) What 'x' can't be. ie if the denominator is x-3, then x can't be 3, so this is the first boundary point.

2) Solve for 'x' with an equality sign instead of an inequality. Whatever 'x' equals is your second point.

Just use a test point to figure out which way the inequality sign goes.

 

[Magical Kebab]

The two boundary points are:

1) What 'x' can't be. ie if the denominator is x-3, then x can't be 3, so this is the first boundary point.

2) Solve for 'x' with an equality sign instead of an inequality. Whatever 'x' equals is your second point.

Just use a test point to figure out which way the inequality sign goes.

Sounds ok. It's a really small part of the work, only one question per year comes up, if any.

 

[RivalryofTroll]

No offense but that's a waste of time. Square the top (both lhs and rhs) by the denominator than solve for x. After that, test a number between what u got for x and if it is true, draw the lines in and if it is false draw the lines out. Done.

idgaf so yeah, I won't take any offense. As long as I get the right answer it doesn't matter. Making the RHS = 0 and solving it graphically is the quickest and safest option for me. I'm not some god at maffs so yeah.. I just wanted to show OP whatever I use....
You can hate on my methods if you want.. The extra time wasted is very minimal anyways so yeah...

 

[Magical Kebab]

idgaf so yeah, I won't take any offense. As long as I get the right answer it doesn't matter. Making the RHS = 0 and solving it graphically is the quickest and safest option for me. I'm not some god at maffs so yeah.. I just wanted to show OP whatever I use....
You can hate on my methods if you want.. The extra time wasted is very minimal anyways so yeah...

Lol don't take it the wrong way. Anyways u obviously took offense to that, no need to, if ur way works than that's all that matters.

 

[RivalryofTroll]

Lol don't take it the wrong way. Anyways u obviously took offense to that, no need to, if ur way works than that's all that matters.

How so? Actually, no need to answer. Pointless arguing is stupid.
I've always been a slow worker so obv I prefer slower methods. As simple as that.

 

[gorillaz]

whatever method should be okay. besides, its important that you learn and practice all of them anyway.