You can multiply both sides by x-2. You just have to deal with the two cases separately: (a) x-2 is positive, in which case x>2 and x>-3 (therefore x>2) and (b) x-2 is negative, in which case x
the mistake comes from the fact of not considering whether the denominator is negative or not when "multiplying" both sides of the inequality. A simplest way consists of taking that into account so let n=2x+1 and d=x-2 so that n/d>1 if d>0 then n>d else if -d>0 then d>n substituting n and d by their original expression we get if 2
It can be simplified by just saying that the function is not continuous in the domain (-inf,inf). It has a vertical asymptote at x=2 so it is only continuous in the domain (-inf,2)U(2,inf), which means that it should be evaluated for both. I think that the real problem here is that most of the time (if you are not used to many functions) people tend to forget domains, ranges and other attributes, including myself. That is easily solved by plotting the functions so you can grasp what is happening. If you where to have something like (2x+1)/(x^2-2) > 1 you would have to evaluate for 2 asymptotes in x=+-sqrt(2) so the domain would be (-inf,-sqrt(2)) U (-sqrt(2),sqrt(2)) U (sqrt(2),inf), for which the inequality only holds true for the domains (-sqrt(2),-1) U (sqrt(2),3).
@@marilynman indeed considering the domain where the function is not continuous and evaluating the sign of a product(±±>0, +-1 where n=2x+1 and d=x²-2 If d>0 ⟹ x
@@Brad-ey6cz no, you don't. let case 1 be x-2>0. This is only the case if x>2, and therefore the solution is [(x>-3) and (x>2)] which reduces to x>2. Let case 2 be x-2
@@Brad-ey6cz No, you wouldn't. Read what I wrote. For case 1 you get the two conditions: x>2 and x>-3, which together imply that x>2. For case 2 you get the two conditions: x
just split into 2 cases. 1) assume (x-2) > 0 that is: x > 2 2x+1 > x-2 x > -3 and (we assumed) x > 2 ... stronger condition is x > 2 2) assume (x-2) < 0 that is: x < 2 2x+1 < x-2 switch the comparison from ">" to "
@@pgskills The stronger condition is just describing if the range specified by the inequality is more specific about what parts are encapsulated if you know that x < 2 and that x < -3 you can simplify that to x < -3 because -3 < 2 similarly if you know that x > -3 and that x > 2 then you can simplify it to x > 2 because 2 > -3
@@christopheriman4921 Hey, thank you! Truly appreciate the clarity. I understand it now and actually have some vague recollection about learning this in connection to graphing quadratic and cubic functions. Much appreciated for the response. Thanks again. 👍👍
If you multiply both sides of the inequality by (x-2)^2 then you don’t have to worry about the sign change as (x-2)^2 cannot be negative for all real x. Then you get a quadratic inequality that gives the same set of solutions
I enjoy listening to Mr. H solving problems which I probably had in Algebra class 65 years ago. Rejuvenation of youth, sorta makes me want to turn on some oldies like “The Twist” by Hank Ballard and the Midnighters (Chubby Checker’s was a few years later).😊
thanks for including the demonstration that the original method resulted in incorrect answer. and we can see why. I'm an adult back in college and i recommend your videos to my classmates and my 17 year old son.
The first enswer is true if x-2 is positive. For example you can use this simplification, if the bottom is positive. If you want to use in positive or negative it should be followed by cases. The second method is simple in R. And useful because the multiplication are difficult in comparison to addition in > or < too. But is not needed if we work in N.
@kriswillems5661 You can do the cross multiply as well and get the results,just remember to build cases for X > and < 2 , because depending of sign of x-2 ,on cross multiply sign of inequality will change . Like 5> 3 ,if you multiply both sides with something negative the sign will change to -5 < -3.
1:10. An explanation of why we are setting the numerator and denominator to zero would be helpful here. It otherwise just comes across as a 'trick'. It is presumably because we want both the numerator and demominator to have the same sign for the inequality to be true. So the points where each can change sign are important, and these are at values of x which make each zero. However, if you just state that both the numerator and denominator have to have the same sign at this stage then it becomes obvious that the answer has to be x2.
In response to the post that "An explanation of why we are setting the numerator and denominator to zero would be helpful here." … My perception is that those points are transition points to the validity of the inequality. On one side of one of the zero points the inequality might be true, but on the other it might not. Thus the follow-up check.
It makes no sense setting the denominator to zero because the equation becomes undefined. It is best as pointed out by others to do cases where x-2 is negative and x-2 is positive.
I would *not* set it to zero. I come to the solution, if i think about "when does a fraction is greater than 0? It is the case, if both, enumerator and denominator, are greater than 0 or if both are smaller than 0.
I have been a practicing engineer for 37 years. I guarantee you that 99% of engineers and scientist that I know would multiply both sides by x-2 and solve for x = -3.
This is one way to do it in college-level pre-calculus: We start with (2*x+1)/(x-2) > 1. (1) Assume x-2>0, i.e., x>2. Multiply by (x-2) to get 2*x+1>x-2, (preserve the direction of the comparison sign), i.e., x>-3. The only way that we get x>2 and x>-3, simultaneously, is for [x>2] (2) Then, assume x-2
That was a powerful lesson! By bringing the equation to a comparison with Zero you brought extra numerical inferences into play. A General's approch rather than a Captain's :-)
Though i like your solution, there's nothing (technically) wrong with wanting to multiply by the variable expression '(x-2)' as a first step. The mistake is to assume that you would simply get '2x+1 > x-2' from it. The issue is that the variable expression, you want to multiply with, might be lower than, equal to or greater than zero. Here 'equal to zero' is not a valid option for a denominator 'x-2', therefore the correct equivalent tranformation would be the following: (2x+1)/(x-2) > 1 (x-2 < 0 and 2x+1 < x-2) or (x-2 > 0 and 2x+1 > x-2) (x < 2 and x < -3) or (x > 2 and x > -3) x < -3 or x > 2
@@pgskills A (boolean) expression 'a and b' is true, if and only if a and b are both true. In case we consider the of 'x < 2 and x < -3', then a is 'x < 2' and b is 'x < -3'. Now you could do draw a number line and mark the part for which a = 'x < 2' is true with yellow and the part for which b = 'x < 3' is true with blue. The part(s) of the numberline, marked with both colors is the result, which happens to be 'x < 2'. Another way to do that is to apply rules of logic (and some purely optical changes), in detail: x < 2 and x < -3 | -3 < 2 is always true, so it doesn't change the truthness of any part if we apply it using and x < 2 and x < -3 and -3 < 2 | draw together x < -3 and -3 < 2 x < 2 and x < -3 < 2 | note that x < 2 is contained in x < -3 < 2 and therefore redundant x < -3 < 2 | split x < -3 < 2 x < -3 and -3 < 2 | removing sth that is always true and appended using and doesn't change the truthness of the expression x < -3 A third method is to map the boolean expression to set theory: x < 2 and x < -3 | exchange english 'and' to default math symbol for 'and' x < 2 ∧ x < -3 | and/∧ corresponds to intersect/∩ { x | x ∈ ℝ^(< 2) } ∩ { x | x ∈ ℝ^(< -3) } { x | x ∈ (ℝ^(< 2) ∩ ℝ^(< -3) } { x | x ∈ (ℝ^(< -3) } x < -3 Sidenote: The operator or/∨ corresponds to union/∪, The operator not/¬ corresponds to complement and the complement of a set A ⊆ ℝ is the set A^c := ℝ\A. If you have real numbers p < q and only consider strict inequalities, then you have the following cases (and can create their equivalencies using the above methods): - x < p and x < q x < p and x < q x < p - x < p and x > q x < p and q < x false (no solution) - x > p and x < q p < x and x < q p < x < q (which technically is only an optical change) - x > p and x > q p < x and q < x x > q Sidenote: You can do the same for 'or' ('a or b' is true if and only if a is true, b is true, or both are true) and 'not' ('not a' is true if and only if a is false) and one of the above methods to drive the possible cases.
You can also write (2x+1)/(x-2) = 2 + 5/(x-2) > 1 so you get 5/(x-2) > -1 which is clearly true for x>2, and (graphing the function) you also get a negative range where 5/(x-2) = -1 at x=-3 so there are the two solutions: x2
This is a good method. However at the point where we have the inequality (x+3)/(x-2) > 0, I would recommend, rather than substituting numbers (which is tedious), simply dividing that number line up into three "zones" and check the sign in each zone giving the pattern: - + + for (x-3) and - - + for (x-2); the sign of the quotient is then easily seen to be + - + ... solved. This 'zonal method' has the advantage that it easily generalises to any inequality of the form f(x)/q(x) where f(x) amd q(x) can be factorised into linear factors. I always taught this method and pupils always found it easy to digest. Finally, if some of the factors are "the wrong way round" then multiplication by -1 sorts it ... e.g. (x-5)(x+2)/(x-1)(4-x) > 0 is equivalent to (x-5)(x+2)/(x-1)(x-4) < 0. Having all the factors with positive coefficients of x means that the pattern ALWAYS starts with a - on the leftmost zone, and changes to + as it crosses its own zero; once this simple technique is understood, solving these sorts of inequalities becomes an absolute breeze.
Much easier is this method: (2x+1) / (x-2) > 1 ------> D: x must be different than 2 if we reach to the expression: (x+3) / (x-2) > 0 we can multiply both sides by (x-2)^2 (which is a positive number) (x +3) (x-2) > 0 We don't need to test any numbers. Just draw the number line and a "rough-draft" of the parabola: a > 0 -----> the arms of parabola directed above the x-axis I I I I -----------------o------o--------------> I _ I x - ∞ -3 2 +∞ we instantly can see x-intercepts and the part of parabola which is above the number line (y > 0) so the solution is: x = (-∞ , -3) v (2, +∞ )
You can multiply both sides by (x - 2)^2 which is always non-negative so will never reverse the inequality. x = 2 is already rejected because it sets the denominator to zero. After expanding and simplifications you will get x^2 + x - 6 > 0 (x - 2)(x + 3) > 0 and the rest is the same.
Given: x^2 + 2*x*y - y^2 Equate to (x + a)*(x + b): (x + a)*(x + b) = x^2 + 2*x*y - y^2 Expand the LHS; x^2 + (a + b)*x + a*b = x^2 + 2*x*y - y^2 Equate like coefficients: a + b = 2*y, thus the mean (m) of a & b is m = y a*b = -y^2 Use Po Shen Lo's simplified quadratic formula, with the mean (m) and product (p) of the two solutions: m +/- sqrt(m^2 - p) Thus: a = y + sqrt(y^2 - (-y^2)) b = y - sqrt(y^2 - (-y^2)) Simplifying: a = y*(1 + sqrt(2)) b = y*(1 - sqrt(2)) Thus, the solution is: [x + y*(1 + sqrt(2))]*[x + y*(1 - sqrt(2))]
Why does multiplying both sides NOT work? This is the key to understanding this issue and remembering not to make the mistake. otherwise you must rely on pure rote learning, which is limiting and subject to memory lapses. The issue is that because (x-2) might be positive or negative or zero, and if negative then the inequality sign would need to be reversed (as multiplying both side by a negative number reverses the sign - try it if you don't see it). This issue does not arise with equalities. If zero you end up with 0 > 0, which is not a solution (graphically this is where there is an asymptote). The equality trick still works, but you have to consider the case that x = 2 (where the asymptote is; the graph line might visually disappear off to infinity and come back from minus infinity for example).
First replace > with =. Then add the expression x ≠ 2. Now we can multiply with (x - 2). Solve the equation; that will be x = -3. Make a sign scheme. Thus we find the solution x < -3.
4:00 *WRONG!!!* You don't get the exact same set whether you include or exclude the numbers 2 and 3 in the set! A non-strict inequation (≥) is *not* equivalent to a strict inequation!
you also could have drawn a parabola instead of doing the "yes" "no" "yes" thing because there are 2 xs in the problem, meaning it is a quadratic equation
Explore the critical limit method. Solve denominator = 0 first limit (can’t have this value) Solve as an equation = other limits (inequality determines inclusion of these values) Plot all limits on a number line (open and closed) and test regions. Best thing is this method works for all inequalities covered in the HSC (NSW Australia)
ig, i'm in the other 10% this is literally soo basic, i wonder how anyone could get that wrong, it's soo obvious, a simple way to avoid this is just don't cross multiply if you see inequality sign, i thought this would be something very tricky most of us wouldn't know but it's simple and nothing special lol.
Many ways to do this: 1. Multiply by x - 2 and handle two cases, depending on the sign of x - 2 2. Write the numerator as 2(x - 2) + 5 to get rid of x in the numerator, then take reciprocals (switching inequality sign of course) after simplification 3. Use this method Probably other ways too
x-2 cannot equal to 0 because (x+3)/0 is undefined. The numerator doesn't just disappear, it becomes the numerator on the opposite side. The test makes sense, but for the the obvious flaw in the math.
There is nothing wrong cross multiply. The inequality sign does not change if the denominator is positive, thus, For x>2, cross multiply result x>-3 so the answer is x>2 since 2>-3. For x
Yeah no. Multiple both sides by x-2 but flip the inequality for x < 2 and solve, finding overlap if any. This method is rote memorization that doesn't show understanding
Hmm, I find it easier to manipulate the original inequality, splitting cases depending on the sign of (x - 2). So if x > 2, we get (2x + 1) > (x - 2) so x > -3. Since x > 2 implies x > -3 the inequality is true for all x > 2. If x < 2, we multiply the original inequality by (x - 2) but reverse the inequality. This gives (2x + 1) < (x - 2), so x < -3. We find that x < -3 implies x < 2 so we have found solutions for all x < -3 and x > 2 from the previous case.
0:30 Well, then tell the audience *why* it gets to the wrong answer, i.e. because the term x-2 could be negative, and when multiplying both sides of an inequality with a negative the inequality gets inverted. The presented method is just *one* way of doing this, one could also treat the cases separately. drawing the graph would've also helped visualize what's going on. Presenting this whole solution like some cookbook recipe is not the best way to learn something, let alone understand what's going on. Someone who already understands math can see what's going on, for someone who doesn't yet understand what is happening the presented solution is just arcane math wizardry to be memorized and exactly reproduced without deeper understanding and high likelihood to fail.
my way is different lets add and subtract 4 from the 2x+1 part. it becomes 2x-4+5. then the equation becomes 2 + (5/(x-2)) >1 lets subt 2 from both sides. Then equation seems to be like that : 5/(x-2) >-1 lets divide both sides by 5, then: 1/(x-2) >1/(-5) lets consider only divider parts now (x-2)
I see no reason, why multipling with (x-2) shoulld be a mistake (if we mae a difference for the cases with x-20 and if we excude the case x=2, which aes the eft side of the original excercise undefined)) But et us loo at thhe other possibiity.: (2x+1)/(x--2)>1 ((x-2)+(x+3))/(x--2(>1 (x2)/(x2)+(+3)/(x2)>>1 1+(x+3)/(x2)>1 (x+3)/(x-2)>0 This is the case, if ((x+3)>0 and (x--2)>0) or if ((x+3)
I admit I was lazy. I only thought of reals. Complex might be possible, didnt even check. Lets see if I was right, just from thumnail. X is not between or including -3 & 2
(2x + 1)/(x - 2) > 1 (2x + 1)/(x - 2) - 1 > 0 (2x + 1 - x + 2)/(x - 2) > 0 (x + 3)/(x - 2) > 0 Inflexion/non-existent points are -3 and 2. for x < -3; the fraction is positive for 2 > x > - 3; the fraction is negative for x > 2; the fraction is positive. thus, the equality only holds for x > 2 or x < -3. {x | x < -3}∪{x | x > 2}
Because when you multiply by (x-2) you don’t know if you multiply by a negative or a positive number. It’s ok when you are solving an equation but here this is an inequation so the sign matters
90% of students get this wrong because 100% of high school math teachers cannot even explain or demonstrate examples of using such math in daily life. As a result most of us half-ass it on the final exam and never see it again.
Literally making things complicated for no reason all you need to do is multiply by denominator squared as it is always positive and wont affect the inequality sign then you get a quadratic inequality which is easy to solve by sketching the graph and critical points.
If the quotient has to be greater than 1, the dividend has to be greater than the divisor, so you can rewrite it as 2x+1 > x-2, which is easy to solve.
This is video an example of how to make a mountain out of a molehill. As many have commented, multiply by (x-2) is fine, just accommodate the cases (x-2)>0 and (x-2)
That seems to be a very complicated way to solve this. Just multiply by (x-2) but make sure you split it into two inequalities with 2 different conditions: 2x+1>x-2 if x>2 and 2x+12 and x
You can multiply both sides by x-2. You just have to deal with the two cases separately: (a) x-2 is positive, in which case x>2 and x>-3 (therefore x>2) and (b) x-2 is negative, in which case x
I BELIEVE THIS IS THE CORRECT WAY OF SOLVING THE QUESTION
Exactly, I was about to write the same. This is the obvious way. (The method in the video is much longer.)
You said it first lol
Don't forget the case where x=2 - you are multiplying by zero, so you want to treat that as its own case (if you want to look at limits).
@@Phylaetra x cannot equal zero, zero is undefined in the original fraction.
the mistake comes from the fact of not considering whether the denominator is negative or not when "multiplying" both sides of the inequality. A simplest way consists of taking that into account so let n=2x+1 and d=x-2 so that n/d>1
if d>0 then n>d
else if -d>0 then d>n
substituting n and d by their original expression we get
if 2
Nice! I was wondering why we simply can't multiply, but now I know we can 😁
@@ogi22 indeed :)
It can be simplified by just saying that the function is not continuous in the domain (-inf,inf). It has a vertical asymptote at x=2 so it is only continuous in the domain (-inf,2)U(2,inf), which means that it should be evaluated for both. I think that the real problem here is that most of the time (if you are not used to many functions) people tend to forget domains, ranges and other attributes, including myself. That is easily solved by plotting the functions so you can grasp what is happening.
If you where to have something like (2x+1)/(x^2-2) > 1 you would have to evaluate for 2 asymptotes in x=+-sqrt(2) so the domain would be (-inf,-sqrt(2)) U (-sqrt(2),sqrt(2)) U (sqrt(2),inf), for which the inequality only holds true for the domains (-sqrt(2),-1) U (sqrt(2),3).
@@marilynman indeed considering the domain where the function is not continuous and evaluating the sign of a product(±±>0, +-1 where n=2x+1 and d=x²-2
If d>0 ⟹ x
@@ogi22 But you have to recognize the 2 cases x>2 and x
We cannot cross-multiply the x-2 because we do not know whether it is positive or negative...then direction of inequality sign will be not correct.
But then you can deal with the two cases separately.
@@brendanward2991Instead use the wavey curve method
@@brendanward2991, well you would get x > -3 and x< -3 if you do that
@@Brad-ey6cz no, you don't.
let case 1 be x-2>0. This is only the case if x>2, and therefore the solution is [(x>-3) and (x>2)] which reduces to x>2.
Let case 2 be x-2
@@Brad-ey6cz No, you wouldn't. Read what I wrote. For case 1 you get the two conditions: x>2 and x>-3, which together imply that x>2. For case 2 you get the two conditions: x
A delight as always. More longer and in depth videos please 😊
just split into 2 cases.
1) assume (x-2) > 0 that is: x > 2
2x+1 > x-2
x > -3 and (we assumed) x > 2 ... stronger condition is x > 2
2) assume (x-2) < 0 that is: x < 2
2x+1 < x-2 switch the comparison from ">" to "
How do you determine "stronger condition"? Please explain.
@@pgskills The stronger condition is just describing if the range specified by the inequality is more specific about what parts are encapsulated if you know that x < 2 and that x < -3 you can simplify that to x < -3 because -3 < 2 similarly if you know that x > -3 and that x > 2 then you can simplify it to x > 2 because 2 > -3
@@christopheriman4921
Hey, thank you! Truly appreciate the clarity. I understand it now and actually have some vague recollection about learning this in connection to graphing quadratic and cubic functions. Much appreciated for the response. Thanks again. 👍👍
Yes, this was the old method taught to me many many years ago.
But you can simplify the cases,if you transform first the original excercise to
(x+3)/(x-2)>0
as shown in the video.
Brilliant! I always learn something from Mr. H!
Thank you, thank you, Mr. H. Very, very useful and informative. Appreciate the effort!
If you multiply both sides of the inequality by (x-2)^2 then you don’t have to worry about the sign change as (x-2)^2 cannot be negative for all real x. Then you get a quadratic inequality that gives the same set of solutions
exactly this, by far the easiest method with no cases to consider
Heh heh, why didn't I think of that?
I enjoy listening to Mr. H solving problems which I probably had in Algebra class 65 years ago. Rejuvenation of youth, sorta makes me want to turn on some oldies like “The Twist” by Hank Ballard and the Midnighters (Chubby Checker’s was a few years later).😊
uma excelente aula, Mr H. MuitObrigArigaThanks.
thanks for including the demonstration that the original method resulted in incorrect answer. and we can see why. I'm an adult back in college and i recommend your videos to my classmates and my 17 year old son.
This is one reason people hate math! Teachers are fixated on final answer while reasoning and logic is the important part!
The first enswer is true if x-2 is positive.
For example you can use this simplification, if the bottom is positive.
If you want to use in positive or negative it should be followed by cases.
The second method is simple in R.
And useful because the multiplication are difficult in comparison to addition in > or < too. But is not needed if we work in N.
I really appreciate these videos, for me this was not the intuitive answer all, so thank you for correcting my train of thought on how to solve this.
You are a great teacher!
You tell how to do it, but not why.
@kriswillems5661 You can do the cross multiply as well and get the results,just remember to build cases for X > and < 2 , because depending of sign of x-2 ,on cross multiply sign of inequality will change . Like 5> 3 ,if you multiply both sides with something negative the sign will change to -5 < -3.
@@PankajKumar-hz3oi Yes, that would be the most logical way to it.
It was really a good video. #support from India .
You make it look so easy!
Why shouldn't it be if you paid attention in school
1:10. An explanation of why we are setting the numerator and denominator to zero would be helpful here. It otherwise just comes across as a 'trick'.
It is presumably because we want both the numerator and demominator to have the same sign for the inequality to be true. So the points where each can change sign are important, and these are at values of x which make each zero.
However, if you just state that both the numerator and denominator have to have the same sign at this stage then it becomes obvious that the answer has to be x2.
In response to the post that "An explanation of why we are setting the numerator and denominator to zero would be helpful here." …
My perception is that those points are transition points to the validity of the inequality. On one side of one of the zero points the inequality might be true, but on the other it might not. Thus the follow-up check.
It makes no sense setting the denominator to zero because the equation becomes undefined. It is best as pointed out by others to do cases where x-2 is negative and x-2 is positive.
I would *not* set it to zero. I come to the solution, if i think about "when does a fraction is greater than 0? It is the case, if both, enumerator and denominator, are greater than 0 or if both are smaller than 0.
I have been a practicing engineer for 37 years. I guarantee you that 99% of engineers and scientist that I know would multiply both sides by x-2 and solve for x = -3.
Thanks for the productive input. I agree with you.
We can multiple 2 terms by x-2 if x>2 we get x>2 then multiply 2 terms in x-2 if x
Great stuff.
This is one way to do it in college-level pre-calculus:
We start with (2*x+1)/(x-2) > 1.
(1) Assume x-2>0, i.e., x>2. Multiply by (x-2) to get 2*x+1>x-2,
(preserve the direction of the comparison sign), i.e., x>-3.
The only way that we get x>2 and x>-3, simultaneously, is for [x>2]
(2) Then, assume x-2
case 1:
2x+1>x-2 and x-2>0
x>-3 and x>2
x>2
case 2:
2x+1
That was a powerful lesson! By bringing the equation to a comparison with Zero you brought extra numerical inferences into play. A General's approch rather than a Captain's :-)
Though i like your solution, there's nothing (technically) wrong with wanting to multiply by the variable expression '(x-2)' as a first step.
The mistake is to assume that you would simply get '2x+1 > x-2' from it.
The issue is that the variable expression, you want to multiply with, might be lower than, equal to or greater than zero.
Here 'equal to zero' is not a valid option for a denominator 'x-2', therefore the correct equivalent tranformation would be the following:
(2x+1)/(x-2) > 1
(x-2 < 0 and 2x+1 < x-2) or (x-2 > 0 and 2x+1 > x-2)
(x < 2 and x < -3) or (x > 2 and x > -3)
x < -3 or x > 2
How do you know which to pick from each answer in your last step?
@@pgskills A (boolean) expression 'a and b' is true, if and only if a and b are both true.
In case we consider the of 'x < 2 and x < -3', then a is 'x < 2' and b is 'x < -3'.
Now you could do draw a number line and mark the part for which a = 'x < 2' is true with yellow and the part for which b = 'x < 3' is true with blue.
The part(s) of the numberline, marked with both colors is the result, which happens to be 'x < 2'.
Another way to do that is to apply rules of logic (and some purely optical changes), in detail:
x < 2 and x < -3 | -3 < 2 is always true, so it doesn't change the truthness of any part if we apply it using and
x < 2 and x < -3 and -3 < 2 | draw together x < -3 and -3 < 2
x < 2 and x < -3 < 2 | note that x < 2 is contained in x < -3 < 2 and therefore redundant
x < -3 < 2 | split x < -3 < 2
x < -3 and -3 < 2 | removing sth that is always true and appended using and doesn't change the truthness of the expression
x < -3
A third method is to map the boolean expression to set theory:
x < 2 and x < -3 | exchange english 'and' to default math symbol for 'and'
x < 2 ∧ x < -3 | and/∧ corresponds to intersect/∩
{ x | x ∈ ℝ^(< 2) } ∩ { x | x ∈ ℝ^(< -3) }
{ x | x ∈ (ℝ^(< 2) ∩ ℝ^(< -3) }
{ x | x ∈ (ℝ^(< -3) }
x < -3
Sidenote: The operator or/∨ corresponds to union/∪, The operator not/¬ corresponds to complement and the complement of a set A ⊆ ℝ is the set A^c := ℝ\A.
If you have real numbers p < q and only consider strict inequalities, then you have the following cases (and can create their equivalencies using the above methods):
- x < p and x < q x < p and x < q x < p
- x < p and x > q x < p and q < x false (no solution)
- x > p and x < q p < x and x < q p < x < q (which technically is only an optical change)
- x > p and x > q p < x and q < x x > q
Sidenote: You can do the same for 'or' ('a or b' is true if and only if a is true, b is true, or both are true) and 'not' ('not a' is true if and only if a is false) and one of the above methods to drive the possible cases.
@@derwolf7810 I get it now. Thanks for the detailed response. Very helpful.
Brilliant!
You can also write (2x+1)/(x-2) = 2 + 5/(x-2) > 1 so you get 5/(x-2) > -1 which is clearly true for x>2, and (graphing the function) you also get a negative range where 5/(x-2) = -1 at x=-3 so there are the two solutions: x2
I like that you use a chalk board - reminds me of my high school and college math classes - from the 1970's.
This is a good method.
However at the point where we have the inequality (x+3)/(x-2) > 0, I would recommend, rather than substituting numbers (which is tedious), simply dividing that number line up into three "zones" and check the sign in each zone giving the pattern:
- + + for (x-3) and - - + for (x-2); the sign of the quotient is then easily seen to be + - + ... solved.
This 'zonal method' has the advantage that it easily generalises to any inequality of the form f(x)/q(x) where f(x) amd q(x) can be factorised into linear factors. I always taught this method and pupils always found it easy to digest.
Finally, if some of the factors are "the wrong way round" then multiplication by -1 sorts it ...
e.g. (x-5)(x+2)/(x-1)(4-x) > 0 is equivalent to (x-5)(x+2)/(x-1)(x-4) < 0. Having all the factors with positive coefficients of x means that the pattern ALWAYS starts with a - on the leftmost zone, and changes to + as it crosses its own zero; once this simple technique is understood, solving these sorts of inequalities becomes an absolute breeze.
This one problem could be a test all by itself. It assesses many skills.
I really liked the chalk. I think that was my favorite part.
Much easier is this method:
(2x+1) / (x-2) > 1 ------> D: x must be different than 2
if we reach to the expression:
(x+3) / (x-2) > 0
we can multiply both sides by (x-2)^2 (which is a positive number)
(x +3) (x-2) > 0
We don't need to test any numbers. Just draw the number line and a "rough-draft" of the parabola:
a > 0 -----> the arms of parabola directed above the x-axis
I I
I I
-----------------o------o-------------->
I _ I x
- ∞ -3 2 +∞
we instantly can see x-intercepts and the part of parabola which is above the number line (y > 0)
so the solution is:
x = (-∞ , -3) v (2, +∞ )
You can multiply both sides by (x - 2)^2 which is always non-negative so will never reverse the inequality. x = 2 is already rejected because it sets the denominator to zero. After expanding and simplifications you will get x^2 + x - 6 > 0 (x - 2)(x + 3) > 0 and the rest is the same.
plz factorize the trinomial. x^2 +2xy - y^2
Given: x^2 + 2*x*y - y^2
Equate to (x + a)*(x + b):
(x + a)*(x + b) = x^2 + 2*x*y - y^2
Expand the LHS;
x^2 + (a + b)*x + a*b = x^2 + 2*x*y - y^2
Equate like coefficients:
a + b = 2*y, thus the mean (m) of a & b is m = y
a*b = -y^2
Use Po Shen Lo's simplified quadratic formula, with the mean (m) and product (p) of the two solutions:
m +/- sqrt(m^2 - p)
Thus:
a = y + sqrt(y^2 - (-y^2))
b = y - sqrt(y^2 - (-y^2))
Simplifying:
a = y*(1 + sqrt(2))
b = y*(1 - sqrt(2))
Thus, the solution is:
[x + y*(1 + sqrt(2))]*[x + y*(1 - sqrt(2))]
I had forgotten about this.
What if there is specific domain for x ?
Why does multiplying both sides NOT work? This is the key to understanding this issue and remembering not to make the mistake. otherwise you must rely on pure rote learning, which is limiting and subject to memory lapses.
The issue is that because (x-2) might be positive or negative or zero, and if negative then the inequality sign would need to be reversed (as multiplying both side by a negative number reverses the sign - try it if you don't see it). This issue does not arise with equalities. If zero you end up with 0 > 0, which is not a solution (graphically this is where there is an asymptote).
The equality trick still works, but you have to consider the case that x = 2 (where the asymptote is; the graph line might visually disappear off to infinity and come back from minus infinity for example).
First replace > with =.
Then add the expression x ≠ 2.
Now we can multiply with (x - 2).
Solve the equation; that will be x = -3.
Make a sign scheme.
Thus we find the solution x < -3.
You can multiply by x-2 if you then break it into two problems. If x>2 and if x
This is one of the most basic topics taught to class 11th students in India...😅
We use wavy curve method to solve such inequalities..
Or: for x>2 you get. 2x+1> x-2 rherefore x>-3 conclude
x>2 but if x 2
It was much faster for me just to multiply by (x - 2), and keep track of whether (x - 2) is positive, negative, or zero.
The right way to solve this is to multiply both sides by (x-2)^2. Since (x-2)^2 is always positive, it does not change the inequality sign.
4:00 *WRONG!!!*
You don't get the exact same set whether you include or exclude the numbers 2 and 3 in the set!
A non-strict inequation (≥) is *not* equivalent to a strict inequation!
you also could have drawn a parabola instead of doing the "yes" "no" "yes" thing because there are 2 xs in the problem, meaning it is a quadratic equation
Explore the critical limit method.
Solve denominator = 0 first limit (can’t have this value)
Solve as an equation = other limits (inequality determines inclusion of these values)
Plot all limits on a number line (open and closed) and test regions.
Best thing is this method works for all inequalities covered in the HSC (NSW Australia)
This was great😁😁
As an Asian, Jee Adv. aspirant, This is the Easiest Question I saw in years,
A little exaggerated i guess..
Nice Video Btw
ig, i'm in the other 10% this is literally soo basic, i wonder how anyone could get that wrong, it's soo obvious, a simple way to avoid this is just don't cross multiply if you see inequality sign, i thought this would be something very tricky most of us wouldn't know but it's simple and nothing special lol.
Many ways to do this:
1. Multiply by x - 2 and handle two cases, depending on the sign of x - 2
2. Write the numerator as 2(x - 2) + 5 to get rid of x in the numerator, then take reciprocals (switching inequality sign of course) after simplification
3. Use this method
Probably other ways too
I tried method #2 and only got x < -3.
x-2 cannot equal to 0 because (x+3)/0 is undefined.
The numerator doesn't just disappear, it becomes the numerator on the opposite side.
The test makes sense, but for the the obvious flaw in the math.
I think I must have been in a grumpy mood I actually hated this one. The most confusing one you've ever done I could not follow it at all.
It is also possible to solve using a sign chart.
U don’t need to test every region separately ( assuming there are no brackets with powers). In this case the correct region will alternate
There is nothing wrong cross multiply.
The inequality sign does not change if the denominator is positive, thus,
For x>2, cross multiply result x>-3 so the answer is x>2 since 2>-3.
For x
Yeah no. Multiple both sides by x-2 but flip the inequality for x < 2 and solve, finding overlap if any. This method is rote memorization that doesn't show understanding
Can we multiply by (x-2)^2 on both sides?
Hmm, I find it easier to manipulate the original inequality, splitting cases depending on the sign of (x - 2). So if x > 2, we get (2x + 1) > (x - 2) so x > -3. Since x > 2 implies x > -3 the inequality is true for all x > 2.
If x < 2, we multiply the original inequality by (x - 2) but reverse the inequality. This gives (2x + 1) < (x - 2), so x < -3. We find that x < -3 implies x < 2 so we have found solutions for all x < -3 and x > 2 from the previous case.
0:30 Well, then tell the audience *why* it gets to the wrong answer, i.e. because the term x-2 could be negative, and when multiplying both sides of an inequality with a negative the inequality gets inverted. The presented method is just *one* way of doing this, one could also treat the cases separately. drawing the graph would've also helped visualize what's going on.
Presenting this whole solution like some cookbook recipe is not the best way to learn something, let alone understand what's going on.
Someone who already understands math can see what's going on, for someone who doesn't yet understand what is happening the presented solution is just arcane math wizardry to be memorized and exactly reproduced without deeper understanding and high likelihood to fail.
A) it is not an equation but a relationship
B) you have to perform a case study.
I was trying to learn this in school (9-10-11:th grade?) bu never how to use this in "real life" - please tell me!
Devious, the asker of the question
Thank You though.
You can multiply by x-2, lol...just consider the case where x-2
DONT CROSS MULTIPLY INEQUALITIES!!!!!
my way is different
lets add and subtract 4 from the 2x+1 part. it becomes 2x-4+5.
then the equation becomes 2 + (5/(x-2)) >1
lets subt 2 from both sides. Then equation seems to be like that : 5/(x-2) >-1
lets divide both sides by 5, then: 1/(x-2) >1/(-5)
lets consider only divider parts now (x-2)
When x is two it results in infinity which is greater than zero.
I can hear jee aspirants incoming
"Table of Signs" - this is what you've done although maybe not so obvious to the student.
3/(-2) > 0, I didn't know that, seems that my maths skills are long gone.
I see no reason, why multipling with (x-2) shoulld be a mistake (if we mae a difference for the cases with x-20 and if we excude the case x=2, which aes the eft side of the original excercise undefined))
But et us loo at thhe other possibiity.:
(2x+1)/(x--2)>1
((x-2)+(x+3))/(x--2(>1
(x2)/(x2)+(+3)/(x2)>>1
1+(x+3)/(x2)>1
(x+3)/(x-2)>0
This is the case, if ((x+3)>0 and (x--2)>0) or if ((x+3)
I admit I was lazy.
I only thought of reals.
Complex might be possible, didnt even check.
Lets see if I was right, just from thumnail.
X is not between or including -3 & 2
Just finding domain of a function
I need to watch the video again
And again and again
I just got tutored !
I thought I understood how to do those.
I tried multiplying by (x-2)^2
(2x + 1)/(x - 2) > 1
(2x + 1)/(x - 2) - 1 > 0
(2x + 1 - x + 2)/(x - 2) > 0
(x + 3)/(x - 2) > 0
Inflexion/non-existent points are -3 and 2.
for x < -3; the fraction is positive
for 2 > x > - 3; the fraction is negative
for x > 2; the fraction is positive.
thus, the equality only holds for x > 2 or x < -3.
{x | x < -3}∪{x | x > 2}
x2, i didnt even watch the video. i just clicked on it to type this out. if youre above the age of 14 and still cant solve this youre preposterous.
If most students are getting the wrong answer, then what are teachers doing wrong?
But why is it wrong
Because 1 greater than 1 isn’t possible.
@@benjaminklapproth2913 i didnt understand what did you mean
Because when you multiply by (x-2) you don’t know if you multiply by a negative or a positive number. It’s ok when you are solving an equation but here this is an inequation so the sign matters
Since a greater than sign there is no definitive answer
Thanks
brilliant
i love it
i ❤ Mathematics
Some motivation for this algorithm would be nice...
Mathematician here. I don't recommend this method. See other comments that split into two cases: x>2 and x
Wrong method:- (2x+1)/(x-2) > 1
= 2x+1 > x-2 , this is incorrect as we don't know x-2>0, Correction:-
(2x+1)/(x-2) > 1
(2x+1)/(x-2) - 1 > 0
(2x+1-x+2)/(x-2) > 0
(x+3)/(x-2) > 0
》x < -3 or x > 2 therefore
x belongs to (-infinity,-3)U(2,infinity)
Isn't that *exactly* , the same math, as shown in the video?
👍👍👍😁🤪👋
90% of students get this wrong because 100% of high school math teachers cannot even explain or demonstrate examples of using such math in daily life. As a result most of us half-ass it on the final exam and never see it again.
🎉🎉🎉🎉🎉🎉🎉🎉
Literally making things complicated for no reason all you need to do is multiply by denominator squared as it is always positive and wont affect the inequality sign then you get a quadratic inequality which is easy to solve by sketching the graph and critical points.
😂😂😂😂😂
If the quotient has to be greater than 1, the dividend has to be greater than the divisor, so you can rewrite it as 2x+1 > x-2, which is easy to solve.
This is video an example of how to make a mountain out of a molehill. As many have commented, multiply by (x-2) is fine, just accommodate the cases (x-2)>0 and (x-2)
That seems to be a very complicated way to solve this. Just multiply by (x-2) but make sure you split it into two inequalities with 2 different conditions: 2x+1>x-2 if x>2 and 2x+12 and x
You got your Facebook account stolen. I hope it comes back!
I was about to say same --- it's been hacked!