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
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.
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 :-)
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).😊
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).
Thx for helping me with that. I would have fallen into the trap, though I surmised already that flipping leading signs might change the validity of the inequation.
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.
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.
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 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
@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.
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.
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 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.
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.
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
Here is a slightly more advanced method that does not require checking to find our intervals, but only works if we know x must be real. Multiply both sides by (x - 2)^2. Since x is real, (x - 2)^2 must be positive, so the inequality sign does not change while our denominator is eliminated. That leaves us: (2x + 1)(x - 2) > (x - 2)^2 Next, expand the polynomials, group like terms, and do _some_ of completing the square. 2x^2 - 3x - 2 > x^2 - 4x + 4 x^2 + x > 6 4x^2 + 4x > 24 4x^2 + 4x + 1 > 24 + 1 (2x + 1)^2 > 25 For the next step in this process, rather than putting the ± on the right-hand side, note the left-hand side as an absolute value and use the principal root on the right-hand side: |2x + 1| > 5 Remembering the definition of an absolute value, we solve this as two inequalities: 2x + 1 > 5 2x > 4 x > 2 2x + 1 < -5 2x < -6 x < -3 And there is the answer Mr. H got without having to check values, and without having to notice the graph would be a parabola.
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.
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
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.
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!
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
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
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.
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.
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))]
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)
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)
(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
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.
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
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.
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
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)
