When adding the equations, you can split 12 into 6 + 6, move y to the other side and factor the -1, so you get: x² - 5x + 6 = (y² - 5y + 6)(-1) (x - 2)(x - 3) = (y - 2)(y - 3)(-1)
I started with the assumption that since the two equations are symmetric, surely a solution existed where x and y are the same From that I got (2, 2) and (3, 3) Then I added the equations and landed in a similar spot as you said I managed to get here: If x^2 - 5x + 6 = 0 then so must y^2 - 5y + 6 thus, (2, 3) and (3, 2) The other possibility is x^2 - 5x + 6 = - (y^2 - 5y + 6)but that's difficult to evaluate Then maybe the subtraction method works
Dear Sir, this time I am trying to solve this problem perhaps in a simpler way. This type of equation is known as symmetrical equation, where exchanging x with y values does not alter the properties of the equations, if f(x) = f(y) then x = y. In this case x = y is a solution. Since x = y, let us replace all y as x in either of the two equations. Say the 1st equation: Now we have (x -1)(x²+6) = x (x² + 1). That is x³ - x² + 6x - 6 = x³ + x. This becomes x² - 5x + 6 = 0. The roots of x are 2 and 3. As they are symmetric, they can be arranged in 4 different ways. That is, (x , y) = (2 , 2) and (3, 3), (2 , 3) and (3, 2).
I took this approach as well and, like you, I solved the equation for x and got x = 2 and 3. The problem is that you (and I) derived the equation to be solved, x² - 5x + 6 = 0, by setting x = y in either of the given equations. So the solutions (x , y) = (2 , 2) or (3, 3) are fine. (x , y) = (2 , 3) and (3, 2) are indeed solutions, but while our approach shows that if (x , y) = (2, 3) is a solution, (3 , 2) must also be a solution (and vice-versa), we haven't shown that either one is a solution.
Yeah. If we replace all x as y and solve it, we will get a similar solution y = (2,3) or (3,2). Therefore, we have x = 2 or x = 3 (we have already proved) and y = 2 or 3. Since the problem is symmetrical in x and y i.e., f( x,y) = f(y,x), we have 4 solutions as (x,y) or (y,x) = {(2,2), (3,3), (2,3), (3,2)}.
@@rcnayak_58 you have literally said "since x=y... we get (2,3) as a solution". that does not follow. edit: to illustrate my objection to your reasoning, let me be concrete: (x-2)^2+(y-2)^2 = 4 and (x+y-3)*(x+y-2) = 0 has the same type of symmetry you have cited. neither are the solutions arranged in a rectangle, nor in fact are there any solutions of the form (x,x).
I went about it slightly differently. Expanding the original equations and subtracting the second from the first, gives 5x-y^2-12+5y-x^2=0. Switching signs, rearranging the terms and splitting 12 in 4+4+2+2 gives (y-2)^2+(x-2)^2-(x-2)-(y-2)=0, which can be rewritten as (y-2)(y-3)+(x-2)(x-3)=0. Setting y=2 gives x=2 and x=3, while setting y=3 gives x=2 and x=3.
I watched it yet a third time. Still impressed and liked seeing the graph of the solution at the end: intersecting hyperbolas (or things that look like hyperbolas).
For those wondering, a+b = -4 ==> a^2 + b^2 + 2ab = 16 Combined with a^2 + b^2 = 1/2 , ==> ab = 7.75 Combined with a+b = -4, this clearly isn't possible (a,b need to be same sign) which is why it doesn't give any extra solutions for a,b real; (to be pedantic, this is because a and b are roots of the quadratic: m^2 + 4m + 7.75 = (m+2)^2 + 3.75 > 0 for all m, and hence has complex roots, meaning a and b are complex)
If in first original equation x is replaced by y, we get second original equation. By using this fact after getting x=2, 3 and x=y, we can write (x, y) = (2, 3), (2, 2), (3, 2), (3, 3)
I think i have watched the video two times and solved in 2 ways which for me seem simpler. After addition and subtraction we are left with X^2-5x+6+Y^2-5y+6=0 And 2xy-x-y-7=0 First method takes ento account the equations are symmetric in x and y so: Let S=x+y and P=xy the equations become: S^2-2P-5S+12=0 and 2P-S-7=0 , so 2P = S+7 which we substitute in firs equation and solve a quadratic S^2-6S+5=0 so (S-1)(S-5)=0 which gives soluions (S,P) of (1,4) and (5,6) solving a quadriatic resulting from vietas formulas yields the real solutions for x,y (2,3) and (3,2) My second solution was to let the first equation become -(x-3)(x-3)-(y-2)(y-3)=0 And the second be X=(7+y)/(2y-1) Substituting in the original equation and factoring we get (Y-2)(Y-3)(-1-15/(2y-1)^2)=0 Last term is always the sum of two negative terms which is never zero I think this problem has something which baits you into solving it and you sir have some really clever manipulations sometimes.
Great video. Keep up the good work. A nice side problem is to show that x - y always divides p( x , y ) - p( y , x ) where p is a polynomial in two variables.
In the equation x^2+y^2-5x-5y+12=0, the x and y coefficients are equal, 1. Moreover, the coefficient of xy is 0. So, it's easy to say without even doing math that it's the equation of a circle.
Hey sir, I have a faster solution for -2xy+x+y=7. When you have x^2+y^2-5x-5y+12=0, then x^2+y^2+2xy-x-y-7-5x-5y+12=0. This becomes (x+y)^2-6(x+y)+5=0 and now we have (x+y-1)(x+y-5)=0, so x+y=1 or x+y=5. When you have x+y, you can get xy and you con solve for x,y with Viète’s theorem. Thank you and have a nice day sir!
Perhaps i can help. At first we take 1/2 to the other side so sqrt(log(x)) = log(sqrt(x)) + 1/2 Then we raise both sides to the power of two so logx = (log(sqrt(x)) +1/2)² = log(sqrt(x))² + log(sqrt(x)) + 1/4 Now we can substitute log(sqrt(x)) as y: Let y = log(sqrt(x)) Using the rules of logarithm log(x) would be equal to 2y 2y = y² + y + 1/4 0=y² - y + 1/4 Using the quadratic formula we have y = 1/2 Note that the equation has only one root as the delta would be .equal to zero log(sqrt(x)) = 1/2 0.5log(x) = 0.5 log(x) = 1 x = 10 And that is our answer
@@richardbraakman7469 Yep you exclude in R, but it's interesting that the equation is 6th degree globally, so admits 6 solutions in C. 4 are real, 2 are not ... the 2 complex solutions are not so difficult to find too ...
I would go a completely different route: Since the equations are exactly the same except the switched variables, you can say, that one valid solution is x = y. This means, you can have one equation with just one variable: (x - 1)(x² + 6) = x(x² + 1) x³ + 6x - x² - 6 = x³ + x |-x³ -x -x² + 5x - 6 = 0 |*-1 x² - 5x + 6 = 0 (x - 2)(x - 3) = 0 x ∈ {2, 3} and therefore also y ∈ {2, 3} You have 2 solutions for 2 variables, that is 4 solutions in total: (x, y) ∈ {(2, 2), (2, 3), (3, 2), (3, 3)}
Since both equations are symmetrical in x and y, it follows that there must exist a solution that satisfies x = y, thus (x - 1)(x² + 6) = x(x² + 1) => x³ - x² + 6x - 6 = x³ + x => x² - 5x + 6 = 0 Or x = {2, 3} = y We already know (2, 2) and (3, 3) are solutions, and if we plug in and check the remaining combinations of (2, 3) and (3, 2) we find that they satisfy the system.
I'm wondering if I'm missing something. After multiplying out the original equations and adding them, you got x^2-5x+y^2-5y+12=0. My immediate thought was to turn that into (x^2-5x+6)+(y^2-5y+6)=0, which becomes (x-2)(x-3)+(y-2)(y-3)=0. That leads you to (2,2), (2,3), (3,2), and (3,3). The only other possible solutions would be when (x-2)(x-3)=-(y-2)(y-3). I'm not quite sure how to prove that's impossible when x is between 2 and 3 or y is between 2 and 3 ... the only ways to generate negative products.
"The only other possible solutions would be when (x-2)(x-3)=-(y-2)(y-3)" It's not "other", it's equivalent to (x-2)(x-3)+(y-2)(y-3)=0. It also has an infinite number of other possible solutions, other than (2,2), (2,3), (3,2), and (3,3). You seem to be trying to prove that (x-2)(x-3)+(y-2)(y-3)=0 on its own is equivalent to the original system, which it's obviously not.
*”Those who stop learning, stop living.”*
When adding the equations, you can split 12 into 6 + 6, move y to the other side and factor the -1, so you get:
x² - 5x + 6 = (y² - 5y + 6)(-1)
(x - 2)(x - 3) = (y - 2)(y - 3)(-1)
I started with the assumption that since the two equations are symmetric, surely a solution existed where x and y are the same
From that I got (2, 2) and (3, 3)
Then I added the equations and landed in a similar spot as you said
I managed to get here:
If x^2 - 5x + 6 = 0 then so must y^2 - 5y + 6
thus, (2, 3) and (3, 2)
The other possibility is x^2 - 5x + 6 = - (y^2 - 5y + 6)but that's difficult to evaluate
Then maybe the subtraction method works
Dear Sir, this time I am trying to solve this problem perhaps in a simpler way. This type of equation is known as symmetrical equation, where exchanging x with y values does not alter the properties of the equations, if f(x) = f(y) then x = y. In this case x = y is a solution. Since x = y, let us replace all y as x in either of the two equations. Say the 1st equation: Now we have (x -1)(x²+6) = x (x² + 1). That is x³ - x² + 6x - 6 = x³ + x. This becomes x² - 5x + 6 = 0. The roots of x are 2 and 3. As they are symmetric, they can be arranged in 4 different ways. That is, (x , y) = (2 , 2) and (3, 3), (2 , 3) and (3, 2).
I took this approach as well and, like you, I solved the equation for x and got x = 2 and 3. The problem is that you (and I) derived the equation to be solved, x² - 5x + 6 = 0, by setting x = y in either of the given equations. So the solutions (x , y) = (2 , 2) or (3, 3) are fine. (x , y) = (2 , 3) and (3, 2) are indeed solutions, but while our approach shows that if (x , y) = (2, 3) is a solution, (3 , 2) must also be a solution (and vice-versa), we haven't shown that either one is a solution.
Yeah. If we replace all x as y and solve it, we will get a similar solution y = (2,3) or (3,2). Therefore, we have x = 2 or x = 3 (we have already proved) and y = 2 or 3. Since the problem is symmetrical in x and y i.e., f( x,y) = f(y,x), we have 4 solutions as (x,y) or (y,x) = {(2,2), (3,3), (2,3), (3,2)}.
@@rcnayak_58 you have literally said "since x=y... we get (2,3) as a solution". that does not follow.
edit: to illustrate my objection to your reasoning, let me be concrete:
(x-2)^2+(y-2)^2 = 4 and (x+y-3)*(x+y-2) = 0 has the same type of symmetry you have cited. neither are the solutions arranged in a rectangle, nor in fact are there any solutions of the form (x,x).
Good approach❤. I have another question.
x^2+y^2=a
x^3+y^3=b
@@theupsonnayak sahab was wrong
I went about it slightly differently. Expanding the original equations and subtracting the second from the first, gives 5x-y^2-12+5y-x^2=0. Switching signs, rearranging the terms and splitting 12 in 4+4+2+2 gives (y-2)^2+(x-2)^2-(x-2)-(y-2)=0, which can be rewritten as (y-2)(y-3)+(x-2)(x-3)=0. Setting y=2 gives x=2 and x=3, while setting y=3 gives x=2 and x=3.
I started by just watching math videos and enjoying the cleverness but now I've graduated to trying the problems myself first :)
This is the way. Just wait until you unironically buy (or download) a textbook and work through it from cover to cover.
Stem redux, just later in life.
I watched this video twice. Algebraic magic.
U gives better education than our schools
"U gives" is no English.
@@azztekehe's african
you give
he/she/it gives
this is the correct grammar
kinda weird to criticise school when you don't know how grammar works
@@luladrgn9155 Ehh, if he is ESL he jind of gets a pass.
Really a beautiful explanation, congratulations from Italy.
I watched it yet a third time. Still impressed and liked seeing the graph of the solution at the end: intersecting hyperbolas (or things that look like hyperbolas).
For those wondering,
a+b = -4 ==> a^2 + b^2 + 2ab = 16
Combined with a^2 + b^2 = 1/2 ,
==> ab = 7.75
Combined with a+b = -4, this clearly isn't possible (a,b need to be same sign) which is why it doesn't give any extra solutions for a,b real;
(to be pedantic, this is because a and b are roots of the quadratic:
m^2 + 4m + 7.75
= (m+2)^2 + 3.75 > 0 for all m, and hence has complex roots, meaning a and b are complex)
Thank you for the explanation.... I understood everything
If in first original equation x is replaced by y, we get second original equation. By using this fact after getting x=2, 3 and x=y, we can write (x, y) = (2, 3), (2, 2), (3, 2), (3, 3)
I think i have watched the video two times and solved in 2 ways which for me seem simpler.
After addition and subtraction we are left with
X^2-5x+6+Y^2-5y+6=0
And
2xy-x-y-7=0
First method takes ento account the equations are symmetric in x and y so:
Let S=x+y and P=xy the equations become:
S^2-2P-5S+12=0 and
2P-S-7=0 , so 2P = S+7 which we substitute in firs equation and solve a quadratic S^2-6S+5=0 so (S-1)(S-5)=0 which gives soluions (S,P) of (1,4) and (5,6) solving a quadriatic resulting from vietas formulas yields the real solutions for x,y (2,3) and (3,2)
My second solution was to let the first equation become
-(x-3)(x-3)-(y-2)(y-3)=0
And the second be
X=(7+y)/(2y-1)
Substituting in the original equation and factoring we get
(Y-2)(Y-3)(-1-15/(2y-1)^2)=0
Last term is always the sum of two negative terms which is never zero
I think this problem has something which baits you into solving it and you sir have some really clever manipulations sometimes.
Great video. Keep up the good work. A nice side problem is to show that x - y always divides p( x , y ) - p( y , x ) where p is a polynomial in two variables.
Respected Sir,
I am pleased, because I create simular tasks for mi students in Serbia. It even bigger when they do it themselves. Wonderful.
It suffices to get two new equations by adding and then subtracting the original equations. Solving the system, one gets x + y = 1 and x + y = 5.
Well done and interesting, I proceeded a little differently after several failed attempts.
In the equation x^2+y^2-5x-5y+12=0, the x and y coefficients are equal, 1. Moreover, the coefficient of xy is 0. So, it's easy to say without even doing math that it's the equation of a circle.
What he showed at last(graph), is my first attempt to answer the question 🫣
From where are you sir.
Straight from the heavens
He is a professional no matter where he is from
from the school of our dreams (which only exists in our dreams)
I started watching your videos just for your beautiful smile and really got impressed with your style @PrimeNewtons
This will save much time. Correct me if i am wrong
Hey sir, I have a faster solution for -2xy+x+y=7. When you have x^2+y^2-5x-5y+12=0, then x^2+y^2+2xy-x-y-7-5x-5y+12=0. This becomes (x+y)^2-6(x+y)+5=0 and now we have (x+y-1)(x+y-5)=0, so x+y=1 or x+y=5. When you have x+y, you can get xy and you con solve for x,y with Viète’s theorem. Thank you and have a nice day sir!
xy2 + 6x - y2 - 6 = x2y + y
x2y + 6y - x2 - 6 = xy2 + x
(i)+(ii)
5(x+y) - (x2+y2) - 12 = 0
(x+y)2 - 2xy - 5(x+y) + 12 = 0 (iii)
will be useful later ...
(i)-(ii)
xy(y-x) + 6(x-y) + (x2-y2) = xy(x-y) + (y-x)
-2xy(x-y) + 7(x-y) +(x+y)(x-y) = 0
(x-y) (x+y-2xy+7) = 0
(a) x=y => in (1) or (ii) : x3 + 6x - x2 - 6 = x3 + x
x2 - 5x + 6 = (x-2)(x+3) = 0
=> sol (x,y) = (2,2) or (3,3)
(b) x+y-2xy+7 = 0
2xy = x+y+7
in (iii) : (x+y)2 - (x+y) -7 - 5(5+y) + 12 = 0
(x+y)2 - 6(x+y) + 5 = 0
(x+y-1)(x+y-5) = 0
(b1) x+y=5 => xy = 6 => (x,y) = (2,3) or (3,2)
(b2) x+y=1 => xy = 4 => x(1-x) = 4 => x2 - x + 4 = 0 no real solution
So 4 real solutions : (2,2) (3,3) (2,3) (3,2)
and probably 2 other complex solutions as global equation is 6th degree :
x2 - x + 4 = 0 => x = (1+/-iV15)/2
so ((1+iV15)/2,(1-iV15)/2) and ((1-iV15)/2,(1+iV15)/2))
By the way, nice curves if you plot them
man! this must have taken you forever to write down...
Exactly how I solved it.
15:12 lol it took me a minute to figure out that your 7\frac{1}{2} meant {\tt 7-1/2} not 7*1/2 lol, who uses that notation any more?
Hello can you solve this question
Sqrt(log x) - 1/2=log sqrt(x)
It was on my math exam and i didnt know how to solve it
Perhaps i can help.
At first we take 1/2 to the other side so
sqrt(log(x)) = log(sqrt(x)) + 1/2
Then we raise both sides to the power of two so
logx = (log(sqrt(x)) +1/2)² = log(sqrt(x))² + log(sqrt(x)) + 1/4
Now we can substitute log(sqrt(x)) as y:
Let y = log(sqrt(x))
Using the rules of logarithm log(x) would be equal to 2y
2y = y² + y + 1/4
0=y² - y + 1/4
Using the quadratic formula we have y = 1/2
Note that the equation has only one root as the delta would be .equal to zero
log(sqrt(x)) = 1/2
0.5log(x) = 0.5
log(x) = 1
x = 10
And that is our answer
@vafasadrif12 ty
@@vafasadrif12 ty
Why are we not considering a+b+4=0?
It means b = 4 - a. Substituting into a^2 + b^2 = 1/2 and expanding gives 2a^2 - 8a + 16 = 1/2 which has no real solutions
@@richardbraakman7469 Yep you exclude in R, but it's interesting that the equation is 6th degree globally, so admits 6 solutions in C. 4 are real, 2 are not ... the 2 complex solutions are not so difficult to find too ...
quicker. fast is velocity and quick is time
Awesome!!!! Simple awesome!
Amazing
Teacher thnks
Lol. I watch every video at 2x speed.
Вторая половина решения (после рассмотрения случая х=у) страшно длинна и не рациональна.
This is my country :o
Resolution too complicated. Adding the equations and you will solve it in 5 minutes.
I would go a completely different route:
Since the equations are exactly the same except the switched variables, you can say, that one valid solution is x = y.
This means, you can have one equation with just one variable:
(x - 1)(x² + 6) = x(x² + 1)
x³ + 6x - x² - 6 = x³ + x |-x³ -x
-x² + 5x - 6 = 0 |*-1
x² - 5x + 6 = 0
(x - 2)(x - 3) = 0
x ∈ {2, 3} and therefore also y ∈ {2, 3}
You have 2 solutions for 2 variables, that is 4 solutions in total:
(x, y) ∈ {(2, 2), (2, 3), (3, 2), (3, 3)}
And (2, 3) and (3, 2) satisfy x+y-2xy+7=0
Since both equations are symmetrical in x and y, it follows that there must exist a solution that satisfies x = y, thus
(x - 1)(x² + 6) = x(x² + 1)
=> x³ - x² + 6x - 6 = x³ + x
=> x² - 5x + 6 = 0
Or x = {2, 3} = y
We already know (2, 2) and (3, 3) are solutions, and if we plug in and check the remaining combinations of (2, 3) and (3, 2) we find that they satisfy the system.
Sir would you make a video on Darboux's theorem?
You sir are a Maths Super Action Hero.
Very nice. 150% speed is a good pace.
I'm wondering if I'm missing something. After multiplying out the original equations and adding them, you got x^2-5x+y^2-5y+12=0. My immediate thought was to turn that into (x^2-5x+6)+(y^2-5y+6)=0, which becomes (x-2)(x-3)+(y-2)(y-3)=0. That leads you to (2,2), (2,3), (3,2), and (3,3). The only other possible solutions would be when (x-2)(x-3)=-(y-2)(y-3). I'm not quite sure how to prove that's impossible when x is between 2 and 3 or y is between 2 and 3 ... the only ways to generate negative products.
"The only other possible solutions would be when (x-2)(x-3)=-(y-2)(y-3)"
It's not "other", it's equivalent to (x-2)(x-3)+(y-2)(y-3)=0. It also has an infinite number of other possible solutions, other than (2,2), (2,3), (3,2), and (3,3).
You seem to be trying to prove that (x-2)(x-3)+(y-2)(y-3)=0 on its own is equivalent to the original system, which it's obviously not.