Prove that the equations
$\displaystyle x^2 - 3xy + 2y^2 + x - y = 0$
and$\displaystyle x^2 - 2xy + y^2 - 5x + 7y = 0$
imply the equation
$\displaystyle xy - 12x + 5y = 0$.
Please help.
Keep Smiling
Malay
Hello, Malay!
I'm baffled . . . (more than usual)
Prove that the equations:
$\displaystyle x^2 - 3xy + 2y^2 + x - y = 0$ and $\displaystyle x^2 - 2xy + y^2 - 5x + 7y = 0$
imply the equation: $\displaystyle xy - 12x + 5y = 0$
I thought this would be simple (perhaps imaginative) algebra,
. . but I've made absolutely no progress.
Then it occured to me that I have never done a problem like this.
Two equations imply another? . . . Do I understand what that means?
(I don't think I do . . . )
I can find the intersections of the two graphs, but that's not the mission.
The first equation represents a pair of intersecting lines.
The second is a parabola (tipped 45°).
Somehow these two imply the hyperbola: $\displaystyle y\:=\:\frac{12x}{x+5}$
How does that work? .I have no idea of what I'm doing . . .
It means, that if $\displaystyle x,y$ are real numbers that satisfy those equations then (implies) they satisfy a new equation.Originally Posted by Soroban
For example, if $\displaystyle x+2y=1$ and $\displaystyle 2x+y=1$ implies that they satisfy, $\displaystyle 3x+3y=2$
Hello, TPHacker!
Thank you for the explanation.
You're saying: If $\displaystyle (x,y)$ satisfies the two given equations,
. . then $\displaystyle (x,y)$ satisfies $\displaystyle xy - 12x + 5y\,=\,0$
I suspected that that was an interpretation of "implies"
. . but it doesn't work.
I found three points that satisfy the two given equations:
. . $\displaystyle (0,0),\;(3,2),\;(5,3)$
But only $\displaystyle (0,0)$ satisfies: $\displaystyle xy - 12x + 5y\,=\,0$
Originally Posted by SorobanGeneralizing TPH's reply, this problem can be seen one of vector spaces and linear algebra.Originally Posted by ThePerfectHacker
We are given three equations in the vector space V of equations of the form $\displaystyle a_1 x^2 + a_2 xy + a_3 y^2 + a_4 x + a_5 y = 0.$ V is a vector space because linear combinations of equations in V are again in V.
The three given equations
$\displaystyle x^2 - 3xy + 2y^2 + x - y = 0$
$\displaystyle x^2 - 2xy + y^2 - 5x + 7y = 0$
$\displaystyle xy - 12x + 5y = 0$
can be represented as coefficient vectors $\displaystyle [a_1, \ldots, a_5].$ Putting those 3 vectors into a matrix
$\displaystyle \bmatrix{
1 & -3 & 2 &1 &-1 \\
1 & 2 & 1 &-5 &7 \\
0 & 1 & 0 &-12 &5\end{bmatrix}$
the problem becomes determining whether or not the third row (equation) is a linear combination of (implied by) the first two rows (equations). The answer is no because there is no way to take a non-trivial linear combination of the first and second rows and get zeroes in both the first and third columns as in the third row. Put more abstractly, the third equation is not in the subspace spanned by the first two equations, so the third equation is not implied by the first two equations.
Back at TPH's example, the third equation is a linear combination (the sum) of the first two equations, so the third equation is implied by the first two equations.
Well, this was supposed to make the answer more obvious, but it turned out to be a long-winded way to do that! At the least I hope it makes the problem a familiar one from linear algebra.
JakeD, I just luv your approach, but I do not understand what you are doing. Can you in steps explain what is going on. Begin by defining the vector space....
Also, does the fact
$\displaystyle
\bmatrix{1 & -3 & 2 &1 &-1 \\1 & 2 & 1 &-5 &7 \\0 & 1 & 0 &-12 &5\end{bmatrix}
$
Should be,
$\displaystyle
\bmatrix{1 & -3 & 2 &1 &-1 \\1 & -2 & 1 &-5 &7 \\0 & 1 & 0 &-12 &5\end{bmatrix}
$
Make a difference in your proof?
Let me try once.
We have two equations
$\displaystyle (x-y)(x-2y+1)=0$
$\displaystyle x^2 - 2xy + y^2 -5x + 7y = 0$
If x=y, we have the case x=y=0 which the third equation satisfies.
$\displaystyle Second Case$
$\displaystyle x-2y+1=0$
Multiplying both sides by $\displaystyle (2x-y)$
We have$\displaystyle 2x^2 - 5xy + y^2 +2x -y=0$
This equation and the second equation$\displaystyle (x^2 - 2xy + y^2 -5x + 7y = 0)$
produces the third equation.
Is there any other method?
Keep Smiling
Malay
Unfortunately my approach was incorrect. The error was in these two statements.Originally Posted by ThePerfectHacker
That the third equation is not a linear combination of the first two equations does not determine that the third equation is not implied by them. I found this generally only works for linear equations.Originally Posted by JakeD
Hello, all!
I found another approach . . . but hit a wall.
. . Maybe someone can carry on . . .
We want: .$\displaystyle \begin{pmatrix}[1]\;x^2-3xy + 2y^2 + x - y \:=\:0\\ [2]\;x^2-2xy + y^2 - 5x + 7y \:= \:0\end{pmatrix}$ $\displaystyle \;\Longrightarrow \;[3]\; xy - 12x + 15y \:=\:0$
$\displaystyle [1]\;x^2 - 3xy + 2y^2 + x - y \:=\:0\quad\Rightarrow\quad x^2$$\displaystyle - 2xy + y^2 - xy + y^2 + x - y \:= \:0$
. . $\displaystyle \Rightarrow\quad (x-y)^2 - y(x - y) + (x - y)\:=\:0\quad\Rightarrow\quad (x-y)^2$$\displaystyle - (x-y)(y-1)\:=\:0$
. . $\displaystyle \Rightarrow\quad(x - y)^2 \:=\x-y)(y-1)$ [4]
$\displaystyle [2]\;x^2-2xy+y^2-5x+7y\:=\:0\quad\Rightarrow\quad (x-y)^2$ $\displaystyle \:=\:-(5x - 7y) \:=\:0$
. . $\displaystyle \Rightarrow\quad (x - y)^2\:=\:5x - 7y$ [5]
Equate [4] and [5]: .$\displaystyle (x - y)(y - 1) \:=\:5x - 7y\quad\Rightarrow\quad xy$ $\displaystyle -\,6x + 8y - y^2\:=\:0$
This is not $\displaystyle [3]$, of course, but it's closer than anything I've found so far.
If only I could show that $\displaystyle y^2 \,= \,6x-7y$, but no luck yet.
Anyone? .Anyone?
I will put in my 2 cents, for what it's worth. I doubt if it amounts to anything, but we'll see.
We have $\displaystyle x^{2}-3xy+2y^{2}+x-y=0$....[/1]
$\displaystyle x^{2}-2xy+y^{2}-5x+7y=0$......[2]
$\displaystyle xy-12x+15y=0$...........[3]
Solve [1] for x and we have $\displaystyle x=2y-1\,\ and\,\ x=y$
Sub $\displaystyle x=2y-1$ into [2] and simplify:
$\displaystyle y^{2}-5y+6=(y-3)(y-2)$.....[4]
Sub into [3] and simplify:
$\displaystyle 2y^{2}-10y+12=2(y-3)(y-2)$.........[5]
Note that [5] is twice [4]
Solutions are (3,2) and (5,3).
(5,3) is the only case which produces 0 in all equations, though.
Well, except for (0,0).
Just a thought.
I put the equations into a spreadsheet and found (3,2), (5,3) and (0,0) are all solutions to all the equations. So galactus it seems your algebra is correct but the arithmetic is not.Originally Posted by galactus
Soroban found these these same solutions before the error in the third equation was revealed.
Originally Posted by Soroban
Watch your final equation.Originally Posted by Soroban
it is just the differnce of first two equations. But your work is not for wastebasket.
You can break the wall for the case x is not equal to y using your 4th and 5th equations.
If x is equal to y, then it is a trivial case since then x=y=0.
Keep Smiling
Malay