Prove that the equations
and
imply the equation
.
Please help.
Keep Smiling
Malay
Hello, Malay!
I'm baffled . . . (more than usual)
Prove that the equations:
and
imply the equation:
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:
How does that work? .I have no idea of what I'm doing . . .
Hello, TPHacker!
Thank you for the explanation.
You're saying: If satisfies the two given equations,
. . then satisfies
I suspected that that was an interpretation of "implies"
. . but it doesn't work.
I found three points that satisfy the two given equations:
. .
But only satisfies:
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 V is a vector space because linear combinations of equations in V are again in V.
The three given equations
can be represented as coefficient vectors Putting those 3 vectors into a matrix
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.
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: .
. .
. . x-y)(y-1)" alt="\Rightarrow\quad(x - y)^2 \:=\x-y)(y-1)" /> [4]
. . [5]
Equate [4] and [5]: .
This is not , of course, but it's closer than anything I've found so far.
If only I could show that , 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 ....[/1]
......[2]
...........[3]
Solve [1] for x and we have
Sub into [2] and simplify:
.....[4]
Sub into [3] and simplify:
.........[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