Mixing
Solve quadratic equation with two variables
$begingroup$
I can't find a solution for this equation
$$frac{(x-2016)(y-2017)}{(x-2016)^2 + (y-2017)^2}= -frac{1}{2}$$
Any help?
linear-algebra quadratics
edited Jan 18 at 22:40
user1337
46110
asked Jan 18 at 22:29
Mai NagahMai Nagah
34
$endgroup$
add a comment |
$begingroup$
I can't find a solution for this equation
$$frac{(x-2016)(y-2017)}{(x-2016)^2 + (y-2017)^2}= -frac{1}{2}$$
Any help?
linear-algebra quadratics
edited Jan 18 at 22:40
user1337
46110
asked Jan 18 at 22:29
Mai NagahMai Nagah
34
$endgroup$
add a comment |
$begingroup$
I can't find a solution for this equation
$$frac{(x-2016)(y-2017)}{(x-2016)^2 + (y-2017)^2}= -frac{1}{2}$$
Any help?
linear-algebra quadratics
edited Jan 18 at 22:40
user1337
46110
asked Jan 18 at 22:29
Mai NagahMai Nagah
34
$endgroup$
I can't find a solution for this equation
$$frac{(x-2016)(y-2017)}{(x-2016)^2 + (y-2017)^2}= -frac{1}{2}$$
Any help?
linear-algebra quadratics
linear-algebra quadratics
edited Jan 18 at 22:40
user1337
46110
asked Jan 18 at 22:29
Mai NagahMai Nagah
34
edited Jan 18 at 22:40
user1337
46110
asked Jan 18 at 22:29
Mai NagahMai Nagah
34
edited Jan 18 at 22:40
user1337
46110
edited Jan 18 at 22:40
user1337
46110
edited Jan 18 at 22:40
user1337
46110
46110
asked Jan 18 at 22:29
Mai NagahMai Nagah
34
asked Jan 18 at 22:29
Mai NagahMai Nagah
34
asked Jan 18 at 22:29
Mai NagahMai Nagah
34
34
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If you try to solve as always you will find:
$$2(xy-2016y-2017x+(2016cdot 2017))=(-1)(x^2-4032x+2016^2+y^2-4034y+2017^2)$$
That is:
$$x^2+2xy+y^2-8066(x+y)+ 4033^2=0$$
$$(x+y)^2-8066(x+y)+4033^2=0$$
$$((x+y) - 4033)^2=0$$
Hence $x+y=4033$.
You can see that last step by doing $t=x+y$ and solving $t^2-8066t +4033^2=0$.
answered Jan 18 at 22:46
idriskameniidriskameni
743321
$endgroup$
add a comment |
$begingroup$
HINT
In the first place, let us set that $a = x - 2016$ and $b = y - 2017$, from whence we conclude that
begin{align*}
frac{ab}{a^{2}+b^{2}} = -frac{1}{2} Longleftrightarrow 2ab = -a^{2} - b^{2} Longleftrightarrow (a + b)^{2} = 0 Longleftrightarrow a + b = 0
end{align*}
Can you proceed from here?
answered Jan 18 at 22:42
user1337user1337
46110
$endgroup$
add a comment |
$begingroup$
$$ frac{(x-2016)(y-2017)}{(x-2016)^2 + (y-2017)^2} $$
$ Let: t=x-2016, s=y-2017 $, then:
$$ frac{ts}{t^2+s^2} = -frac{1}{2} $$
If $t=0$ we have no solution, so multiply left side by $frac{t^2}{t^2} $ getting:
$$ frac{frac{s}{t}}{(frac{s}{t})^2 + 1} = -frac{1}{2} $$
Now substituting $ z=frac{s}{t} $ we finally get:
$$frac{z}{z^2+1} = -frac{1}{2}$$
$ -2z = z^2 + 1 $
$ (z+1)^2 = 0 $
so $ z=-1 $, that is $t = -s$, in terms of $x,y$ it means:
$x-2016 = -y + 2017$
$x+y = 4033 $
answered Jan 18 at 22:45
Dominik KutekDominik Kutek
3345
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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active
oldest
votes
$begingroup$
If you try to solve as always you will find:
$$2(xy-2016y-2017x+(2016cdot 2017))=(-1)(x^2-4032x+2016^2+y^2-4034y+2017^2)$$
That is:
$$x^2+2xy+y^2-8066(x+y)+ 4033^2=0$$
$$(x+y)^2-8066(x+y)+4033^2=0$$
$$((x+y) - 4033)^2=0$$
Hence $x+y=4033$.
You can see that last step by doing $t=x+y$ and solving $t^2-8066t +4033^2=0$.
answered Jan 18 at 22:46
idriskameniidriskameni
743321
$endgroup$
add a comment |
$begingroup$
If you try to solve as always you will find:
$$2(xy-2016y-2017x+(2016cdot 2017))=(-1)(x^2-4032x+2016^2+y^2-4034y+2017^2)$$
That is:
$$x^2+2xy+y^2-8066(x+y)+ 4033^2=0$$
$$(x+y)^2-8066(x+y)+4033^2=0$$
$$((x+y) - 4033)^2=0$$
Hence $x+y=4033$.
You can see that last step by doing $t=x+y$ and solving $t^2-8066t +4033^2=0$.
answered Jan 18 at 22:46
idriskameniidriskameni
743321
$endgroup$
add a comment |
$begingroup$
If you try to solve as always you will find:
$$2(xy-2016y-2017x+(2016cdot 2017))=(-1)(x^2-4032x+2016^2+y^2-4034y+2017^2)$$
That is:
$$x^2+2xy+y^2-8066(x+y)+ 4033^2=0$$
$$(x+y)^2-8066(x+y)+4033^2=0$$
$$((x+y) - 4033)^2=0$$
Hence $x+y=4033$.
You can see that last step by doing $t=x+y$ and solving $t^2-8066t +4033^2=0$.
answered Jan 18 at 22:46
idriskameniidriskameni
743321
$endgroup$
If you try to solve as always you will find:
$$2(xy-2016y-2017x+(2016cdot 2017))=(-1)(x^2-4032x+2016^2+y^2-4034y+2017^2)$$
That is:
$$x^2+2xy+y^2-8066(x+y)+ 4033^2=0$$
$$(x+y)^2-8066(x+y)+4033^2=0$$
$$((x+y) - 4033)^2=0$$
Hence $x+y=4033$.
You can see that last step by doing $t=x+y$ and solving $t^2-8066t +4033^2=0$.
answered Jan 18 at 22:46
idriskameniidriskameni
743321
answered Jan 18 at 22:46
idriskameniidriskameni
743321
answered Jan 18 at 22:46
idriskameniidriskameni
743321
answered Jan 18 at 22:46
idriskameniidriskameni
743321
743321
add a comment |
add a comment |
$begingroup$
HINT
In the first place, let us set that $a = x - 2016$ and $b = y - 2017$, from whence we conclude that
begin{align*}
frac{ab}{a^{2}+b^{2}} = -frac{1}{2} Longleftrightarrow 2ab = -a^{2} - b^{2} Longleftrightarrow (a + b)^{2} = 0 Longleftrightarrow a + b = 0
end{align*}
Can you proceed from here?
answered Jan 18 at 22:42
user1337user1337
46110
$endgroup$
add a comment |
$begingroup$
HINT
In the first place, let us set that $a = x - 2016$ and $b = y - 2017$, from whence we conclude that
begin{align*}
frac{ab}{a^{2}+b^{2}} = -frac{1}{2} Longleftrightarrow 2ab = -a^{2} - b^{2} Longleftrightarrow (a + b)^{2} = 0 Longleftrightarrow a + b = 0
end{align*}
Can you proceed from here?
answered Jan 18 at 22:42
user1337user1337
46110
$endgroup$
add a comment |
$begingroup$
HINT
In the first place, let us set that $a = x - 2016$ and $b = y - 2017$, from whence we conclude that
begin{align*}
frac{ab}{a^{2}+b^{2}} = -frac{1}{2} Longleftrightarrow 2ab = -a^{2} - b^{2} Longleftrightarrow (a + b)^{2} = 0 Longleftrightarrow a + b = 0
end{align*}
Can you proceed from here?
answered Jan 18 at 22:42
user1337user1337
46110
$endgroup$
HINT
In the first place, let us set that $a = x - 2016$ and $b = y - 2017$, from whence we conclude that
begin{align*}
frac{ab}{a^{2}+b^{2}} = -frac{1}{2} Longleftrightarrow 2ab = -a^{2} - b^{2} Longleftrightarrow (a + b)^{2} = 0 Longleftrightarrow a + b = 0
end{align*}
Can you proceed from here?
answered Jan 18 at 22:42
user1337user1337
46110
answered Jan 18 at 22:42
user1337user1337
46110
answered Jan 18 at 22:42
user1337user1337
46110
answered Jan 18 at 22:42
user1337user1337
46110
46110
add a comment |
add a comment |
$begingroup$
$$ frac{(x-2016)(y-2017)}{(x-2016)^2 + (y-2017)^2} $$
$ Let: t=x-2016, s=y-2017 $, then:
$$ frac{ts}{t^2+s^2} = -frac{1}{2} $$
If $t=0$ we have no solution, so multiply left side by $frac{t^2}{t^2} $ getting:
$$ frac{frac{s}{t}}{(frac{s}{t})^2 + 1} = -frac{1}{2} $$
Now substituting $ z=frac{s}{t} $ we finally get:
$$frac{z}{z^2+1} = -frac{1}{2}$$
$ -2z = z^2 + 1 $
$ (z+1)^2 = 0 $
so $ z=-1 $, that is $t = -s$, in terms of $x,y$ it means:
$x-2016 = -y + 2017$
$x+y = 4033 $
answered Jan 18 at 22:45
Dominik KutekDominik Kutek
3345
$endgroup$
add a comment |
$begingroup$
$$ frac{(x-2016)(y-2017)}{(x-2016)^2 + (y-2017)^2} $$
$ Let: t=x-2016, s=y-2017 $, then:
$$ frac{ts}{t^2+s^2} = -frac{1}{2} $$
If $t=0$ we have no solution, so multiply left side by $frac{t^2}{t^2} $ getting:
$$ frac{frac{s}{t}}{(frac{s}{t})^2 + 1} = -frac{1}{2} $$
Now substituting $ z=frac{s}{t} $ we finally get:
$$frac{z}{z^2+1} = -frac{1}{2}$$
$ -2z = z^2 + 1 $
$ (z+1)^2 = 0 $
so $ z=-1 $, that is $t = -s$, in terms of $x,y$ it means:
$x-2016 = -y + 2017$
$x+y = 4033 $
answered Jan 18 at 22:45
Dominik KutekDominik Kutek
3345
$endgroup$
add a comment |
$begingroup$
$$ frac{(x-2016)(y-2017)}{(x-2016)^2 + (y-2017)^2} $$
$ Let: t=x-2016, s=y-2017 $, then:
$$ frac{ts}{t^2+s^2} = -frac{1}{2} $$
If $t=0$ we have no solution, so multiply left side by $frac{t^2}{t^2} $ getting:
$$ frac{frac{s}{t}}{(frac{s}{t})^2 + 1} = -frac{1}{2} $$
Now substituting $ z=frac{s}{t} $ we finally get:
$$frac{z}{z^2+1} = -frac{1}{2}$$
$ -2z = z^2 + 1 $
$ (z+1)^2 = 0 $
so $ z=-1 $, that is $t = -s$, in terms of $x,y$ it means:
$x-2016 = -y + 2017$
$x+y = 4033 $
answered Jan 18 at 22:45
Dominik KutekDominik Kutek
3345
$endgroup$
$$ frac{(x-2016)(y-2017)}{(x-2016)^2 + (y-2017)^2} $$
$ Let: t=x-2016, s=y-2017 $, then:
$$ frac{ts}{t^2+s^2} = -frac{1}{2} $$
If $t=0$ we have no solution, so multiply left side by $frac{t^2}{t^2} $ getting:
$$ frac{frac{s}{t}}{(frac{s}{t})^2 + 1} = -frac{1}{2} $$
Now substituting $ z=frac{s}{t} $ we finally get:
$$frac{z}{z^2+1} = -frac{1}{2}$$
$ -2z = z^2 + 1 $
$ (z+1)^2 = 0 $
so $ z=-1 $, that is $t = -s$, in terms of $x,y$ it means:
$x-2016 = -y + 2017$
$x+y = 4033 $
answered Jan 18 at 22:45
Dominik KutekDominik Kutek
3345
answered Jan 18 at 22:45
Dominik KutekDominik Kutek
3345
answered Jan 18 at 22:45
Dominik KutekDominik Kutek
3345
answered Jan 18 at 22:45
Dominik KutekDominik Kutek
3345
3345
add a comment |
add a comment |
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